Op Amp Calculations

[RST Engineering \(jw\)]

A lot of years ago (38 to be exact) a wise old "rule of thumb" engineer
taught me a slick way of getting maximum gain out of an opamp without
resorting to very high or very low values of resistors.

As we all know, for an inverting opamp, the gain is given simply by Rf/Ri,
where Rf is the feedback resistor from output to inverting input and Ri is
the resistor between signal and inverting input. The DC level of the output
may be set anywhere you choose by an appropriate bias level on the
noninverting input. For AC amplifiers from a single supply, this is
generally Vcc/2 with capacitive coupling between Ri and the signal.

However, for very large AC gains, either Rf must be rather large or Ri
rather small. Rf being rather large makes the input voltage/current errors
become significant as regards quiescent DC output point and Ri being rather
small requires large capacitors for coupling and loading errors from the
signal source.

So, sez old wily rule-of-thumb, just break Rf into two reasonable sized
equal value resistors equal to Rf/2 and run them in series from output back
to (-) input. And, from the midpoint tap on these two resistors run a
series RC circuit to ground. Bingo, the AC gain improves greatly.

And guess what, it works. How do I calculate the R in the series RC circuit
I asks old wily. The answer comes back "Tweak it until you get the gain you
want." (Assume that C can be made appropriately large to get the
low-frequency gain you want.)

I haven't used that trick in an awfully long time, but I've got an
application that needs it. And, if I want to use Diddle's constant in a
simulation program I can fool around (ahem, heuristically experiment) to get
the gain I need.

However, I can't convince myself that I can mathematically come up with the
resistor value. I have googled the problem and come up short. Anybody got
a pointer to a URL that goes through the math of how this configuration
works? And what I'm doing to my phase margin?

Jim

 

[John Woodgate]

However, I can't convince myself that I can mathematically come up with
the resistor value. I have googled the problem and come up short.
Anybody got a pointer to a URL that goes through the math of how this
configuration works? And what I'm doing to my phase margin?

I believe Win has sort-of proprietary rights on this circuit. But it's
easy if you re-draw it. Why do these things come up in s.e.d. instead of
a.b.s.e?

ASCII art; use Courier font.

_______R4_______
| - |
o----R1--+---|\ |
| \_____ out |
| / | |
o------------|/ R2 |
+ |----'
R3
|
o--------------------+------ Gnd


You can see then that R2 and R3 just form a potential divider across the
output, allowing R4 and R2 to be much smaller for a given gain:

Gain = (R4/R1) x {(R2 + R3)}/R3, if R4 is very much larger than R3. If
it isn't, replace R4 by R4 + (R2R3/(R2 + R3).

The circuit is highly redundant, which means that you get to choose
highly inappropriate values for all the resistors and then wonder why it
doesn't work, even though the gain calculation works out.(;-)

You can probably avoid that by making R1 something in the 1 k to 10 k
range, but don't take bets. I remember one of my fellow students making
a 'see-saw' [1](what we now call an inverting amplifier) with 1 kohm and
1 Mohm around a 6J5 triode and wondering why the gain wasn't 1000.

[1] I think that's a 'teeter-totter' in US English, but I wonder if the
circuit was called that, or 'see-saw'.

 

[Noway2]

Adding an RC circuit to the feedback path would cause the feedback
impedance to become frequency dependant, which would produce different
gains depending on the frequency. I am having a bit of trouble
understanding how this configuration would increase the gain (Zf / Zi)
as opposed to reducing it (apparently I am not envisioning the circuit
correctly). Wouldn't placing the RC combination in the feedback path
cause the effective Rf (parralllel combination) to be lower, in which
case the gain would go down?

I have seen similar ideas used in common emitter transistor amplifier
circuits, where large AC gain is desired around a DC bias point. In
those cases, the capacitor shunts the AC thereby decreasing the emitter
resistance and increasing the gain (Rc / Re). I don't see how this
works for the op-amp case, though.

