Reverse voltage on a diode (forward converter)

[reggie]

Hi all

I have found a formula which confuses me. Its in ST Topologies for
SMPSUs Bu L.Wuidart (AN513/0393)

http://www.stmicroelectronics.com/stonline/books/pdf/docs/3721.pdf

Page 12/18 formula :FREEWHEELING D2:
Vrrm>= [Vinmax.(Vout+Vf)]/ Vinmin . dutymax -----( STs equation)

The link describes the reverse voltage on secondary freewheeling diode
of an isolated forward converter:

D1 conducts during fet ontime
D2 conducts during fet off time (freewheeling diode)

I am trying to understand the above equation but am struggling to
derive it this is what I have so far:

1) The voltage on the output of the LC filter is the time average
voltage (Vout) of the input to the LC filter times the duty factor
2) The input to the LC filter is a square wave voltage with a peak
voltage
Vpeak = Vout / dutyfactor
3) D1s forward voltage drop = Vf
4) The voltage on the secondry turns of the transformer is Vs = Vpeak
+ Vf
5) also Vs = (ns/np) .Vp [Vp = volts on primary]
6) therefore merging 4&5 Vpeak + Vf = (ns/np).Vp
7) Vpeak = [(ns/np).Vp] -Vf
8) now to get worst case reverse voltage on D2 Vpeakmax =[(ns/
np).Vpmax]-Vf

A) Is what I have said in step 1 to 8 above correct?

B) I don’t see how STs equation can have a “+” Vf term in it, I think
the Vo+Vf is the voltage on the secondary turns not the reverse
voltage across D2. Am I wrong?

Please help, I am going out of my tiny little mind trying to figure
this out.

Reggie.

 

[Tom Bruhns]

Hi all

I have found a formula which confuses me. Its in ST Topologies for
SMPSUs Bu L.Wuidart (AN513/0393)

http://www.stmicroelectronics.com/stonline/books/pdf/docs/3721.pdf

Page 12/18 formula :FREEWHEELING D2:
Vrrm>= [Vinmax.(Vout+Vf)]/ Vinmin . dutymax -----( STs equation)

The link describes the reverse voltage on secondary freewheeling diode
of an isolated forward converter:

D1 conducts during fet ontime
D2 conducts during fet off time (freewheeling diode)

I am trying to understand the above equation but am struggling to
derive it this is what I have so far:

1) The voltage on the output of the LC filter is the time average
voltage (Vout) of the input to the LC filter times the duty factor
2) The input to the LC filter is a square wave voltage with a peak
voltage
Vpeak = Vout / dutyfactor
3) D1s forward voltage drop = Vf
4) The voltage on the secondry turns of the transformer is Vs = Vpeak
+ Vf
5) also Vs = (ns/np) .Vp [Vp = volts on primary]
6) therefore merging 4&5 Vpeak + Vf = (ns/np).Vp
7) Vpeak = [(ns/np).Vp] -Vf
8) now to get worst case reverse voltage on D2 Vpeakmax =[(ns/
np).Vpmax]-Vf

A) Is what I have said in step 1 to 8 above correct?

B) I don’t see how STs equation can have a “+” Vf term in it, I think
the Vo+Vf is the voltage on the secondary turns not the reverse
voltage across D2. Am I wrong?

Please help, I am going out of my tiny little mind trying to figure
this out.

Reggie.

Without having examined it in excruciating detail, your logic looks
right to me. People who publish ap notes (and data sheets!) do make
mistakes; I find them all the time. Sometimes the author is to blame,
and sometimes the people who pretty up the equations and figures for
publication are the culprits. When I report errors I even
occasionally get a thank-you back.

You do want to allow for the inevitable nonidealities in the system,
of course--things like leakage inductance, the effects of switches
that have significant on resistance and lots of off capacitance, the
effects of resistance in the output choke, the output ripple
voltage, ... I've found that it's quite possible to build a very
decent SPICE model of a switcher if you spend a bit of time
characterizing the components for their parasitic effects; I've gotten
quite respectable predictions of actual ringing frequencies,
amplitudes and damping of leakage inductances with intended and
parasitic capacitances, for example.

You could toss together a Spice model of the idealized circuit in the
ST ap note quite easily and, using a low voltage so that Vf is a
significant fraction of the secondary peak voltage, demonstrate that
the ap note is either incorrect or correct. Suggest you give that a
try.

Cheers,
Tom

 

[Joerg]

Hi all

I have found a formula which confuses me. Its in ST Topologies for
SMPSUs Bu L.Wuidart (AN513/0393)

http://www.stmicroelectronics.com/stonline/books/pdf/docs/3721.pdf

Page 12/18 formula :FREEWHEELING D2:
Vrrm>= [Vinmax.(Vout+Vf)]/ Vinmin . dutymax -----( STs equation)

The link describes the reverse voltage on secondary freewheeling diode
of an isolated forward converter:

D1 conducts during fet ontime
D2 conducts during fet off time (freewheeling diode)

I am trying to understand the above equation but am struggling to
derive it this is what I have so far:

1) The voltage on the output of the LC filter is the time average
voltage (Vout) of the input to the LC filter times the duty factor
2) The input to the LC filter is a square wave voltage with a peak
voltage
Vpeak = Vout / dutyfactor
3) D1s forward voltage drop = Vf
4) The voltage on the secondry turns of the transformer is Vs = Vpeak
+ Vf
5) also Vs = (ns/np) .Vp [Vp = volts on primary]
6) therefore merging 4&5 Vpeak + Vf = (ns/np).Vp
7) Vpeak = [(ns/np).Vp] -Vf
8) now to get worst case reverse voltage on D2 Vpeakmax =[(ns/
np).Vpmax]-Vf

A) Is what I have said in step 1 to 8 above correct?

