Engineering Differential amplifier confusion (BJTs + Operational Amp)

[cnh1995]

Homework Statement

Find the Op-amp output voltage Vo in the following circuit.

Relevant Equations

Vout= Vin(1+Rf/R1) for non-inverting input.

Here is the circuit diagram provided in the book.
In the solution, the book has used the following approach (red markings in the image):
Input to the left transistor is 2V. Considering base-emitter junction drop to be 0.7V, the emitters are at 1.3V (left red arrow). Now, using the "virtual short" concept, the base of the right transistor is also pulled to 2V.
As the non-inverting terminal of op-amp is at 2V, using Kirchhoff's rules and the feedback equation, Vout=4V.

But how is the base of the right transistor pulled to 2V? Is the concept of "virtual short" really applicable here? As per my understanding, the right transistor should be cut off.

What is wrong here?

 

[Baluncore]

Here is the circuit diagram provided in the book.

I would like a reference to the circuit, or to see the original page from the book.

As per my understanding, the right transistor should be cut off.

That is correct for the circuit as shown.

What is wrong here?

Everything. It seems like a useless circuit because there is no feedback from the op-amp to bias the transistor base on the RHS.

 

[cnh1995]

I would like a reference to the circuit, or to see the original page from the book.

Will try to get the image of the problem. It is not a standard textbook, but a collection of problems in analog electronics (by a local publishing house). So the author/publisher might have carelessly printed this answer. I was helping a friend with his assignment and this problem is copied from his notebook.

Thanks for the response. My analog electronics is a bit rusty, so I posted here for confirmation.

 

[alan123hk]

If the two transistors are matched, the DC bias voltage, and the magnitude and direction of the current source are set correctly, then we should get the desired effect, which is the same base voltage for both transistors.

 

[Baluncore]

If the two transistors are matched, the DC bias voltage, and the magnitude and direction of the current source are set correctly, then we should get the desired effect, which is the same base voltage for both transistors.

Where does the base current for Q2 come from?

 

[LvW]

To cnh1995:
"Now, using the "virtual short" concept, the base of the right transistor is also pulled to 2V."

Why do you expect that the base of T2 will be "pulled" to 2V ?
There is absolutely no reason for such an effect. The "virtual short" effect applies to active devices with very large gain and negative DC feedback only.

 

[alan123hk]

Where does the base current for Q2 come from?

Yes, you are right, I missed that, the base current of Q2 can be provided by the ground resistor, but the input voltage must be kept below -0.7V.
Since this circuit requires an input voltage of 2V, a voltage source greater than 2V is required to supply current to the base of Q2.

 

[Baluncore]

I believe there must be three or more errors in the schematic.
The circuit would be more useful, easier to model and understand if;
The voltage source for the base of Q1 was negative two volts.
Or the transistors were N-channel FETs.
And the polarity of the conventional current source was reversed to become a sink.

 

[cnh1995]

Why do you expect that the base of T2 will be "pulled" to 2V ?

I don't.

But how is the base of the right transistor pulled to 2V? Is the concept of "virtual short" really applicable here? As per my understanding, the right transistor should be cut off.

 

[LvW]

In your circuit, for a suitable bias point, the DC current through each transistor should be app. 5mA.
That means: A current source of 10mA in the common emitter lags.
In this case, the unsymmetrical voltage gain (collector T2) would be A=-0.01*10³=-10.
Hence, the input signal at T1 must not be larger than app. 100mV.

Question: What is the purpose of the opamp? To provide additional gain or something else? Perhaps it should be connected to the collector of T2?

 

[cnh1995]

Question: What is the purpose of the opamp? To provide additional gain or something else? Perhaps it should be connected to the collector of T2?

The problem is not from any standard textbook. It is from a collection (booklet) of analog electronics problems for certain national/state level entrance tests in my country. So the purpose of the op-amp could just be to increase the difficulty level of the problem.

I ran a simutaion of this circuit on my phone and it shows that Q2 (right transistor) is indeed cut off.

 

[LvW]

So the purpose of the op-amp could just be to increase the difficulty level of the problem.

Please, could you explain again? What is the purpose of the opamp?
Are you sure that it is not connected to the collector of T2?

 

[cnh1995]

Are you sure that it is not connected to the collector of T2?

Yes, it is connected to the base and not the collector of T2.
The purpose of the op-amp is not mentioned in the problem. Seems the book's solution is incorrect.

 

[Baluncore]

What is the purpose of the opamp?

The only purpose of the op-amp is to be part of a problem. That problem appears to have been set by someone who did not understand electronics, or maybe a student copied the schematic wrongly.

There are too many possible reconstructions in the search space.
We need to see the original.