 

[[email protected]]

A lot of years ago (38 to be exact) a wise old "rule of thumb" engineer
taught me a slick way of getting maximum gain out of an opamp without
resorting to very high or very low values of resistors.

From a Burr Brown ap note,
http://focus.ti.com/lit/an/sboa061/sboa061.pdf, ....

• A very high resistance feedback resistor is MUCH better
than a low resistance in a T network. See Figure 5. Although
transimpedance gain (eOUT/iSIGNAL) is equivalent, the T network
will sacrifice performance. The low feedback resistance
will generate higher current noise (iN) and the voltage
divider formed by R1/R2 multiply input offset voltage, drift,
and amplifier voltage noise by the ratio of 1+
R1/R2. In most electrometer amplifiers, these input specifications
are not very good to start with. Multiplying an already
high offset and drift (sometimes as high as 3mV and 50mV/
°C) by use of a T network becomes impractical. By using a
far better amplifier, such as the OPA128, moderate T network
ratios can be accommodated and the resulting multiplied
errors will be far smaller. Although a single very-high
resistance will give better performance, the T network can
overcome such problems as gain adjustment and difficulty in
finding a large value resistor.

Jim "The other one." Meyer

 

[John Fields]

I believe Win has sort-of proprietary rights on this circuit. But it's
easy if you re-draw it. Why do these things come up in s.e.d. instead of
a.b.s.e?

ASCII art; use Courier font.

_______R4_______
| - |
o----R1--+---|\ |
| \_____ out |
| / | |
o------------|/ R2 |
+ |----'
R3
|
o--------------------+------ Gnd

---

I believe the circuit he described looks more like this:

Vcc------------+
|
[R1]
|
+----------|+\
| | >-----+-->Vout
Vin>------[R3]-------+----|-/ |
| | |
[R2] +-[R4]-+-[R5]-+
| |
GND>-----------+-[C1]--[R6]-+


Where, for DC,


R4 + R5 Vcc R2
Vout = -Vin --------- + ---------
R3 R1 + R2

For AC, however, (assuming Vin and Vout are perfect voltage sources)
we wind up with Vout trying to force the - input of the opamp to
stay at whatever voltage the + input is set to by R1 and R2.

As the frequency of Vin increases, the reactance of C1 will
decrease, diverting some of the signal from the output of the opamp
to ground, with the result that the output voltage will have to rise
in order to keep the voltage on the - input of the opamp the same as
the voltage on the + input.


Looking at R5R6C1 as a lowpass filter with single pole, we have:


Vout
|
[R5]
|
+----Vfb
|
[R6]
|
[C1]
|
GND

And,suddenly, it's too close to dinner (Thai beef salad and a nice
white for me and Thai beef salad and a nice red for she) and too
complicated to get into right now.

Mañana, talvez...

 

[The Phantom]

I believe Win has sort-of proprietary rights on this circuit. But it's
easy if you re-draw it. Why do these things come up in s.e.d. instead of
a.b.s.e?

ASCII art; use Courier font.

_______R4_______
| - |
o----R1--+---|\ |
| \_____ out |
| / | |
o------------|/ R2 |
+ |----'
R3
|
o--------------------+------ Gnd

---

I believe the circuit he described looks more like this:

Vcc------------+
|
[R1]
|
+----------|+\
| | >-----+-->Vout
Vin>------[R3]-------+----|-/ |
| | |
[R2] +-[R4]-+-[R5]-+
| |
GND>-----------+-[C1]--[R6]-+


Where, for DC,


R4 + R5 Vcc R2
Vout = -Vin --------- + ---------
R3 R1 + R2

For AC, however, (assuming Vin and Vout are perfect voltage sources)
we wind up with Vout trying to force the - input of the opamp to
stay at whatever voltage the + input is set to by R1 and R2.