B) I don’t see how STs equation can have a “+” Vf term in it, I think
the Vo+Vf is the voltage on the secondary turns not the reverse
voltage across D2. Am I wrong?

Please help, I am going out of my tiny little mind trying to figure
this out.

Reggie.

Without having examined it in excruciating detail, your logic looks
right to me. People who publish ap notes (and data sheets!) do make
mistakes; I find them all the time. Sometimes the author is to blame,
and sometimes the people who pretty up the equations and figures for
publication are the culprits. When I report errors I even
occasionally get a thank-you back.

So did I. Occasionally people where even shocked when I pointed out a
major booboo. However, most of the time a datasheet or app note was not
corrected. I don't get that. It's almost as if you tell someone that
there is smoke coming from under the hood and they just keep on driving.

 

[Tom Bruhns]

Hi all

I have found a formula which confuses me. Its in ST Topologies for
SMPSUs Bu L.Wuidart (AN513/0393)

http://www.stmicroelectronics.com/stonline/books/pdf/docs/3721.pdf

Page 12/18 formula :FREEWHEELING D2:
Vrrm>= [Vinmax.(Vout+Vf)]/ Vinmin . dutymax -----( STs equation)

The link describes the reverse voltage on secondary freewheeling diode
of an isolated forward converter:

D1 conducts during fet ontime
D2 conducts during fet off time (freewheeling diode)

I am trying to understand the above equation but am struggling to
derive it this is what I have so far:

1) The voltage on the output of the LC filter is the time average
voltage (Vout) of the input to the LC filter times the duty factor
2) The input to the LC filter is a square wave voltage with a peak
voltage
Vpeak = Vout / dutyfactor
3) D1s forward voltage drop = Vf
4) The voltage on the secondry turns of the transformer is Vs = Vpeak
+ Vf
5) also Vs = (ns/np) .Vp [Vp = volts on primary]
6) therefore merging 4&5 Vpeak + Vf = (ns/np).Vp
7) Vpeak = [(ns/np).Vp] -Vf
8) now to get worst case reverse voltage on D2 Vpeakmax =[(ns/
np).Vpmax]-Vf

A) Is what I have said in step 1 to 8 above correct?

B) I don’t see how STs equation can have a “+” Vf term in it, I think
the Vo+Vf is the voltage on the secondary turns not the reverse
voltage across D2. Am I wrong?

Please help, I am going out of my tiny little mind trying to figure
this out.

Reggie.

Having now examined it in a bit more detail (forced by looking at a
simulation output ;-), I can suggest that you consider your point 2
more carefully. Realize that it is indeed a square wave, but it goes
negative by Vf when the switch is off--that is, when D2 is forward
biased. That means that to get the average voltage over a cycle at
the inductor input to be equal to the output voltage, the peak voltage
at that point must go higher than Vout/dutyfactor. In fact, it must
go higher by Vf*(dutyfactor-1), I believe...so then it spends
dutyfactor at Vout/dutyfactor + Vf*(dutyfactor-1) and (dutyfactor-1)
at -Vf. Does that help?

Cheers,
Tom

 

[Tom Bruhns]

Hi all

I have found a formula which confuses me. Its in ST Topologies for
SMPSUs Bu L.Wuidart (AN513/0393)

http://www.stmicroelectronics.com/stonline/books/pdf/docs/3721.pdf

Page 12/18 formula :FREEWHEELING D2:
Vrrm>= [Vinmax.(Vout+Vf)]/ Vinmin . dutymax -----( STs equation)

The link describes the reverse voltage on secondary freewheeling diode
of an isolated forward converter:

D1 conducts during fet ontime
D2 conducts during fet off time (freewheeling diode)

I am trying to understand the above equation but am struggling to
derive it this is what I have so far:

1) The voltage on the output of the LC filter is the time average
voltage (Vout) of the input to the LC filter times the duty factor
2) The input to the LC filter is a square wave voltage with a peak
voltage
Vpeak = Vout / dutyfactor
3) D1s forward voltage drop = Vf
4) The voltage on the secondry turns of the transformer is Vs = Vpeak
+ Vf
5) also Vs = (ns/np) .Vp [Vp = volts on primary]
6) therefore merging 4&5 Vpeak + Vf = (ns/np).Vp
7) Vpeak = [(ns/np).Vp] -Vf
8) now to get worst case reverse voltage on D2 Vpeakmax =[(ns/
np).Vpmax]-Vf

A) Is what I have said in step 1 to 8 above correct?

B) I don’t see how STs equation can have a “+” Vf term in it, I think
the Vo+Vf is the voltage on the secondary turns not the reverse
voltage across D2. Am I wrong?

Please help, I am going out of my tiny little mind trying to figure
this out.

Reggie.

Having now examined it in a bit more detail (forced by looking at a
simulation output ;-), I can suggest that you consider your point 2
more carefully. Realize that it is indeed a square wave, but it goes
negative by Vf when the switch is off--that is, when D2 is forward
biased. That means that to get the average voltage over a cycle at
the inductor input to be equal to the output voltage, the peak voltage
at that point must go higher than Vout/dutyfactor. In fact, it must
go higher by Vf*(dutyfactor-1), I believe...so then it spends
dutyfactor at Vout/dutyfactor + Vf*(dutyfactor-1) and (dutyfactor-1)
at -Vf. Does that help?