 

[LvW]

When the shown circuit is correctly redrawn the only purpose of the opamp could be to measure/verify the voltage at the base of T2 (it acts only as a buffer with a gain of two )

 

[alan123hk]

The first-stage two-transistor differential amplifier can be used as a wide-band high-impedance input follower, followed by a gain-of-two op amp in the second stage.
In any case, the original designer of the circuit may need to be consulted to understand its true function and purpose.

 

[LvW]

..., followed by a gain-of-two op amp in the second stage.

To me, "to follow" a stage means "to be connected to the output of that stage."

 

[alan123hk]

To me, "to follow" a stage means "to be connected to the output of that stage."

If the transistor difference amplifier of the front stage is only used as a wide-band high impedance input follower , it does not need any gain, so it does not need to be connected to the collector, but there is still a problem with my statement, if the output is only connected to the base, as shown in the circuit diagram , then its output impedance must be very high, and there is no driving force at all.

 

[LvW]

If the transistor difference amplifier of the front stage is only used as a wide-band high impedance input follower , it does not need any gain, so it does not need to be connected to the collector....

So what is the output node of the follower?
I must admit that I never have seen any BJT-based circuit where the base is used as an output.

 

[alan123hk]

So what is the output node of the follower?
I must admit that I never have seen any BJT-based circuit where the base is used as an output.

I've also never seen a follower that uses the base as the output, and to be honest, the so-called follower is just my own assumptions or imagination, I may be wrong.
So I think if we need to find out the truth, maybe we need to find the original author.

 

[alan123hk]

I think what I actually meant was that the original author probably wanted to design a simple and accurate follower, but didn't think the transistor base could not be used as an output because it didn't have any driving force.

 

[Baluncore]

I try to avoid the abyss of silly circuits that were clearly designed by monsters.
"He who fights with monsters should look to it that he himself does not become a monster.
And if you gaze long into an abyss, the abyss also gazes into you."
Friedrich Nietzsche - Beyond Good and Evil - Aphorism 146.

 

[alan123hk]

I have to agree with you. The more I looked at this circuit diagram, the more it felt incomplete, incomprehensible, and pointless.

 

[DaveE]

Where does the base current for Q2 come from?

This! It's actually a very simple circuit for inputs above 0V. It does nothing.

 

[LvW]

I think it could be interesting to discuss the properties of the shown circuit - in particular when it is compared with the classical four-resistor opamp based differential amplifier (advantages, disadvantages).

 

[Baluncore]

To close the loop, in the rational analysis of an irrational circuit, required that I also become crazy. To avoid furthering that process, I looked at the characteristic patterns in the OP circuit, while ignoring the detail and connectivity. Then I applied an "inverse acid-trip transform" to see what clean circuit concepts might have been present in the ancestral circuit, and how they might interrelate. The result was a neat and functional circuit, that features several different teaching concepts.

I think it could be interesting to discuss the properties of the shown circuit - in particular when it is compared with the classical four-resistor opamp based differential amplifier (advantages, disadvantages).

Go ahead, I hope to learn something.

 

[LvW]

Go ahead, I hope to learn something.

Well - it was YOU who has proposed this circuit. Therefore, I think that - at first - you could say something about the usability of the circuit (a gain-of-two differential gain stage).
(I think, the circuit will only be stable for a relatively small opamp feedback factor (k<0.1) - depending, of course, on the opamp used.)

 

[DaveE]

I think it could be interesting to discuss the properties of the shown circuit - in particular when it is compared with the classical four-resistor opamp based differential amplifier (advantages, disadvantages).

First, I assume that you mean the diff amp circuit that should have been shown, one that actually works. Look at @Baluncore's schematics for this.

Second, That diff amp stage IS Link Removed, in various incarnations. So I would reinterpret your question as what are the pros and cons of having a separate discrete input stage vs. integrating it with additional amplifier bits.

BTW, it sounds obvious that you would put it on the IC silicon, but there are times, like low noise circuits, when you mighty actually do what @Baluncore showed.

 

[LvW]

First, I assume that you mean the diff amp circuit that should have been shown, one that actually works. Look at @Baluncore's schematics for this.

Second, That diff amp stage IS Link Removed, in various incarnations. So I would reinterpret your question as what are the pros and cons of having a separate discrete input stage vs. integrating it with additional amplifier bits.

BTW, it sounds obvious that you would put it on the IC silicon, but there are times, like low noise circuits, when you mighty actually do what @Baluncore showed.

1.) First: Yes, of course. I don`t like to discuss circuits that do not work.
2.) Second: What I see in Baluncore`s contribution, is a differential amplifier with a single low-resistive opamp output. Such a differential amplifier can also be realized using an opamp and 4 resistors. That means - I like to compare two stand-alone units with each other.
In this context, I do not have any IC integration in mind.

 

[DaveE]

a single low-resistive opamp output

Not sure what those words mean. But it doesn't matter since we both have the schematic to look at.

