Somebody explaining this design?

T

Tony Williams

Jan 1, 1970
0
Vout = [ (Vs-Vx)/R2a - (I + Vin/R1) ]*R3a
- [ (Vs-Vx)/R2b - (I - Vin/R1) ]*R3b

Now make R2a=R2b=R2 and R3a=R3b=R3, and cancel things.

Vout = -2.Vin*(R3/R1).
[/QUOTE]
It's simpler to look at the collector circuits as Norton current
sources with shunt impedances the R2's. Then each is driven by a
differential current of +/-Vin/R1 which is an equivalent voltage
drive of +/-Vin*R2/R1 in series with R2 into the DA. The DA only
responds to the equivalent differential input, which is then
2*Vin*R2/R1, with a gain of R3/R2, for a Vout=2Vin*R3/R1.

Probably. Except that I find it easier to get to
it via that big sum for Vout, which can then be
expanded-out to show the effects of the various
mismatches, etc..... which affect the Gain, DC
Offset or PSRR, or whatever.

Note that whilst the R2's do cancel out, their
value has a marked effect on the front end Gain,
and therefore on the equivalent input noise.
 
J

Jim Thompson

Jan 1, 1970
0
Pullup-R's DON'T cancel.

Well, let's do some sums. Admittedly on a simplified and
idealised circuit (to make life easy for me).
R2a and R2b are the Pull-up R's in question.

-------+-------------+---Vs
| |
\ \ R3b
/R2a R2b/ +-/\/\---+-->Vout
\ \ | |
| | | _ |
| +---------+--|- \ |
| | | >-+
Vx---> +-------------|---------+--|+_/
| | |
\|/I1 I2\|/ \
| | /R3a
|/ \| \
Vin ---|npn npn|----. |
|\e e/| | |
0v--- | R1 | ---+--+---0v
+----/\/\-----+
| |
\|/ I I \|/
Constant current

I1 = (I + Vin/R1) and I2 = (I - Vin/R1).

First find out where I1 sets the opamp +ve input (Vx).

I(R3a) = (Vs-Vx)/R2a - I1 = (Vs-Vx)/R2a - (I + Vin/R1).

Vx = I(R3a)*R3a = [ (Vs-Vx)/R2a. - (I + Vin/R1) ]*R3a

The opamp negative feedback also sets the -ve input at Vx.

So I(R3b) = (Vs-Vx)/R2b - I2 = (Vs-Vx)/R2b - (I - Vin/R1).

So V(R3b) = I(R3b)*R3b = [ (Vs-Vx/R2b) - (I - Vin/R1) ]*R3b.

Voltage out, Vout = Vx - V(R3b).

Vout = [ (Vs-Vx)/R2a - (I + Vin/R1) ]*R3a
- [ (Vs-Vx)/R2b - (I - Vin/R1) ]*R3b

Now make R2a=R2b=R2 and R3a=R3b=R3, and cancel things.

Vout = -2.Vin*(R3/R1).

That's a different circuit than the original.

...Jim Thompson
 
E

Eeyore

Jan 1, 1970
0
Jim said:
Tony Williams said:
Jim said:
Tony Williams wrote:
Isn't it just the Pullup-R divided by the Tail-R,
times the final Feedback-R divided by the Pullup-R,
probably with a x2 in there somewhere?
So the Pullup-R cancels, leaving the overall Gain
as the ratio of final Feedback-R/Tail-R.
Pullup-R's DON'T cancel.

Well, let's do some sums. Admittedly on a simplified and
idealised circuit (to make life easy for me).
R2a and R2b are the Pull-up R's in question.

-------+-------------+---Vs
| |
\ \ R3b
/R2a R2b/ +-/\/\---+-->Vout
\ \ | |
| | | _ |
| +---------+--|- \ |
| | | >-+
Vx---> +-------------|---------+--|+_/
| | |
\|/I1 I2\|/ \
| | /R3a
|/ \| \
Vin ---|npn npn|----. |
|\e e/| | |
0v--- | R1 | ---+--+---0v
+----/\/\-----+
| |
\|/ I I \|/
Constant current

I1 = (I + Vin/R1) and I2 = (I - Vin/R1).

First find out where I1 sets the opamp +ve input (Vx).

I(R3a) = (Vs-Vx)/R2a - I1 = (Vs-Vx)/R2a - (I + Vin/R1).