Cheers,
Tom

Cripes, belay that. The last part should read (barring more slips on
the mental gymnastics bar):

high voltage at inductor input = (Vout + Vf*(1-dutyfactor))/dutyfactor
= D2 peak reverse voltage
low voltage at inductor input = -Vf
Average voltage at inductor input
= high * dutyfactor + low * (1-dutyfactor)
= Vout + Vf*(1-dutyfactor) - Vf*(1-dutyractor)
= Vout
as expected.

Cheers,
Tom

 

[reggie]

Hi all
I have found a formula which confuses me. Its in ST Topologies for
SMPSUs Bu L.Wuidart (AN513/0393)
http://www.stmicroelectronics.com/stonline/books/pdf/docs/3721.pdf
Page 12/18 formula :FREEWHEELING D2:
Vrrm>= [Vinmax.(Vout+Vf)]/ Vinmin . dutymax     -----( STs equation)
The link describes the reverse voltage on secondary freewheeling diode
of an isolated forward converter:
D1 conducts during fet ontime
D2 conducts during fet off time (freewheeling diode)
I am trying to understand the above equation but am struggling to
derive it this is what I have so far:
1) The voltage on the output of the LC filter is the time average
voltage (Vout) of the input to the LC filter times the duty factor
2) The input to the LC filter is a square wave voltage with a peak
voltage
Vpeak = Vout / dutyfactor
3) D1s forward voltage drop = Vf
4) The voltage on the secondry turns of the transformer is Vs = Vpeak
+ Vf
5)  also Vs  = (ns/np) .Vp           [Vp = volts on primary]
6) therefore merging  4&5   Vpeak + Vf = (ns/np).Vp
7) Vpeak = [(ns/np).Vp] -Vf
8) now to get worst case reverse voltage on D2   Vpeakmax =[(ns/
np).Vpmax]-Vf
A) Is what I have said in step 1 to 8 above correct?
B) I don’t see how STs equation can have a “+” Vf term in it, I think
the Vo+Vf is the voltage on the secondary turns not the reverse
voltage across D2.  Am I wrong?
Please help, I am going out of my tiny little mind trying to figure
this out.
Reggie.

Having now examined it in a bit more detail (forced by looking at a
simulation output ;-), I can suggest that you consider your point 2
more carefully.  Realize that it is indeed a square wave, but it goes
negative by Vf when the switch is off--that is, when D2 is forward
biased.  That means that to get the average voltage over a cycle at
the inductor input to be equal to the output voltage, the peak voltage
at that point must go higher than Vout/dutyfactor.  In fact, it must
go higher by Vf*(dutyfactor-1), I believe...so then it spends
dutyfactor at Vout/dutyfactor + Vf*(dutyfactor-1) and (dutyfactor-1)
at -Vf.  Does that help?

Cheers,
Tom

Cripes, belay that.  The last part should read (barring more slips on
the mental gymnastics bar):

high voltage at inductor input = (Vout + Vf*(1-dutyfactor))/dutyfactor
   = D2 peak reverse voltage
low voltage at inductor input = -Vf
Average voltage at inductor input
   = high * dutyfactor + low * (1-dutyfactor)
   = Vout + Vf*(1-dutyfactor) - Vf*(1-dutyractor)
   = Vout
as expected.

Cheers,
Tom- Hide quoted text -

- Show quoted text -

HI Tom,

Thanks for your previous post. I can see what you are doing and agree
that the Vf should be included for completeness.

I have looked at your workings and can manage to verify them to be
correct from first principles only when Vf is negative. eg
Average=area under curve/length of base etc...

When I tried to start from first principles myself I get something
that concurs with your result eg Vpeak = [[vout+vf]/dutyfactor]-vf
(ignoring lekage L spikes)

However your line "high voltage at inductor input = (Vout + Vf*(1-
dutyfactor))/dutyfactor" I believe is a little confusing as you are
adding a negative term there fore you are subtracting.

In effect you are reducing the on area by the off area.

I think it should read "high voltage at inductor input = (Vout - Vf*(1-
dutyfactor))/dutyfactor" for clarity, but I am picking at straws!

This result would suggest that the max reverse voltage on D2 occurs
when the duty factor = min, when duty factor is min primary voltage is
at max, so this concurs with step 8 of mine above.

Now if duty factor = 0.01 the term (Vout - Vf*(1-dutyfactor))/
dutyfactor becomes very large, practically too large for the input
voltage and the turns ratio to produce.

Say vout = 25V Vf=0.3 the term above = [25-(0.3x0.01)]/ 0.01 = 2499.7V
so perhaps the application note tries to put a limiting factor of
vinmax/vinmin.duty max to cap the max voltage to real realistic
values? What do you think?

Practically I don’t know what to specify for the min duty factor in a
real circuit so I couldn’t work out max reverse voltage on the diode
using your method.

Am I talking twaddle or am I making sense?

Reggie.

