a strange structure about operational amplifier

[mendenz]

this op is used as a error amplifier, how can i calculate the gain ?
and what are thr merits of this structure?
3ks

 

[hardings]

The picture is not clear.
Pls attach a better one.

 

[jordan76]

Since the picture is not very clear, my best guess: it is a folded cascode structure implemented in bipolar with a level-shift input stage and common-emitter output stage.

regards,
jordan76

 

[mendenz]

here is the clear version

 

[hardings]

My first guess
the gain before the CE stage is 4*Gm*(rds(x1)||rds(Q29))

Let's talk about this circuit, it is beautiful , right?

Added after 4 minutes:

It is a folded cascode, of course.
X2,X3,Q32,Q33 is for what ?
We can just add a fixed voltage at the base of X1,X4,Q31 and Q34

 

[aryajur]

What functions do transistors N27 and P24 do????

 

[hardings]

For better bias voltage

 

[aryajur]

I didn't understand, how?? How are they working? Their Source Drain Bulk nodes are shorted, what does this mean? Its just like a capacitance, thats all? Is it how they are used?

 

[hardings]

It is a capacitor.

mendenz,can you give more application information,
For example, gain, gm, offset, sr?

 

[sutapanaki]

Here is my interpretation of the circuit. Q32 Q33 X3 and X2 are mainly for biasing. Yes, you can put a fixed voltage at the bases of Q31, Q34,X4 and X1, but then you have to worry about producing this voltage, and here this is done automatically by the diode connected transistors. The gain can be found like this (if we disregard the input resistor divider):

The current delivered by the input followers into Q33-34 and Q31-32 is

i= (gm1gm2)/(gm1+2gm2)Vdi

gm1 - transconductance of X16-17
gm2 - transconductance of Q31-34
Vdi - input differential voltage

Only half of this current goes to the output of the first stage and into the load. The load is the parallel combination of the resistor loads and the equivelent resistance at the emitters of the PNP stage. The resistive load is RL=R39//2R43 and the equivelent PNP stage resistance is Req=1/gm3, where gm3 is the transconductance of X1-4. So, the output voltage at the output of the folded cascode stage is:

V1=[Ro(x1)//Ro(Q29)//Rpi(30)]*RL/(1/gm3+RL)*0.5(gm1gm2)/(gm1+2gm2)*Vdi

Finally, the output voltage of the amplifier is:

Vo=|gm4(Ro30//Ro22)|V1

where gm4 is the transconductance of Q30

 

[hardings]

Hi,sutapanaki

I have different view about the gain calculation. If somethint is wrong ,please point out.

1. Input follower
as your equation point out, Q32,Q33,X2,X3 is not only for bias , but also for load of input follower.So the Gm is
2*gm1*gm2/(gm2+2gm1), then the gain of the follower is
2*gm1*gm2/(gm2+2gm1)*(gm2+2gm1))

2. Sencond Stage
Gain is gm2*(0.5R43||0.5(1/gm3))
I think the load of this stage is 0.5*R43 parallel with 1/gm3, can you explain your RL more clearly?

3. Third Stage
gm3*(Ro(x1)//Ro(Q29))

Please try to get more investment.

 

[sutapanaki]

OK, investment it is. Although I'd like to see some return on investment Just kidding.

1. Input follower
as your equation point out, Q32,Q33,X2,X3 is not only for bias , but also for load of input follower.

Yes, I agree with you that those transistors influence the loading of the followers. But if we take for example P22, it also influences the loading of Q30 and yet I hope you'll agree that it is primarily for biasing and not in the signal path. In the same way Q32-33 are used to bias Q31 Q34 and also they allow for the voltage at the emitters of X16-17 to be exactly where it should be with respect to the input common-mode voltage. Transistors Q32-33 also waste half of the signal current of the input followers, exactly because they influence the loading.

So the Gm is
2*gm1*gm2/(gm2+2gm1), then the gain of the follower is
2*gm1*gm2/(gm2+2gm1)*(gm2+2gm1))

Here, I haven't tried to verify if you are right or not simply because in my derivation I didn't care about the voltage transfer from the differential input to the emitters of the followers. I was interested in the signal current out of these followers. If we assume that the circuit is symetric, it is easy to see that the differential impedance the two input followers see at their emitters is 1/gm2. Now, if you load the input differential emitter follower structure with this differential inpedance (which means it is connected between the emitteres) and disregard 1/Rπ of the follower transistors with respect to gm1 and gm2 you'll see that the current going into the differential load of 1/gm2 is:

i=Vdi* (gm1gm2)/(gm1+2gm2)

If we assume that betta is very big and disregard the base current, half of the current above is going to the collectors of Q31 Q34 and into the load there.

2. Sencond Stage
Gain is gm2*(0.5R43||0.5(1/gm3))
I think the load of this stage is 0.5*R43 parallel with 1/gm3, can you explain your RL more clearly?

That collector current sees the load of the resistors R39, R42 and R43 and the load of the PNP transistors - note I'm talking about differential load here. For RL see the attached picture


In this case RL=(2*R43)//R39. In parallel to RL you have the load presented by the emitteres of the PNP transistors, which is 1/gm3. We are interested only in the current that goes in this 1/gm3 which is RL/(1/gm3+RL)*i. Half of this current is again wasted in X2-3, but then we add the collector currents of X1 and Q29 at the output, so for the output current we don't have to devide by 2 as before.
The output current now goes into a load which is the parallel combination of
Ro(x1), Ro(Q29) and the input impedance of Q30 (we disregard the compensation network, because we talk DC here only). The output current into this load will give you the input voltage of the output stage and the output voltage of the amplifier now is easy to find - just
gm(Q30)*(Ro(Q30)//Ro(P22)).
That's all for the investment.

 

[rajath]

I'm a bit confused here:
The input stage(X16/17) is an emitter follower - which implies that it will just replicate the input voltage at it emitter terminal. So why does the gain calculation of the first stage include the gm of the input transistor? As far as I can see, the input signal passes through this emitter follower, and is then inserted at the second stage by the common base stage (Q34/Q32). Can someone clarify?

 

[sutapanaki]

Ideally, emitter followers do repeat the input signal at the emitter terminal - especially when they are loaded with ideal current sourse in the emitters. However if the emitter is loaded with some impedance Re, then the transfer from base to emitter to a first order (disregarding any impedance in the base) is Re/(Re+1/gm).
The smaller the Re, the more 1/gm enters into the picture. In the circuit under discussion the load to the input followers is in the order of Re=1/gm2, so relatively small.

 

[hardings]

Hi,sutapanaki
I have checked the two method.Your equation is right and more intuitive,so your investment is valuable, hehe.
But we should get same result with the two method. I found if I fixed some error of my equation, the only difference between my equation and your equation is the output resistance of the 1st stage, your equation imply that it is 1/(2*gm2),my equation imply that it is 1(gm1+2gm2). I am not very sure which one is right. How do you think of that?

 

[sutapanaki]

I think that this is not really a difference. You see, if you ar interested in the voltage gain from input to the emitters of the input followers, then you need to take the full load at the emitters - and this is composed of the loading presented by the Q31-34 structure, which is 1/gm2 in parallel with the output impedance of the followers 2/gm1 (looking differentially). Because I work with currents and don't calculate the voltage transfer, I'm interested in the current that goes into the load of 1/gm2 and even half of this current. But irrespective of this, the two approaches should finally result in the same thing. I just think that when analyzing folded cascode structures it is easier to manage currents than voltages.