Vx = I(R3a)*R3a = [ (Vs-Vx)/R2a. - (I + Vin/R1) ]*R3a

The opamp negative feedback also sets the -ve input at Vx.

So I(R3b) = (Vs-Vx)/R2b - I2 = (Vs-Vx)/R2b - (I - Vin/R1).

So V(R3b) = I(R3b)*R3b = [ (Vs-Vx/R2b) - (I - Vin/R1) ]*R3b.

Voltage out, Vout = Vx - V(R3b).

Vout = [ (Vs-Vx)/R2a - (I + Vin/R1) ]*R3a
- [ (Vs-Vx)/R2b - (I - Vin/R1) ]*R3b

Now make R2a=R2b=R2 and R3a=R3b=R3, and cancel things.

Vout = -2.Vin*(R3/R1).

That's a different circuit than the original.

...Jim Thompson

It is in fact just like the one I posted in abse.

Graham
 
T

Tony Williams

Jan 1, 1970
0
That's a different circuit than the original.

I was trying to avoid even looking at those silly
resistors in series with the opamp inputs.
But lets drop them in. R4a and R4b below.

-------+-------------+---Vs
| |
\ \ R3b
/R2a R2b/ +-/\/\---+-->Vout
\ \ | |
| | R4b | _ |
| +--/\/\---+--|- \ |
| | Vx | >-+
+-------------|--/\/\---+--|+_/
| | R4a |
\|/I1 I2\|/ \
| | /R3a
|/ \| \
Vin ---|npn npn|----. |
|\e e/| | |
0v--- | R1 | ---+--+---0v
+----/\/\-----+
| |
\|/ I I \|/
Constant current

Doing similar sums as before, I get...

Vout/Vin = -2(R3/R1)*( R2/(R2+R4) ).

Now put some spin on it..... Call R4 = K*R2.

Vout/Vin = -2(R3/R1)*( 1/(1+K) ). :)
 
C

colin

Jan 1, 1970
0
Tony Williams said:
I was trying to avoid even looking at those silly
resistors in series with the opamp inputs.
But lets drop them in. R4a and R4b below.

-------+-------------+---Vs
| |
\ \ R3b
/R2a R2b/ +-/\/\---+-->Vout
\ \ | |
| | R4b | _ |
| +--/\/\---+--|- \ |
| | Vx | >-+
+-------------|--/\/\---+--|+_/
| | R4a |
\|/I1 I2\|/ \
| | /R3a
|/ \| \
Vin ---|npn npn|----. |
|\e e/| | |
0v--- | R1 | ---+--+---0v
+----/\/\-----+
| |
\|/ I I \|/
Constant current

Doing similar sums as before, I get...

Vout/Vin = -2(R3/R1)*( R2/(R2+R4) ).

Now put some spin on it..... Call R4 = K*R2.

Vout/Vin = -2(R3/R1)*( 1/(1+K) ). :)

nope, r2a,b now soak up some of the differential current.

Colin =^.^=
 
E

Eeyore

Jan 1, 1970
0
Tony said:
Now put some spin on it..... Call R4 = K*R2.

Vout/Vin = -2(R3/R1)*( 1/(1+K) ). :)

Nice way of showing the effect !

Graham
 
C

colin

Jan 1, 1970
0
colin said:
nope, r2a,b now soak up some of the differential current.

oops missed the fact that r2 is included via K,
didnt realise youd changed your mind about them cancelling.
 
T

Tony Williams

Jan 1, 1970
0
nope, r2a,b now soak up some of the differential current.
[/QUOTE]
oops missed the fact that r2 is included via K,
didnt realise youd changed your mind about them cancelling.

I didn't change my mind as such because in the first
instance I didn't realise that anyone was considering
those series resistors seriously.

I must admit though that the slippery quickstep to the
1/(1+K) was in the hope that JT would read it with his
bad eye and not realise what was going on. :)
 
W

Winfield Hill

Jan 1, 1970
0
( Repost, apparently the first one didn't make it!? )


Right, it's a slick single-stage amplifier,
compared to Phil's two-stage amplifier.
I was trying to avoid even looking at those
silly resistors in series with the opamp inputs.
But lets drop them in. R4a and R4b below.

-------+-------------+---Vs
| |
\ \ R3b
/R2a R2b/ +-/\/\---+-->Vout
\ \ | |
| | R4b | _ |
| +--/\/\---+--|- \ |
| | Vx | >-+
+-------------|--/\/\---+--|+_/
| | R4a |
\|/I1 I2\|/ \
| | /R3a
|/ \| \
Vin ---|npn npn|----. |
|\e e/| | |
0v--- | R1 | ---+--+---0v
+----/\/\-----+
| |
\|/ I I \|/
Constant current

Doing similar sums as before, I get...