 

[Tom Bruhns]

Hi all
I have found a formula which confuses me. Its in ST Topologies for
SMPSUs Bu L.Wuidart (AN513/0393)
http://www.stmicroelectronics.com/stonline/books/pdf/docs/3721.pdf
Page 12/18 formula :FREEWHEELING D2:
Vrrm>= [Vinmax.(Vout+Vf)]/ Vinmin . dutymax -----( STs equation)
The link describes the reverse voltage on secondary freewheeling diode
of an isolated forward converter:
D1 conducts during fet ontime
D2 conducts during fet off time (freewheeling diode)
I am trying to understand the above equation but am struggling to
derive it this is what I have so far:
1) The voltage on the output of the LC filter is the time average
voltage (Vout) of the input to the LC filter times the duty factor
2) The input to the LC filter is a square wave voltage with a peak
voltage
Vpeak = Vout / dutyfactor
3) D1s forward voltage drop = Vf
4) The voltage on the secondry turns of the transformer is Vs = Vpeak
+ Vf
5) also Vs = (ns/np) .Vp [Vp = volts on primary]
6) therefore merging 4&5 Vpeak + Vf = (ns/np).Vp
7) Vpeak = [(ns/np).Vp] -Vf
8) now to get worst case reverse voltage on D2 Vpeakmax =[(ns/
np).Vpmax]-Vf
A) Is what I have said in step 1 to 8 above correct?
B) I don’t see how STs equation can have a “+” Vf term in it, I think
the Vo+Vf is the voltage on the secondary turns not the reverse
voltage across D2. Am I wrong?
Please help, I am going out of my tiny little mind trying to figure
this out.
Reggie.
Having now examined it in a bit more detail (forced by looking at a
simulation output ;-), I can suggest that you consider your point 2
more carefully. Realize that it is indeed a square wave, but it goes
negative by Vf when the switch is off--that is, when D2 is forward
biased. That means that to get the average voltage over a cycle at
the inductor input to be equal to the output voltage, the peak voltage
at that point must go higher than Vout/dutyfactor. In fact, it must
go higher by Vf*(dutyfactor-1), I believe...so then it spends
dutyfactor at Vout/dutyfactor + Vf*(dutyfactor-1) and (dutyfactor-1)
at -Vf. Does that help?
Cheers,
Tom

Cripes, belay that. The last part should read (barring more slips on
the mental gymnastics bar):

high voltage at inductor input = (Vout + Vf*(1-dutyfactor))/dutyfactor
= D2 peak reverse voltage
low voltage at inductor input = -Vf
Average voltage at inductor input
= high * dutyfactor + low * (1-dutyfactor)
= Vout + Vf*(1-dutyfactor) - Vf*(1-dutyractor)
= Vout
as expected.

Cheers,
Tom- Hide quoted text -

- Show quoted text -

HI Tom,

Thanks for your previous post. I can see what you are doing and agree
that the Vf should be included for completeness.

I have looked at your workings and can manage to verify them to be
correct from first principles only when Vf is negative. eg
Average=area under curve/length of base etc...

Huh?? Vf is positive; typically about half a volt for Schottkys near
their max current. That is, during the time that D2 is conducting,
the voltage at the input to the inductor is NEGATIVE by Vf. Then when
the transformer is supplying energy, the voltage at the inductor input
must be more positive by an amount to make up for that, if you will.

When I tried to start from first principles myself I get something
that concurs with your result eg Vpeak = [[vout+vf]/dutyfactor]-vf
(ignoring lekage L spikes)

However your line "high voltage at inductor input = (Vout + Vf*(1-
dutyfactor))/dutyfactor" I believe is a little confusing as you are
adding a negative term there fore you are subtracting.

Nope, adding. The voltage must be HIGHER because during the
transformer "off" time the voltage to the inductor input was negative.

In effect you are reducing the on area by the off area.

I think it should read "high voltage at inductor input = (Vout - Vf*(1-
dutyfactor))/dutyfactor" for clarity, but I am picking at straws!

This result would suggest that the max reverse voltage on D2 occurs
when the duty factor = min, when duty factor is min primary voltage is
at max, so this concurs with step 8 of mine above.

Now if duty factor = 0.01 the term (Vout - Vf*(1-dutyfactor))/
dutyfactor becomes very large, practically too large for the input
voltage and the turns ratio to produce.

Say vout = 25V Vf=0.3 the term above = [25-(0.3x0.01)]/ 0.01 = 2499.7V
so perhaps the application note tries to put a limiting factor of
vinmax/vinmin.duty max to cap the max voltage to real realistic
values? What do you think?

Practically I don’t know what to specify for the min duty factor in a
real circuit so I couldn’t work out max reverse voltage on the diode
using your method.

Well, another way to look at it is that the voltage out of the
transformer, neglecting effects of resonances and leakage inductances
and the like, is the primary voltage times the turns ratio, and the
reverse voltage across the freewheeling diode is that voltage minus
the forward drop of the diode connected from the transformer to the
inductor. Start with the basics: what's the input voltage range?
What output voltage and current do you need? You can then play with
different turns ratios to check out max and min duty factors, and from
those, figure the peak voltages and currents. For an ideal case, you
can do all that with a spreadsheet. Or--use something like LTSpice.

Am I talking twaddle or am I making sense?

Reggie.

Cheers,
Tom

 

[reggie]