For AC, assuming that C1 is very large so that it can be treated as an AC short, and
the op amp gain is infinite, the gain is:

-((r5*r6 + r4*(r5 + r6))/(r3*r6))

If A is the op amp gain and C1 is included in the calculation, the gain is :

-A*(r4 + r5) - A*c1*(r5*r6 + r4*(r5 + r6))*s
-------------------------------------------------------------------------
r3 + A*r3 + r4 + r5 + c1*(r5*r6 + r4*(r5 + r6) + r3*(r5 + r6 + A*r6))*s

As the frequency of Vin increases, the reactance of C1 will
decrease, diverting some of the signal from the output of the opamp
to ground, with the result that the output voltage will have to rise
in order to keep the voltage on the - input of the opamp the same as
the voltage on the + input.


Looking at R5R6C1 as a lowpass filter with single pole, we have:


Vout
|
[R5]
|
+----Vfb
|
[R6]
|
[C1]
|
GND

And,suddenly, it's too close to dinner (Thai beef salad and a nice
white for me and Thai beef salad and a nice red for she) and too
complicated to get into right now.

Mañana, talvez...

 

[Roger Lascelles]

http://focus.ti.com/lit/an/sboa061/sboa061.pdf, ....

. A very high resistance feedback resistor is MUCH better
than a low resistance in a T network. See Figure 5. Although
transimpedance gain (eOUT/iSIGNAL) is equivalent, the T network
will sacrifice performance. The low feedback resistance
will generate higher current noise (iN) and the voltage
divider formed by R1/R2 multiply input offset voltage, drift,
and amplifier voltage noise by the ratio of 1+
R1/R2. In most electrometer amplifiers, these input specifications
are not very good to start with. Multiplying an already
high offset and drift (sometimes as high as 3mV and 50mV/
°C) by use of a T network becomes impractical. By using a
far better amplifier, such as the OPA128, moderate T network
ratios can be accommodated and the resulting multiplied
errors will be far smaller. Although a single very-high
resistance will give better performance, the T network can
overcome such problems as gain adjustment and difficulty in
finding a large value resistor.

Jim "The other one." Meyer

ASCII art; use Courier font.

_______R4_______
| - |
o----R1--+---|\ |
| \_____ out |
| / | |
o------------|/ R2 |
+ |----'
R3
|
o--------------------+------ Gnd

The input signal at the opamp minus input is reduced by a voltage divider
formed by R1 and R4 + (R2 || R3). The *least* reduction of the input signal
occurs when you use a straight feedback resistor, not the R2, R3, R4
combination.

Since the input signal is reduced but the overall circuit has the same gain,
the opamp must be making up the deficit from its reserve of loop gain. In
other words, more noise, offset, drift, less bandwidth.

People who want variable gain balanced amps use a pot or single gain setting
resistor. Not too bad for a small range of gain increase, but you often see
R2 < R1.

An example : overall gain of 1:

1. Omit R3, R2 = 0, R4 = R1 . Input signal is reduced to 0.5 . This is the
optimum case with singe feedback resistance.

2. R2 = R4 = 0.25 R1 and R3= 0.125 R1. Input signal is reduced to 0.25 .
Now you need 6db more gain from the opamp.

Roger Lascelles

 

[Winfield Hill]

The Phantom wrote...
Hmm, a bottle of red for you, a bottle of white for her, plus a
little Thai beef salad on the side...

 

[Pooh Bear]

From a Burr Brown ap note,
http://focus.ti.com/lit/an/sboa061/sboa061.pdf, ....