Vout/Vin = -2(R3/R1)*( R2/(R2+R4) ).

Now put some spin on it..... Call R4 = K*R2.

Vout/Vin = -2(R3/R1)*( 1/(1+K) ). :)

Tony, thanks loads, for taking the time to put up
an ASCII drawing, where everyone can be on the same
page in the discussion. This business of posting
web page links, or placing pdfs on a.b.s.e., etc.,
simply does not lend itself well to a conversation.

I imagine one of Jim's criticisms of the circuit,
I'm guessing since he refused to spell it out, is
that the R4 resistors take away from the potential
gain of the overall amplifier, or, for a set gain,
take away loop gain that could be used to reduce
distortion, etc. Your first (top) drawing without
the R4 resistors is better, but it still suffers
from the fact that the R2 resistors will have to
be smaller than the R3 resistors, shunting current.
Even though R2 doesn't appear in the gain equation,
they are degrading the performance. One solution
is to replace R2 with current sources, Ix. These
would have to be servo'd to match I1, with another
opamp looking at the average value of Vx compared
to some bias voltage, etc.

servo'd constant current
| | R3b
\|/ Ix \|/ ,-/\/\---+-->Vout
| | | _ |
| +---------+--|- \ |
| | | >-'
Vx---> +-------------|---------+--|+_/
| | |
\|/I1 I2\|/ \
| | /R3a
|/ \| \
Vin ---| npn npn |----. |
|\e e/| | |
0v--- | R1 | ---+--+---0v
+----/\/\-----+
| |
\|/ I I \|/
Constant current

This would be a bit of a mess, for an arguable
improvement, but possibly not something an IC
designer, who generally has lots of silicon
available, would shy away from.

Now, to change the subject to an item I'm not at
all happy with, the huge ugly electrolytic in the
input-pair emitter path, see C1 below.

-------+-------------+---Vs
| | G = 2 R3 / R1
\ \
/R2a R2b/ R3b
\ \ ,-/\/\---+-->Vout
| | | _ |
| +---------+--|- \ |
| | | >-'
Vx---> +-------------|---------+--|+_/
| | |
| | \
| Q1 Q2 | /R3a
|/ \| \
--+ --| npn npn |--. |
Vin | |\e e/| | |
--+-|-- | R1 C1 | --' -+---0v
| | +--/\/\--||---+
\ \ | | Q1 Q2 composite
/ / \|/ I I \|/ Sziklai pairs
| | Constant currents
--+-+-0V

C1 is deemed necessary to prevent the offset voltage
of the input NPN transistors, which could be 25mV or
even 50mV, from saturating the output at G = 1000.
R1 is only 22 ohms at the highest gain, so C1 has to
be large, 1000uF for a 7Hz rolloff, in Phil's design,
http://sound.westhost.com/project66.htm

But C1 is trouble for several reasons. One problem
is the distortion created from the high dielectric
absorption for electrolytics. I've read that this
is an issue for frequencies up to 10x away from the
-3dB frequency, which implies degradation up to 70Hz
at the highest gain. I'd like to eliminate C1, and
rely on the set of C2 capacitors at the opamp input
to deal with the offset voltage. Phil used 47uF
electrolytics, which gives a 1.5Hz rolloff compared
to his R2 = 2.2k resistors, which nicely moves the
10x distortion region down to 15Hz.

And as a bonus, the 100ms time constant is fixed,
independent of gain. This compares to C1 R1, which
varied from 22ms to 10 seconds over the gain range.

-------+-------------+---Vs
| | G = 2 R3 / R1
\ \
/R2a R2b/ R3b
\ \ ,-/\/\---+-->Vout
| | C2b | _ |
| Vy--> +----||---+--|- \ |
| | C2a | >-'
Vx---> +-------------|----||---+--|+_/
| | + - |
| | \
| Q1 Q2 | /R3a
|/ \| \
--+ --| npn npn |--. |
Vin | |\e e/| | |
--+-|-- | R1 | --' -+---0v
| | +----/\/\-----+
\ \ | | Q1 Q2 composite
/ / \|/ I I \|/ Sziklai pairs,
| | Constant currents matched offsets
--+-+-0V

I'd solve the dc-offset problem simply by selecting
Q1 and Q2 for a maximum offset voltage of say 5mV.
The input transistors are cheap, easily allowing
for a small pile to choose from, and they should be
checked for noise anyway, before use.* If we assume
a 10mV worst-case input offset, this results in a
modest +/- 1-volt deviation from nominal at points
Vx and Vy. With C2 in place, that's acceptable.