Hi all
I have found a formula which confuses me. Its in ST Topologies for
SMPSUs Bu L.Wuidart (AN513/0393)
http://www.stmicroelectronics.com/stonline/books/pdf/docs/3721.pdf
Page 12/18 formula :FREEWHEELING D2:
Vrrm>= [Vinmax.(Vout+Vf)]/ Vinmin . dutymax     -----( STs equation)
The link describes the reverse voltage on secondary freewheeling diode
of an isolated forward converter:
D1 conducts during fet ontime
D2 conducts during fet off time (freewheeling diode)
I am trying to understand the above equation but am struggling to
derive it this is what I have so far:
1) The voltage on the output of the LC filter is the time average
voltage (Vout) of the input to the LC filter times the duty factor
2) The input to the LC filter is a square wave voltage with a peak
voltage
Vpeak = Vout / dutyfactor
3) D1s forward voltage drop = Vf
4) The voltage on the secondry turns of the transformer is Vs = Vpeak
+ Vf
5)  also Vs  = (ns/np) .Vp           [Vp = voltson primary]
6) therefore merging  4&5   Vpeak + Vf = (ns/np).Vp
7) Vpeak = [(ns/np).Vp] -Vf
8) now to get worst case reverse voltage on D2   Vpeakmax =[(ns/
np).Vpmax]-Vf
A) Is what I have said in step 1 to 8 above correct?
B) I don’t see how STs equation can have a “+” Vf term in it, I think
the Vo+Vf is the voltage on the secondary turns not the reverse
voltage across D2.  Am I wrong?
Please help, I am going out of my tiny little mind trying to figure
this out.
Reggie.
Having now examined it in a bit more detail (forced by looking at a
simulation output ;-), I can suggest that you consider your point 2
more carefully.  Realize that it is indeed a square wave, but it goes
negative by Vf when the switch is off--that is, when D2 is forward
biased.  That means that to get the average voltage over a cycle at
the inductor input to be equal to the output voltage, the peak voltage
at that point must go higher than Vout/dutyfactor.  In fact, it must
go higher by Vf*(dutyfactor-1), I believe...so then it spends
dutyfactor at Vout/dutyfactor + Vf*(dutyfactor-1) and (dutyfactor-1)
at -Vf.  Does that help?
Cheers,
Tom
Cripes, belay that.  The last part should read (barring more slips on
the mental gymnastics bar):
high voltage at inductor input = (Vout + Vf*(1-dutyfactor))/dutyfactor
   = D2 peak reverse voltage
low voltage at inductor input = -Vf
Average voltage at inductor input
   = high * dutyfactor + low * (1-dutyfactor)
   = Vout + Vf*(1-dutyfactor) - Vf*(1-dutyractor)
   = Vout
as expected.
Cheers,
Tom- Hide quoted text -
- Show quoted text -

HI Tom,

Thanks for your previous post. I can see what you are doing and agree
that the Vf should be included for completeness.

I have looked at your workings and can manage to verify them to be
correct from first principles only when Vf is negative. eg
Average=area under curve/length of base etc...

Huh??  Vf is positive; typically about half a volt for Schottkys near
their max current.  That is, during the time that D2 is conducting,
the voltage at the input to the inductor is NEGATIVE by Vf.  Then when
the transformer is supplying energy, the voltage at the inductor input
must be more positive by an amount to make up for that, if you will.

When I tried to start from first principles myself I get something
that concurs with your result eg Vpeak = [[vout+vf]/dutyfactor]-vf
(ignoring lekage L spikes)

However your line "high voltage at inductor input = (Vout + Vf*(1-
dutyfactor))/dutyfactor" I believe is a little confusing as you are
adding a negative term there fore you are subtracting.

Nope, adding.  The voltage must be HIGHER because during the
transformer "off" time the voltage to the inductor input was negative.

In effect you are reducing the on area by the off area.

I think it should read "high voltage at inductor input = (Vout - Vf*(1-
dutyfactor))/dutyfactor" for clarity, but I am picking at straws!

This result would suggest that the max reverse voltage on D2 occurs
when the duty factor = min, when duty factor is min primary voltage is
at max, so this concurs with step 8 of mine above.

Now if duty factor = 0.01 the term  (Vout - Vf*(1-dutyfactor))/
dutyfactor becomes very large, practically too large for the input
voltage and the turns ratio to produce.

Say vout = 25V Vf=0.3 the term above = [25-(0.3x0.01)]/ 0.01 = 2499.7V
so perhaps the application note tries to put a limiting factor of
vinmax/vinmin.duty max to cap the max voltage to real realistic
values? What do you think?

Practically I don’t know what to specify for the min duty factor in a
real circuit so I couldn’t work out max reverse voltage on the diode
using your method.

Well, another way to look at it is that the voltage out of the
transformer, neglecting effects of resonances and leakage inductances
and the like, is the primary voltage times the turns ratio, and the
reverse voltage across the freewheeling diode is that voltage minus
the forward drop of the diode connected from the transformer to the
inductor.  Start with the basics:  what's the input voltage range?
What output voltage and current do you need?  You can then play with
different turns ratios to check out max and min duty factors, and from
those, figure the peak voltages and currents.  For an ideal case, you
can do all that with a spreadsheet.  Or--use something like LTSpice.

Am I talking twaddle or am I making sense?

Reggie.

Cheers,
Tom- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

Hi Tom,

I have made a mistake! I have looked again at what you have said and I
do agree with your explanation (I copied your maths down incorrectly
from the web) sorry. What I had copied down was wrong not your
response to my original question.

However my original workings still stand, just the way I went about it
was different from you.
Imagine the actual waveform across D2 with negative value during toff
and positive waveform during ton. I averaged the waveform as it is,
but using a negative sign for the
area Vf.(1-dutyfactor)T

Eg:

Average voltage = vout = area under curve / length of base

Average voltage = vout = [high area – low area] / length of base

(high voltage at inductor input = Vpkdiode)

Average voltage = vout = [ton. Vpkdiode – toff . Vf] / length of base

Average voltage = vout = [Dutyfactor.T. Vpkdiode – (1-
dutyfactor)T.Vf] / T

Solve for Vpkdiode

Vpkdiode = ((Vout+Vf)/dutyfactor) - Vf

This gives the same result as your :

high voltage at inductor = [Vout + Vf (1-dutyfactor)]/dutyfactor

The important points that you covered are to maintain the same Volt
seconds across the coil over one period (otherwise it will saturate),
which is the same thing as saying the input average voltage has to
equal the output average voltage (which is of course vout).