• A very high resistance feedback resistor is MUCH better
than a low resistance in a T network. See Figure 5. Although
transimpedance gain (eOUT/iSIGNAL) is equivalent, the T network
will sacrifice performance. The low feedback resistance
will generate higher current noise (iN) and the voltage
divider formed by R1/R2 multiply input offset voltage, drift,
and amplifier voltage noise by the ratio of 1+
R1/R2. In most electrometer amplifiers, these input specifications
are not very good to start with. Multiplying an already
high offset and drift (sometimes as high as 3mV and 50mV/
°C) by use of a T network becomes impractical. By using a
far better amplifier, such as the OPA128, moderate T network
ratios can be accommodated and the resulting multiplied
errors will be far smaller. Although a single very-high
resistance will give better performance, the T network can
overcome such problems as gain adjustment and difficulty in
finding a large value resistor.

Jim "The other one." Meyer

If the R to ground is simply ac coupled ther's no degradation of DC
performance though.

I used a similar trick once with a non-inverting configuration on the input
to produce a 'bootstrapped' Very high input Z.

Graham

 

[The Phantom]

The Phantom wrote...

Hmm, a bottle of red for you, a bottle of white for her, plus a
little Thai beef salad on the side...

Actually, it was John Fields who wrote this.

 

[Roger Lascelles]

ASCII art; use Courier font.

_______R4_______
| - |
o----R1--+---|\ |
| \_____ out |
| / | |
o------------|/ R2 |
+ |----'
R3
|
o--------------------+------ Gnd

This is the formula you are looking for:

If R4 = R2 and R3 = K R2

Av = 2 ( 1 + (1/(2K)) ) R2 / R1

derived from the balanced amp case given in Toby, Graeme, Huelsman,
"Operational Amplifiers", Burr-Brown.

I hope "wiley old engineer" mentioned the drawbacks - this ciruit will *not*
give you the max gain possible from an opamp, because it wastes some of the
available gain.

Looking at the circuit open loop, for a high gain amp, you want R4 >> R1, so
most of the input signal hits the opamp minus input, but you end up with the
opposite.

I think the amplifier can be made quite good by having non-equal values for
R2 and R4. Select R4 > 10 R1 to fix the gain wastage problem. Make R2 ||
R3 much less than R4. Select the ratio of R2 and R3 for the gain you want.
The gain is then approx :

Av = ( R4 / R1 ) ( R2 + R3 ) / ( R3 )


Roger Lascelles

 

[Pooh Bear]

This is the formula you are looking for:

If R4 = R2 and R3 = K R2

Av = 2 ( 1 + (1/(2K)) ) R2 / R1

derived from the balanced amp case given in Toby, Graeme, Huelsman,
"Operational Amplifiers", Burr-Brown.

I hope "wiley old engineer" mentioned the drawbacks - this ciruit will *not*
give you the max gain possible from an opamp, because it wastes some of the
available gain.

How can it 'waste it' ? Max gain = Avol. End of story.

Graham

 

[Fred Bartoli]

I read in sci.electronics.design that "RST Engineering (jw)"
about 'Op Amp Calculations', on Wed, 21 Sep 2005:

However, I can't convince myself that I can mathematically come up with
the resistor value. I have googled the problem and come up short.
Anybody got a pointer to a URL that goes through the math of how this
configuration works? And what I'm doing to my phase margin?

I believe Win has sort-of proprietary rights on this circuit. But it's
easy if you re-draw it. Why do these things come up in s.e.d. instead of
a.b.s.e?

ASCII art; use Courier font.

_______R4_______
| - |
o----R1--+---|\ |
| \_____ out |
| / | |
o------------|/ R2 |
+ |----'
R3
|
o--------------------+------ Gnd

---

I believe the circuit he described looks more like this:

Vcc------------+
|
[R1]
|
+----------|+\
| | >-----+-->Vout
Vin>------[R3]-------+----|-/ |
| | |
[R2] +-[R4]-+-[R5]-+
| |
GND>-----------+-[C1]--[R6]-+ A


Where, for DC,


R4 + R5 Vcc R2
Vout = -Vin --------- + ---------
R3 R1 + R2

For AC, however, (assuming Vin and Vout are perfect voltage sources)
we wind up with Vout trying to force the - input of the opamp to
stay at whatever voltage the + input is set to by R1 and R2.

