Opto-couplers in smps - part 3 (final chapter)

[krw]

Nerr, nerr, nerr. I said that first.

I'll take that as an admission that you're an illiterate dumb
donkey. Moved goal posts noted.

 

[Eeyore]

[email protected] says...

I'll take that as an admission that you're an illiterate dumb
donkey. Moved goal posts noted.

I take that as an admission you need to lighten up.

Not playing any more now.

Graham

 

[krw]

I take that as an admission you need to lighten up.

Why should I lighten up. It's fun whacking dumb donkeys. It makes
'em look even dumber when they squirm.

Not playing any more now.

You will. You can't resist making an ass out of yourself.

 

[Terry Given]

Terry,

I've uploaded an improved schematic.
All components are now given a value, including "primary" side opto
circuit and 2525 internal error amp, though is used only as a buffer.

Best regards
Stefan

OK.

Firstly, I wonder what the effect of measuring the voltage across both
outputs is. I look at that and think that Ltotal = 160uH and Ctotal =
10mF. So a quick energy balance:

I+ = I- = I (balanced load)

E_L1 = 0.5*80uH*I^2

E_L2 = 0.5*80uH*I^2

E_L = 0.5*160uH*I^2. OK, the L's sum.

likewise for the caps:

E_C1 = 0.5*20mF*(V/2)^2,

E_C2 = 0.5*20mF*(V/2)^2,

E_c = 20mF*(V/2)^2 = 0.5*10mF*V^2


so a re-drawn circuit is:

---] xfmr [---[160uH]--[ESR_L]--+---- +100V
40T 40T |
[10mF]
|
[ESR_C]
|
0V

DC gain = (Vin/Vramp)*(Ns/Np) = (320/2.6)*(40/40) = 123


note: if you regulated only one output, then you would reflect the other
LC circuit (easy, same no. of turns) to the regulated output, and
calculate the parallel equivalent, giving 40uH + 40mF, and the DC gain
would be 61.5

Note that the effective inductor ESR is the sum of the two; for the
caps, it turns out to be that of a single 10mF cap, as each 20mF cap is
2 in parallel (ESR_20mF = 0.5*ESR_10mF) so the sum is ESR_10mF


You can usually ignore the choke ESR. Because it has the full output
current flowing thru it, the choke is (usually, not always) designed for
extremely low losses (hence low ESR). if its not quite negligible, the
next approximation is to add it to the cap ESR. Or you can just use the
actual transfer function:

The AC transfer function of the power stage is:


Vout(s)/Vc(s) = Xc(s)/[Xc(s) + Xl(s)]

Xc(s) = ESR_C = 1/(s*C)

Xl(s) = ESR_L + s*L

(s = j*w)


the zener resistor current is 11mA c.f. a shunt circuit maximum of

Ishunt_max = (12 - 1.2 - 2.5)/2k2 = 8.3mA

this is a lot higher than your 5mA figure. I guesstimated Vled = 1.2V.
The TL431 cant pull lower than 2.5V.

regardless, 8.3mA < 11mA, so the zener will clamp the opto supply at ~
12V, and it wont vary by much.

the LED will see a maximum of 8.3mA - 1.2mA (the 1k resistor) = 7.1mA.


OK, we've established the zener shunt regulator works.

The TL431 circuit transfer function (from Vout to Vk) is:

Vk(s)/Vout(s) = -Zf(s)/Zin(s)

// = parallel

Zf(s) = [1/(s*220pF)]//(47k + 1/(s*22nF))

Zin(s) = 47k + 47k//[1/(s*22nF)]


the transfer function from Vk to Iled is simply -1/2k2

so Iled(s)/Vout(s) = (1/2k2)*Zf(s)/Zin(s)

the transfer function from the opto transistor to led is:

Ibjt(s)/Iled(s) = CTR*1/(1+s/w_opto)

where w_opto is the corner frequency of the opto itself. You probably
cant ignore this.


and to get from the opto bjt to the control voltage, the transfer
function is Acl(s)*(-8k2)

its -8k2 because the voltage is (5V - 8k2*Ibjt), and the 5V (being
constant) disappears when you take partial derivatives.


where Acl(s) is the error amplifier closed-loop gain. for a shitty old
smps controller, these error amplifiers can have quite low GBW. look it
up in the datasheet, but I suspect you can ignore it as the gain is one.

Acl(s) = 1/(1+s/(2*pi*GBW))

OK, I looked it up, its 1MHz so you can safely ignore the controller
error amp, and set

Acl(s) = 1.

Putting it all together, you get:

Vc(s)/Vout(s) = 1*(-8k2)*CTR*[1/(1+s/w_opto)]*(1/2k2)*Zf(s)/Zin(s)

simplifying:

Vc(s)/Vout(s) = -3.73*CTR*[1/(1+s/w_opto)]*Zf(s)/Zin(s)


and thats all you need to close your loop.