Correct me if I am wrong but I think you have averaged both the high
area and the low area (with their corresponding times), added them
together to give a total increased average. You then divide this
average by the dutyfactor to give the (high voltage at inductor input
= Vpkdiode)

Your taking into account the negative voltage at the input to the
inductor and compensating for it theoretically is accurate, but in
practice wont the duty factor compensate for the small inaccuracies
introduced by neglecting this negative value? I know if the output
voltage is smaller the errors will be a bigger percentage and more
problematic.

Also when you are modelling leakage inductance to predict spike
frequency and amplitude, how do you take into account track
inductances / capacitances before you have designed the PCB? Or do you
assume good practice PCB design rules to minimise track parasitics?

Say for a 24V output will the error on the diode reverse voltage
specification be larger or smaller than the leakage inductance spike?
I am not sure, but I bet the leakage inductance spike will be bigger
than the error introduced by neglecting the negative voltage across
D2s effect on reverse voltage, what do you think?

I take your point about min duty and doing a spreadsheet.

Once again sorry for my mistake,

I hope I have explained myself clearly this time.


Reggie.

 

[reggie]

Hi all
I have found a formula which confuses me. Its in ST Topologies for
SMPSUs Bu L.Wuidart (AN513/0393)
http://www.stmicroelectronics.com/stonline/books/pdf/docs/3721.pdf
Page 12/18 formula :FREEWHEELING D2:
Vrrm>= [Vinmax.(Vout+Vf)]/ Vinmin . dutymax -----( STs equation)
The link describes the reverse voltage on secondary freewheeling diode
of an isolated forward converter:
D1 conducts during fet ontime
D2 conducts during fet off time (freewheeling diode)
I am trying to understand the above equation but am struggling to
derive it this is what I have so far:
1) The voltage on the output of the LC filter is the time average
voltage (Vout) of the input to the LC filter times the duty factor
2) The input to the LC filter is a square wave voltage with a peak
voltage
Vpeak = Vout / dutyfactor
3) D1s forward voltage drop = Vf
4) The voltage on the secondry turns of the transformer is Vs = Vpeak
+ Vf
5) also Vs = (ns/np) .Vp [Vp = volts on primary]
6) therefore merging 4&5 Vpeak + Vf = (ns/np).Vp
7) Vpeak = [(ns/np).Vp] -Vf
8) now to get worst case reverse voltage on D2 Vpeakmax =[(ns/
np).Vpmax]-Vf
A) Is what I have said in step 1 to 8 above correct?
B) I don't see how STs equation can have a "+" Vf term in it, I think
the Vo+Vf is the voltage on the secondary turns not the reverse
voltage across D2. Am I wrong?
Please help, I am going out of my tiny little mind trying to figure
this out.
Reggie.
Having now examined it in a bit more detail (forced by looking at a
simulation output ;-), I can suggest that you consider your point 2
more carefully. Realize that it is indeed a square wave, but it goes
negative by Vf when the switch is off--that is, when D2 is forward
biased. That means that to get the average voltage over a cycle at
the inductor input to be equal to the output voltage, the peak voltage
at that point must go higher than Vout/dutyfactor. In fact, it must
go higher by Vf*(dutyfactor-1), I believe...so then it spends
dutyfactor at Vout/dutyfactor + Vf*(dutyfactor-1) and (dutyfactor-1)
at -Vf. Does that help?
Cheers,
Tom
Cripes, belay that. The last part should read (barring more slips on
the mental gymnastics bar):
high voltage at inductor input = (Vout + Vf*(1-dutyfactor))/dutyfactor
= D2 peak reverse voltage
low voltage at inductor input = -Vf
Average voltage at inductor input
= high * dutyfactor + low * (1-dutyfactor)
= Vout + Vf*(1-dutyfactor) - Vf*(1-dutyractor)
= Vout
as expected.
Cheers,
Tom- Hide quoted text -
- Show quoted text -

HI Tom,

Thanks for your previous post. I can see what you are doing and agree
that the Vf should be included for completeness.

I have looked at your workings and can manage to verify them to be
correct from first principles only when Vf is negative. eg
Average=area under curve/length of base etc...

Huh?? Vf is positive; typically about half a volt for Schottkys near
their max current. That is, during the time that D2 is conducting,
the voltage at the input to the inductor is NEGATIVE by Vf. Then when
the transformer is supplying energy, the voltage at the inductor input
must be more positive by an amount to make up for that, if you will.

When I tried to start from first principles myself I get something
that concurs with your result eg Vpeak = [[vout+vf]/dutyfactor]-vf
(ignoring lekage L spikes)

However your line "high voltage at inductor input = (Vout + Vf*(1-
dutyfactor))/dutyfactor" I believe is a little confusing as you are
adding a negative term there fore you are subtracting.

Nope, adding. The voltage must be HIGHER because during the
transformer "off" time the voltage to the inductor input was negative.

In effect you are reducing the on area by the off area.

I think it should read "high voltage at inductor input = (Vout - Vf*(1-
dutyfactor))/dutyfactor" for clarity, but I am picking at straws!

This result would suggest that the max reverse voltage on D2 occurs
when the duty factor = min, when duty factor is min primary voltage is
at max, so this concurs with step 8 of mine above.