For AC, assuming that C1 is very large so that it can be treated as an AC short, and
the op amp gain is infinite, the gain is:

-((r5*r6 + r4*(r5 + r6))/(r3*r6))

If A is the op amp gain and C1 is included in the calculation, the gain is :

-A*(r4 + r5) - A*c1*(r5*r6 + r4*(r5 + r6))*s
----------------------------------------------------------------------- --
r3 + A*r3 + r4 + r5 + c1*(r5*r6 + r4*(r5 + r6) + r3*(r5 + r6 + A*r6))*s

Well, you've forgotten the GBW in all that
If we set WT=2.pi.GBW then we have

-A - B p
----------------- with
1 + C p + D p^2

A0(R4+R5)
A = ----------------
R3(1+A0)+R4+R5

A0.C1(R4 R5 + (R4 + R5) R6)
B = ----------------------------
R3(1+A0)+R4+R5

(A0(R3 + R4 + R5 ) + C1.WT((R3 + R4)R5 + R6(R3 + A0.R3 + R4 + R5))
C = --------------------------------------------------------------------
(R3 (1 + A0) + R4 + R5 ) WT


A0.C1 ((R3 + R4) R5 + (R3 + R4 + R5) R6)
D = ------------------------------------------
(R3 + A0 R3 + R4 + R5) WT


Less than 2 min to work this out from scratch, incl. sign error correction.
Isn't that Mathematica lovely? ;-)

 

[Roger Lascelles]

available gain.

How can it 'waste it' ? Max gain = Avol. End of story.

Graham

Look at this circuit :

_______R4_______
| - gnd
--R1--+---|\
| \_____
| / out
|/
+

The amplifier is open loop, with no feedback voltage into R4. The inverting
input voltage is divided by the R4, R1 attenuator. The amplifier signal is
attenuated, and is now smaller in relation to the amplifier's own noise,
offset and drift.

Now let the opamp have an open loop gain of Avol - I think that is what you
mean by Avol. The maximum gain our overall amplifier can have is :

Atotal = Avol (R4 / (R1 + R2 ) )

which is less than Avol .

So the input circuit has "wasted" some gain. A high gain opamp makes up for
it - but not really, because we actually have "less feedback" in order to
get the required level of output. In effect we have an input attenuator
followed by a higher gain opamp amplifier.

This is why the divided feedback circuit is bad - it can heavily attenuate
the input signal.

Roger Lascelles













It is common to speak of a reserve of gain - the


If you open the loop and do a gain analsis, you see that the



An opamp is a high gain amplifier with feedback around it. The "extra gain"
or loop gain is the differe

If you look at the amplifier

 

[John Fields]

The Phantom wrote...

Hmm, a bottle of red for you, a bottle of white for her, plus a
little Thai beef salad on the side...

 

[John Woodgate]

I read in sci.electronics.design that John Fields

I believe the circuit he described looks more like this:

Vcc------------+
|
[R1]
|
+----------|+\
| | >-----+-->Vout
Vin>------[R3]-------+----|-/ |
| | |
[R2] +-[R4]-+-[R5]-+
| |
GND>-----------+-[C1]--[R6]-+

The bias on the + input is irrelevant in the context of the thread, and
C6 is normally so large that it has no effect within the operating
bandwidth.

 

[John Woodgate]

I read in sci.electronics.design that Roger Lascelles

Looking at the circuit open loop, for a high gain amp, you want R4 >>
R1, so most of the input signal hits the opamp minus input, but you end
up with the opposite.

The inverting op-amp circuit works by having NO signal on the - input.
That's what 'virtual earth' means.

 

[Roger Lascelles]

The inverting op-amp circuit works by having NO signal on the - input.
That's what 'virtual earth' means.

Hello John - also Graham (Pooh Bear). You people are making me defend my
not very academic grasp of opamps!