What's more, thanks to the use of Sziklai pairs,
the operating current of each input transistor is
unaffected, because the current change is taken up
by the second transistor in each Sziklai pair.

So that's three parts taken from Phil's microphone
amplifier, arguably for an improved performance.

* Reading in the forum suggested by Martin Griffith
http://www.prodigy-pro.com/forum/viewtopic.php?t=20038
some in the studio pro-audio crowd like the 2n4403
pnp transistor for low-noise input stages (most of
us instrumentation engineers prefer other parts).
Apparently Motorola made very quiet ones in the good
old days, and Fairchild's parts are pretty good now.
But the 2n4403 isn't specified or suggested for low
noise, and surely each one should be individually
vetted for such use.
 
E

Eeyore

Jan 1, 1970
0
Winfield said:
* Reading in the forum suggested by Martin Griffith
http://www.prodigy-pro.com/forum/viewtopic.php?t=20038
some in the studio pro-audio crowd like the 2n4403
pnp transistor for low-noise input stages (most of
us instrumentation engineers prefer other parts).
Apparently Motorola made very quiet ones in the good
old days, and Fairchild's parts are pretty good now.
But the 2n4403 isn't specified or suggested for low
noise, and surely each one should be individually
vetted for such use.

They're nuts if they like the 2N4403. I have it used myself back around 1980
when there was simply nothing better.

Nice quite devices of that ilk but purpose designed became popular for a while
but are hard to come by these days.

The 2SA1083/4/5 seems to be the most widely available still. Other candidates
are the 2SA970 and 2SA1316.

Rohm's 2SB737 seems to have the lowest intrinsic resistnces I've ever seen but
is no longer manufactured. It also has a slightly higher Cob and I've seen its
influence on higher frequency THD btw.

Last time you posted on the subject, you said you had a stash of 2SB737s btw
Win. Do you still have them ?

Graham
 
T

Tony Williams

Jan 1, 1970
0
Winfield Hill said:
-------+-------------+---Vs
| | G = 2 R3 / R1
\ \
/R2a R2b/ R3b
\ \ ,-/\/\---+-->Vout
| | C2b | _ |
| Vy--> +----||---+--|- \ |
| | C2a | >-'
Vx---> +-------------|----||---+--|+_/
| | + - |
| | \
| Q1 Q2 | /R3a
|/ \| \
--+ --| npn npn |--. |
Vin | |\e e/| | |
--+-|-- | R1 | --' -+---0v
| | +----/\/\-----+
\ \ | | Q1 Q2 composite
/ / \|/ I I \|/ Sziklai pairs,
| | Constant currents matched offsets
--+-+-0V
I'd solve the dc-offset problem simply by selecting
Q1 and Q2 for a maximum offset voltage of say 5mV.
The input transistors are cheap, easily allowing
for a small pile to choose from, and they should be
checked for noise anyway, before use.* If we assume
a 10mV worst-case input offset, this results in a
modest +/- 1-volt deviation from nominal at points
Vx and Vy. With C2 in place, that's acceptable.

I still don't like the C2's where they are.

My first thought for solving the DC offset problem
for an AC amplifier was to sense the output offset,
integrate, and use it to unbalance the CC sources
in the input tails. As sketched below.

-------+-------------+---Vs
| | G = 2 R3 / R1
\ \
/R2a R2b/ R3b
\ \ .-/\/\---+-->Vout
| | | _ |
| +---------+--|- \ |
| | | >-+
Vx---> +-------------|---------+--|+_/ |
| | | |
| | \ \
| Q1 Q2 | /R3a /
|/ \| \ \
--+---| npn npn |--. | |
Vin | |\e e/| +----+---0v |
--+-|-- | R1 | |
| | +----/\/\-----+ .----||----+
\ \ | | | _ |
/ / \|/ I I \|/ | / -|---'
| | Constant currents <----+--< |
--+-+-0V \_+|--0v
 
E

Eeyore

Jan 1, 1970
0
Tony said:
I still don't like the C2's where they are.

But you don't need the Cs.

See my post in abse "a better mic amplifier".