Now if duty factor = 0.01 the term (Vout - Vf*(1-dutyfactor))/
dutyfactor becomes very large, practically too large for the input
voltage and the turns ratio to produce.

Say vout = 25V Vf=0.3 the term above = [25-(0.3x0.01)]/ 0.01 = 2499.7V
so perhaps the application note tries to put a limiting factor of
vinmax/vinmin.duty max to cap the max voltage to real realistic
values? What do you think?

Practically I don't know what to specify for the min duty factor in a
real circuit so I couldn't work out max reverse voltage on the diode
using your method.

Well, another way to look at it is that the voltage out of the
transformer, neglecting effects of resonances and leakage inductances
and the like, is the primary voltage times the turns ratio, and the
reverse voltage across the freewheeling diode is that voltage minus
the forward drop of the diode connected from the transformer to the
inductor. Start with the basics: what's the input voltage range?
What output voltage and current do you need? You can then play with
different turns ratios to check out max and min duty factors, and from
those, figure the peak voltages and currents. For an ideal case, you
can do all that with a spreadsheet. Or--use something like LTSpice.

Am I talking twaddle or am I making sense?

Reggie.

Cheers,
Tom- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

Hi Tom,

I have made a mistake! I have looked again at what you have said and I
do agree with your explanation (I copied your maths down incorrectly
from the web) sorry. What I had copied down was wrong not your
response to my original question.

However my original workings still stand, just the way I went about it
was different from you.

Imagine the actual waveform across D2 with negative value during toff
and positive waveform during ton. I averaged the waveform as it is,
but using a negative sign for the area
Vf.(1-dutyfactor).T

Eg:

Average voltage = vout = area under curve / length of base

Average voltage = vout = [high area - low area] / length of base

(high voltage at inductor input = Vpkdiode)

Average voltage = vout = [ton. Vpkdiode - toff . Vf] / length of base

Average voltage = vout = [dutyfactor.T . Vpkdiode - (1 -
dutyfactor).Tƒn. Vf] / T

Solve for Vpkdiode

Vpkdiode = ((Vout+Vf)/ƒn dutyfactor)-Vf

This gives the same result as your :

high voltage at inductor = [Vout + Vf (1 - dutyfactor)] / dutyfactor

The important points that you covered are to maintain the same Volt
seconds across the coil over one period (otherwise it will saturate),
which is the same thing as saying the input average voltage has to
equal the output average voltage (which is of course vout).

Correct me if I am wrong but I think you have averaged both the high
area and the low area (with their corresponding times), added them
together to give a total increased average. You then divide this
average by the dutyfactor to give the (high voltage at inductor input
= Vpkdiode)

Your taking into account the negative voltage at the input to the
inductor and compensating for it theoretically is accurate, but in
practice wont the duty factor compensate for the small inaccuracies
introduced by neglecting this negative value? I know if the output
voltage is smaller the errors will be a bigger percentage and more
problematic.

Also when you are modelling leakage inductance to predict spike
frequency and amplitude, how do you take into account track
inductances / capacitances before you have designed the PCB? Or do you
assume good practice PCB design rules to minimise track parasitics?

Say for a 24V output will the error on the diode reverse voltage
specification be larger or smaller than the leakage inductance spike?
I am not sure, but I bet the leakage inductance spike will be bigger
than the error introduced by neglecting the negative voltage across
D2s effect on reverse voltage, what do you think?

Once again sorry for my mistake,

I hope I have explained myself clearly this time.

I take your point about min duty and doing a spreadsheet.

Reggie.

 

[reggie]

Hi all
I have found a formula which confuses me. Its in ST Topologies for
SMPSUs Bu L.Wuidart (AN513/0393)
http://www.stmicroelectronics.com/stonline/books/pdf/docs/3721.pdf
Page 12/18 formula :FREEWHEELING D2:
Vrrm>= [Vinmax.(Vout+Vf)]/ Vinmin . dutymax -----( STs equation)
The link describes the reverse voltage on secondary freewheeling diode
of an isolated forward converter:
D1 conducts during fet ontime
D2 conducts during fet off time (freewheeling diode)
I am trying to understand the above equation but am struggling to
derive it this is what I have so far:
1) The voltage on the output of the LC filter is the time average
voltage (Vout) of the input to the LC filter times the duty factor
2) The input to the LC filter is a square wave voltage with a peak
voltage
Vpeak = Vout / dutyfactor
3) D1s forward voltage drop = Vf
4) The voltage on the secondry turns of the transformer is Vs = Vpeak
+ Vf
5) also Vs = (ns/np) .Vp [Vp = volts on primary]
6) therefore merging 4&5 Vpeak + Vf = (ns/np).Vp
7) Vpeak = [(ns/np).Vp] -Vf
8) now to get worst case reverse voltage on D2 Vpeakmax =[(ns/
np).Vpmax]-Vf
A) Is what I have said in step 1 to 8 above correct?
B) I don't see how STs equation can have a "+" Vf term in it, I think
the Vo+Vf is the voltage on the secondary turns not the reverse
voltage across D2. Am I wrong?
Please help, I am going out of my tiny little mind trying to figure
this out.
Reggie.
Having now examined it in a bit more detail (forced by looking at a
simulation output ;-), I can suggest that you consider your point 2
more carefully. Realize that it is indeed a square wave, but it goes
negative by Vf when the switch is off--that is, when D2 is forward
biased. That means that to get the average voltage over a cycle at
the inductor input to be equal to the output voltage, the peak voltage
at that point must go higher than Vout/dutyfactor. In fact, it must
go higher by Vf*(dutyfactor-1), I believe...so then it spends
dutyfactor at Vout/dutyfactor + Vf*(dutyfactor-1) and (dutyfactor-1)
at -Vf. Does that help?
Cheers,
Tom
Cripes, belay that. The last part should read (barring more slips on
the mental gymnastics bar):
high voltage at inductor input = (Vout + Vf*(1-dutyfactor))/dutyfactor
= D2 peak reverse voltage
low voltage at inductor input = -Vf
Average voltage at inductor input
= high * dutyfactor + low * (1-dutyfactor)
= Vout + Vf*(1-dutyfactor) - Vf*(1-dutyractor)
= Vout
as expected.
Cheers,
Tom- Hide quoted text -
- Show quoted text -

HI Tom,

Thanks for your previous post. I can see what you are doing and agree
that the Vf should be included for completeness.