I can see people don't like me talking about the voltage divider at the
input, but it is there when you open the feedback loop and therefore affects
noise, offset and drift. I first saw it when calculating signal to noise
ratios for audio amplifiers using opamp stages.

For example :


_______R2______
| - |
o----R1--+---|\ |
| \_________|__ out
| /
Vn --|/
+

R1 = R2 for gain = 1. Vn is noise, offset & drift referred to input and has
gain of (1 + R2/R1) = 2. So signal to noise is half as good as the opamp can
do with the signal going straight in. Now, short the output to ground and
look - the signal IS halved by the R1 R2 divider in exactly that 2 to 1
ratio.

The ratio holds for whatever R1 and R2 you choose. R1 and R2 are dividing
down the input signal before it hits the opamp input terminal. I know this
continues to be correct when I close the loop, because I continue to see the
same signal to noise ratio at the amplifier output - although gain changes.

Makes sense really - simple feedback does not change the S/N performance the
network had before you added feedback.

Certainly you don't see much voltage at the virtual earth point when you
close the feedback loop, and I see that I needed to explain what I meant by
'input attenuation".

So when I see R2 < R1, I immediately think "input attenuation", poor noise,
offset and drift. Just like the T feedback circuit we were discussing. If
you can get R1 >> R2 you can really tap into available opamp performance.

------------------

The other issue is that the T feedback circuit runs with less loop gain for
the same overall gain than a simple two resistor amplifier. This means less
bandwidth and accuracy. If anyone is interested, I can post the maths, but
the 'input attenuation" line of thought gives it to you for free - the input
signal is attenuated, but you get the same overall gain, so you must have
sacrificed loop gain.

Roger Lascelles

 

[Jim Thompson]

On Fri, 23 Sep 2005 13:29:48 +1000, "Roger Lascelles"

[snip]

------------------

The other issue is that the T feedback circuit runs with less loop gain for
the same overall gain than a simple two resistor amplifier. This means less
bandwidth and accuracy. If anyone is interested, I can post the maths, but
the 'input attenuation" line of thought gives it to you for free - the input
signal is attenuated, but you get the same overall gain, so you must have
sacrificed loop gain.

Roger Lascelles

Roger,

I take issue with your comment on "T" feedback circuits. If you
choose the more common situation where you looking to reduce the
feedback impedance, say by a factor of 10X, the difference is barely
noticeable.

I simulated two configurations both with GBW=10MHz and GDC=100K. The
difference in high-end roll-off response is less than 1dB.

If you add a capacitor in series with the R to ground, in the "T"
configuration, you can knock down Vos sensitivity by nearly a factor
of 10X.

I will post the results on A.B.S.E shortly.

...Jim Thompson

 

[Jim Thompson]

On Fri, 23 Sep 2005 13:29:48 +1000, "Roger Lascelles"

[snip]

------------------

The other issue is that the T feedback circuit runs with less loop gain for
the same overall gain than a simple two resistor amplifier. This means less
bandwidth and accuracy. If anyone is interested, I can post the maths, but
the 'input attenuation" line of thought gives it to you for free - the input
signal is attenuated, but you get the same overall gain, so you must have
sacrificed loop gain.

Roger Lascelles

Roger,

I take issue with your comment on "T" feedback circuits. If you
choose the more common situation where you looking to reduce the
feedback impedance, say by a factor of 10X, the difference is barely
noticeable.

I simulated two configurations both with GBW=10MHz and GDC=100K. The
difference in high-end roll-off response is less than 1dB.

If you add a capacitor in series with the R to ground, in the "T"
configuration, you can knock down Vos sensitivity by nearly a factor
of 10X.

I will post the results on A.B.S.E shortly.

...Jim Thompson

See...

Newsgroups: alt.binaries.schematics.electronic
Subject: Op Amp Calculations (from S.E.D) - MultiPathFeedback.pdf
Message-ID: <[email protected]>

...Jim Thompson