Graham
 
W

Winfield Hill

Jan 1, 1970
0
Tony said:
I still don't like the C2's where they are.

My first thought for solving the DC offset problem
for an AC amplifier was to sense the output offset,
integrate, and use it to unbalance the CC sources
in the input tails. As sketched below.

-------+-------------+---Vs
| | G = 2 R3 / R1
\ \
/R2a R2b/ R3b
\ \ .-/\/\---+-->Vout
| | | _ |
| +---------+--|- \ |
| | | >-+
Vx---> +-------------|---------+--|+_/ |
| | | |
| | \ \
| Q1 Q2 | /R3a /
|/ \| \ \
--+---| npn npn |--. | |
Vin | |\e e/| +----+---0v |
--+-|-- | R1 | |
| | +----/\/\-----+ .----||----+
\ \ | | | _ |
/ / \|/ I I \|/ | / -|---'
| | Constant currents <----+--< |
--+-+-0V \_+|--0v

That's much better, Tony. I was sure you'd have a
bright idea. BTW, why do you keep drifting back to
a single-ended input? Do you have more to tell us?
 
J

Jim Thompson

Jan 1, 1970
0
oops missed the fact that r2 is included via K,
didnt realise youd changed your mind about them cancelling.

I didn't change my mind as such because in the first
instance I didn't realise that anyone was considering
those series resistors seriously.

I must admit though that the slippery quickstep to the
1/(1+K) was in the hope that JT would read it with his
bad eye and not realise what was going on. :)[/QUOTE]

I couldn't even see the screen with my bad eye. But it's fixed now
;-)

...Jim Thompson
 
T

Tony Williams

Jan 1, 1970
0
-------+-------------+---Vs
| | G = 2 R3 / R1
\ \
/R2a R2b/ R3b
\ \ .-/\/\---+-->Vout
| | | _ |
| +---------+--|- \ |
| | | >-+
Vx---> +-------------|---------+--|+_/ |
| | | |
| | \ \
| Q1 Q2 | /R3a /
|/ \| \ \
--+---| npn npn |--. | |
Vin | |\e e/| +----+---0v |
--+-|-- | R1 | |
| | +----/\/\-----+ .----||----+
\ \ | | | _ |
/ / \|/ I I \|/ | / -|---'
| | Constant currents <----+--< |
--+-+-0V \_+|--0v
[/QUOTE]
That's much better, Tony. I was sure you'd have a
bright idea. BTW, why do you keep drifting back to
a single-ended input? Do you have more to tell us?

I have this long standing medical condition Win, called
CBA, or sometimes BI.

Which stands for Couldn't Be Arsed, or Bone Idle.

In this case I was too BI to do the differential sums. :)
 
W

Winfield Hill

Jan 1, 1970
0
That's much better, Tony. I was sure you'd have a
bright idea. BTW, why do you keep drifting back to
a single-ended input? Do you have more to tell us?

I have this long standing medical condition Win,
called CBA, or sometimes BI.

Which stands for Couldn't Be Arsed, or Bone Idle.

In this case I was too BI to do the differential sums. :)[/QUOTE]

ROFLOL.
 
M

martin griffith

Jan 1, 1970
0
They're nuts if they like the 2N4403. I have it used myself back around 1980
when there was simply nothing better.

Nice quite devices of that ilk but purpose designed became popular for a while
but are hard to come by these days.

The 2SA1083/4/5 seems to be the most widely available still. Other candidates
are the 2SA970 and 2SA1316.

Rohm's 2SB737 seems to have the lowest intrinsic resistnces I've ever seen but
is no longer manufactured. It also has a slightly higher Cob and I've seen its
influence on higher frequency THD btw.

Last time you posted on the subject, you said you had a stash of 2SB737s btw
Win. Do you still have them ?

Graham
ISTR Barry Porter saying that CADAC had a stash of 10,000



martin
 
E

Eeyore

Jan 1, 1970
0
martin said:
ISTR Barry Porter saying that CADAC had a stash of 10,000

Goodness ! Dave Dearden's (formerly of DDA) outfit Audient were still using them a
while back but they were relying on their sub-contractor AIUI.

Graham
 
M

martin griffith

Jan 1, 1970
0
chomp

Goodness ! Dave Dearden's (formerly of DDA) outfit Audient were still using them a
while back but they were relying on their sub-contractor AIUI.

Graham

Dave Dearden, I remember him from the MCI days, nice guy.


martin
 
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