Derive expressions for the voltage gain of this opamp circuit

[Boltzman Oscillation]

Homework Statement

Derive the expressions for the voltage gain (Gv) of the following op amp:

Homework Equations

In = Ip = 0
Vp =Vn

The Attempt at a Solution

I can use KCL, and the fact that In and Ip are both 0, to derive the two equations, one from the top node and the other from the bottom node respectively:

Iri + Ir2 = 0 [1]
Ir3 + Ir4 = 0 [2]

now I can use the following equations:

Vr3 = V(bottom node) - V2

so here I must ask, what is the relationship between the bottom node's voltage and Vp? Usually there is not source between the node and the positive input of the op amp but in this problem there is a voltage source V1. Is the Vp equal to the voltage source and thus also equal to the bottom node?

 

[LvW]

Under the assumption that the opamp works linear (I cannot know because of missing supply voltages; single supply?) you can simply apply the classical method for analyzing feedback amplifiers - based on the feedback factor. The only point to be considered is that the feedback path consists of two parts: Negative and pos. feedback. The circuit is dynamically stable because the negative feedback is the dominating one. For calculation purposes you simly can superimpose/add (with correct sign) both feedback factors.

 

[gneill]

Note that the input is an AC source. The 12 V source is DC so is going to add a DC bias to the AC input. No problem, just carry the input (120 Vrms AC) symbolically through your calculations. Your diagram labels it as V2, so leave it as "V2" in your algebra.

If you're only interested in the AC gain of the circuit you can suppress the 12 V source. Remember superposition.

 

[Boltzman Oscillation]

Note that the input is an AC source. The 12 V source is DC so is going to add a DC bias to the AC input. No problem, just carry the input (120 Vrms AC) symbolically through your calculations. Your diagram labels it as V2, so leave it as "V2" in your algebra.

If you're only interested in the AC gain of the circuit you can suppress the 12 V source. Remember superposition.

So with superposition would I short both the DC and AC source or just the DC source?

 

[gneill]

So with superposition would I short both the DC and AC source or just the DC source?

If you shorted both, what would be left to be amplified?

Suppress the DC source and concentrate on the signal.

 

[Boltzman Oscillation]

If you shorted both, what would be left to be amplified?

Suppress the DC source and concentrate on the signal.

Sorry I am new to this material. What do you mean by suppressing the DC source and concentrating on the signal?

 

[gneill]

Sorry I am new to this material. What do you mean by suppressing the DC source and concentrating on the signal?

Do as you would when applying superposition to analyze a circuit. You can evaluate the effect of each independent source individually by suppressing all others and leaving one active at a time.

If you're studying op-amps you should have already covered basic circuit analysis techniques including the superposition theorem. If somehow you never covered that (dunno how) then google is your friend.

 

[Boltzman Oscillation]

Do as you would when applying superposition to analyze a circuit. You can evaluate the effect of each independent source individually by suppressing all others and leaving one active at a time.

If you're studying op-amps you should have already covered basic circuit analysis techniques including the superposition theorem. If somehow you never covered that (dunno how) then google is your friend.

Oh i thought that when using superposition you turn off all current sources or all voltage sources. So i can turn off one voltage source and leave the other one on?

 

[gneill]

Oh i thought that when using superposition you turn off all current sources or all voltage sources. So i can turn off one voltage source and leave the other one on?

You suppress all but one and analyze the circuit. Then you suppress that source and allow another one to operate and analyze again, and so forth. Go through all the sources one at a time. Then sum the results.

 

[LvW]

Just to avoid confusion and misunderstandings:
"Sum the results" - only as far as the output voltage is concerned (combination of DC and AC)!
However, in the original question you are asking for the voltage gain!
Please note that the gain is always the output-to-input ratio for one single input only!
In the case under discussion you have two voltage sources: A DC source, which serves only one purpose - proper DC operating point.
And - as a second one - a signal voltage V2.
The corresponding DC output voltage (determined by V1) must allow amplification of the signal voltage - but has NOTHING to do with signal gain value!
The signal gain is simply Vout/V2 (and is to be calculated without any superposition procedure).

 

[jim hardy]

I'd say go back to your very basics.
You'll figure out the shortcuts by understanding the basics better.
Once you've figured them out they become intuitive.
After reaching that point most of us forget what it was like to struggle in the beginning.

That's why i like @LvW 's statement

The signal gain is simply Vout/V2 (and is to be calculated without any superposition procedure).

An operational amplifier must drive its inputs equal else it's not succeeding at performing its "operation"
so,,

V- is easy , by voltage divider action it's = 2/3 Vout

to get V+ requires we find volts at junction R3-R4 .
To do that you can write KVL around loop V2 - R3 - R4 - Vout
and then V+ will be, again by by KVL, = (volts at junction R3-R4) - 12

i'll bet you can solve this one with just KVL and a little algebra.

Spoiler: Basic KVL applied to op amp circuit

Difference (Vout - V2) divides between R3 and R4, $\frac {3} {18}ths$ of it across R3 and $\frac {15} {18} ths$ across R4
(Volts at junction R3 - R4) - V2 - $\frac {3} {18}$(Vout - V2) = 0
(Volts at junction R3 - R4) = V2 + (Vout - V2) / 6

V+ then = V2 + Vout/6 - V2/6 -12

if V+ - V-
then
V2 + Vout/6 - V2/6 -12 = 2/3 Vout
multiply both sides by 6 to remove denominators
6V2 +Vout - V2 - 72 = 4Vout
collect terms and rearrange
5V2 - 72 = 3Vout

Vout = $\frac {5}{3}V2 -24$
and that's the sort of y = mx + b linear transfer function you'd expect from a simple opamp circuit like this.

I always have to check my arithmetic, so
.........
Sanity check at V2 = 0:::
Does V+ = V- ?
Vout/6 -12 = 2/3 Vout
Vout - 72 = 4 Vout
3Vout = - 72
Vout = -24
V- = 2/3 Vout = -16
and
V+ = Vout/6 -12 = -4 -12 = -16
yep, they're equal.
.........

second sanity check at Vout = 0
0 = V2 X 5/3 -24
V2 = 24 X3/5 = 72/5
Does V+ = V- when V2 = 72/5 ?
going back to V+ = V2 + Vout/6 - V2/6 -12
V+ = 72/5 + 0/6 - 72/(5X6) -12 = 14.4 + 0 - 2.4 -12 = 0
and V- = 2/3 of Vout = 0
Yep, they're equal.
.........

whew i think it worked out. Gain = 5/3 .
that's the hard way
but it's reliable
and you don't have to remember much.
but you have to practice practice practice !

You'd better check this old guy's arithmetic, too.

Basics Rock ! Trust them.

i hope this helps you gain confidence
it probably crosses the line of "too much help"
but i felt you needed to be pointed back toward the basics.

old jim

ps i hope this wasn't a trick question.
Those voltages would wreck most real opamps.
Typically they can't handle more than 15 volts though i have used some 40 volt ones.

 

[bhobba]

Jim is correct. The gains of op-amps are usually quite high which means when stable the difference between positive and negative is for all practical purposes zero. This should allow you to write simplified equations for the output. If you want a bit of a challenge using that as a start see if you can get a a better one depending on the gain of the amp G. As G → ∞ it should become the equation you got before.

Added Later - just used the approximation method and you end up with two equations in Vout for the voltage at + and - of the amp respectfully. Set them as equal and you can calculate Vout. It's slightly tricky writing the equations, but if you do it step by step there is no problems.

Thanks
Bill