I have looked at your workings and can manage to verify them to be
correct from first principles only when Vf is negative. eg
Average=area under curve/length of base etc...

Huh?? Vf is positive; typically about half a volt for Schottkys near
their max current. That is, during the time that D2 is conducting,
the voltage at the input to the inductor is NEGATIVE by Vf. Then when
the transformer is supplying energy, the voltage at the inductor input
must be more positive by an amount to make up for that, if you will.

When I tried to start from first principles myself I get something
that concurs with your result eg Vpeak = [[vout+vf]/dutyfactor]-vf
(ignoring lekage L spikes)

However your line "high voltage at inductor input = (Vout + Vf*(1-
dutyfactor))/dutyfactor" I believe is a little confusing as you are
adding a negative term there fore you are subtracting.

Nope, adding. The voltage must be HIGHER because during the
transformer "off" time the voltage to the inductor input was negative.

In effect you are reducing the on area by the off area.

I think it should read "high voltage at inductor input = (Vout - Vf*(1-
dutyfactor))/dutyfactor" for clarity, but I am picking at straws!

This result would suggest that the max reverse voltage on D2 occurs
when the duty factor = min, when duty factor is min primary voltage is
at max, so this concurs with step 8 of mine above.

Now if duty factor = 0.01 the term (Vout - Vf*(1-dutyfactor))/
dutyfactor becomes very large, practically too large for the input
voltage and the turns ratio to produce.

Say vout = 25V Vf=0.3 the term above = [25-(0.3x0.01)]/ 0.01 = 2499.7V
so perhaps the application note tries to put a limiting factor of
vinmax/vinmin.duty max to cap the max voltage to real realistic
values? What do you think?

Practically I don't know what to specify for the min duty factor in a
real circuit so I couldn't work out max reverse voltage on the diode
using your method.

Well, another way to look at it is that the voltage out of the
transformer, neglecting effects of resonances and leakage inductances
and the like, is the primary voltage times the turns ratio, and the
reverse voltage across the freewheeling diode is that voltage minus
the forward drop of the diode connected from the transformer to the
inductor. Start with the basics: what's the input voltage range?
What output voltage and current do you need? You can then play with
different turns ratios to check out max and min duty factors, and from
those, figure the peak voltages and currents. For an ideal case, you
can do all that with a spreadsheet. Or--use something like LTSpice.

Am I talking twaddle or am I making sense?

Reggie.

Cheers,
Tom- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

Hi Tom,

I have made a mistake! I have looked again at what you have said and I
do agree with your explanation (I copied your maths down incorrectly
from the web) sorry. What I had copied down was wrong not your
response to my original question.

However my original workings still stand, just the way I went about it
was different from you.

Imagine the actual waveform across D2 with negative value during toff
and positive waveform during ton. I averaged the waveform as it is,
but using a negative sign for the area
Vf.(1-dutyfactor).T

Eg:

Average voltage = vout = area under curve / length of base

Average voltage = vout = [high area - low area] / length of base

(high voltage at inductor input = Vpkdiode)

Average voltage = vout = [ton. Vpkdiode - toff . Vf] / length of base

Average voltage = vout = [dutyfactor.T . Vpkdiode - (1 -
dutyfactor).Tƒn. Vf] / T

Solve for Vpkdiode

Vpkdiode = ((Vout+Vf)/ƒn dutyfactor)-Vf

This gives the same result as your :

high voltage at inductor = [Vout + Vf (1 - dutyfactor)] / dutyfactor

The important points that you covered are to maintain the same Volt
seconds across the coil over one period (otherwise it will saturate),
which is the same thing as saying the input average voltage has to
equal the output average voltage (which is of course vout).

Correct me if I am wrong but I think you have averaged both the high
area and the low area (with their corresponding times), added them
together to give a total increased average. You then divide this
average by the dutyfactor to give the (high voltage at inductor input
= Vpkdiode)

Your taking into account the negative voltage at the input to the
inductor and compensating for it theoretically is accurate, but in
practice wont the duty factor compensate for the small inaccuracies
introduced by neglecting this negative value? I know if the output
voltage is smaller the errors will be a bigger percentage and more
problematic.

Also when you are modelling leakage inductance to predict spike
frequency and amplitude, how do you take into account track
inductances / capacitances before you have designed the PCB? Or do you
assume good practice PCB design rules to minimise track parasitics?

Say for a 24V output will the error on the diode reverse voltage
specification be larger or smaller than the leakage inductance spike?
I am not sure, but I bet the leakage inductance spike will be bigger
than the error introduced by neglecting the negative voltage across
D2s effect on reverse voltage, what do you think?

Once again sorry for my mistake,

I hope I have explained myself clearly this time.

I take your point about min duty and doing a spreadsheet.

Reggie.