I might have made an algebraic error or 6, but the technique is pretty
much there. If I was being paid, I'd be very careful to get it right


Cheers
Terry

 

[Andrew Edge]

Terry,

I've uploaded an improved schematic.
All components are now given a value, including "primary" side opto
circuit and 2525 internal error amp, though is used only as a buffer.

Best regards
Stefan

Where did you upload that ... I still see the old circuit .

Andy

 

[Steve]

I've uploaded an improved schematic.
All components are now given a value, including "primary" side opto
circuit and 2525 internal error amp, though is used only as a buffer.

Best regards
Stefan

OK.

Firstly, I wonder what the effect of measuring the voltage across both
outputs is. I look at that and think that Ltotal = 160uH and Ctotal =
10mF. So a quick energy balance:

I+ = I- = I (balanced load)

E_L1 = 0.5*80uH*I^2

E_L2 = 0.5*80uH*I^2

E_L = 0.5*160uH*I^2. OK, the L's sum.

likewise for the caps:

E_C1 = 0.5*20mF*(V/2)^2,

E_C2 = 0.5*20mF*(V/2)^2,

E_c = 20mF*(V/2)^2 = 0.5*10mF*V^2

so a re-drawn circuit is:

---] xfmr [---[160uH]--[ESR_L]--+---- +100V
40T 40T |
[10mF]
|
[ESR_C]
|
0V

DC gain = (Vin/Vramp)*(Ns/Np) = (320/2.6)*(40/40) = 123

note: if you regulated only one output, then you would reflect the other
LC circuit (easy, same no. of turns) to the regulated output, and
calculate the parallel equivalent, giving 40uH + 40mF, and the DC gain
would be 61.5

Note that the effective inductor ESR is the sum of the two; for the
caps, it turns out to be that of a single 10mF cap, as each 20mF cap is
2 in parallel (ESR_20mF = 0.5*ESR_10mF) so the sum is ESR_10mF

You can usually ignore the choke ESR. Because it has the full output
current flowing thru it, the choke is (usually, not always) designed for
extremely low losses (hence low ESR). if its not quite negligible, the
next approximation is to add it to the cap ESR. Or you can just use the
actual transfer function:

The AC transfer function of the power stage is:

Vout(s)/Vc(s) = Xc(s)/[Xc(s) + Xl(s)]

Xc(s) = ESR_C = 1/(s*C)

Xl(s) = ESR_L + s*L

(s = j*w)

the zener resistor current is 11mA c.f. a shunt circuit maximum of

Ishunt_max = (12 - 1.2 - 2.5)/2k2 = 8.3mA

this is a lot higher than your 5mA figure. I guesstimated Vled = 1.2V.
The TL431 cant pull lower than 2.5V.

regardless, 8.3mA < 11mA, so the zener will clamp the opto supply at ~
12V, and it wont vary by much.

the LED will see a maximum of 8.3mA - 1.2mA (the 1k resistor) = 7.1mA.

OK, we've established the zener shunt regulator works.

The TL431 circuit transfer function (from Vout to Vk) is:

Vk(s)/Vout(s) = -Zf(s)/Zin(s)

// = parallel

Zf(s) = [1/(s*220pF)]//(47k + 1/(s*22nF))

Zin(s) = 47k + 47k//[1/(s*22nF)]

the transfer function from Vk to Iled is simply -1/2k2

so Iled(s)/Vout(s) = (1/2k2)*Zf(s)/Zin(s)

the transfer function from the opto transistor to led is:

Ibjt(s)/Iled(s) = CTR*1/(1+s/w_opto)

where w_opto is the corner frequency of the opto itself. You probably
cant ignore this.

and to get from the opto bjt to the control voltage, the transfer
function is Acl(s)*(-8k2)

its -8k2 because the voltage is (5V - 8k2*Ibjt), and the 5V (being
constant) disappears when you take partial derivatives.

where Acl(s) is the error amplifier closed-loop gain. for a shitty old
smps controller, these error amplifiers can have quite low GBW. look it
up in the datasheet, but I suspect you can ignore it as the gain is one.

Acl(s) = 1/(1+s/(2*pi*GBW))

OK, I looked it up, its 1MHz so you can safely ignore the controller
error amp, and set

Acl(s) = 1.

Putting it all together, you get:

Vc(s)/Vout(s) = 1*(-8k2)*CTR*[1/(1+s/w_opto)]*(1/2k2)*Zf(s)/Zin(s)

simplifying:

Vc(s)/Vout(s) = -3.73*CTR*[1/(1+s/w_opto)]*Zf(s)/Zin(s)

and thats all you need to close your loop.

I might have made an algebraic error or 6, but the technique is pretty
much there. If I was being paid, I'd be very careful to get it right

Cheers
Terry- Dölj citerad text -

- Visa citerad text -

Thanks Terry,

I'll go through your calculations and try to understand.

Again, thanks

Stefan