Such a differential amplifier can also be realized using an opamp and 4 resistors. That means - I like to compare two stand-alone units with each other.
In this context, I do not have any IC integration in mind.

But, in the last 50 years, op-amps are ICs.

Anyway, what features do you want to compare? Cost, gain-bandwidth, input impedance, quiescent current, noise figure, reliability, stability, ... There are lots of specs for amplifiers. My guess is that IC amplifiers win in essentially all categories, which is why that's what everyone specifies. However, there are cases where an external discrete input stage has benefits, like low noise or high voltages. There are also many cases where a larger discrete output stage is needed.

Look at Link Removed. Everyone of those transistors has a purpose. They are put there to make it work better. So maybe your question is, regardless of implementation, why does one circuit have two transistors and the other has 20? My short answer reflects the ideal op-amp assumptions we all use: increased gain, increased input impedance, decreased output impedance. Also greater bandwidth and lower quiescent current. When you have multiple transistors to get the job done you can better optimise each for its job. The output stage is designed differently than the input stage in an IC, but in a two transistor diff amp, one transistor has to be good at both jobs.

 

[Baluncore]

Well - it was YOU who has proposed this circuit. Therefore, I think that - at first - you could say something about the usability of the circuit (a gain-of-two differential gain stage).

My circuit would not be used in reality, but the separate concepts grouped together there are individually rational and educational. It is simply a functional example, of what could be built from the disjointed collection of components available in the OP schematic.

Further extrapolation and analysis has little foundation and would be off-topic, unless the ancestral circuit was first presented and shown to be very similar.

 

[alan123hk]

I'm interested in revisiting the incomprehensible circuit mentioned at the beginning of this thread. As I mentioned before, if you apply a proper bias voltage, you can make the base voltages of the two transistors very close, as shown below.

This differential amplifier has high input impedance over a wide frequency range because the impedance of the current source is infinite.

On the other hand, it can be seen that its output (Q2 Base) remains very close to the input over a wide voltage range up to 100 MHz, although the DC level is slightly shifted and the amplitude is reduced. But unfortunately in the end this electronic circuit is still meaningless and cannot be called a follower because its transistor base output impedance is surprisingly high.

 

[Baluncore]

@alan123hk
What is the purpose of R1 and R2 ?
The circuit would respond even faster if you removed them with their miller capacitance.

 

[alan123hk]

What is the purpose of R1 and R2 ?
The circuit would respond even faster if you removed them with their miller capacitance.

Sorry I didn't think about this, the original circuit had these two resistors, so I didn't specifically think about what they were doing in my new circuit, so they stayed.

My main point is that if you want the base voltage of both transistors to be the same, it's not too hard.
This is one of the issues mentioned in the first post.

 

[Baluncore]

When I remove two resistors, and reduce the bias current from 60 mA to 5 mA, I get;

 

[LvW]

Not sure what those words mean. But it doesn't matter since we both have the schematic to look at.
.....
Anyway, what features do you want to compare? Cost, gain-bandwidth, input impedance, quiescent current, noise figure, reliability, stability, ... There are lots of specs for amplifiers.

Look at Link Removed. Everyone of those transistors has a purpose.

Sorry - it seems That I have not expressed myself in a clear manner.

The situation is as follows:
Wenn I want to realize a differential amplifier there are, of course, several options. Such a situation is quite normal in analog electronics (e.g. active filters, oscillators,...).
* In the case under discussion we have a novel circuit (Baluncore`s circuit, post#22) which provides a single-ended output with a very low output resistance (opamp output).
This one of the advantages if compared with the well-known two-transistor circuit without this opamp (long-tailed pair, first stage of Baluncore`s circuit).
* However, there is a well-known opamp circuit (with 4 external resistors) that also can provide such a low-resistive single-ended output.
* And my question was if it makes sense to compare these two circuits which each other, as far as some important parameters are concerned - for example: Input resistances, linearity, dynamic stability, gain restrictions, CMRR, accuracy requirements (symmetry), ...(just to collect some thoughts of forum members who are interested).
* I have mentioned already one restriction: Due to active feedback (feedback factor can be above unity) the differential gain of Baluncore`s circuit (post#22) must not be too low (a gain of two leads to instability).

 

[Mark Harder]

Dammit, I posted a long answer, but PF seems to have deleted it. The short answer is that all of Is flows through the left transistor because you are trying to keep its base at 2V when it tries to stay at 0.7 V above common (what you mean by ground in your circuit). A transistor is a current-in, current-out device, so the high input current determined by the 2V source will (I guess. It depends on a lot of details.) take up all the current allowed by the Is constant current drain. The rt. transistor will be completely off. Since these are (I assume) ideal NPN devices, there will be no current in or out of the rt. transistor's base and therefore no voltage across its base resistor. Vo is zero volts.

 

[Baluncore]

The short answer is that all of Is flows through the left transistor because you are trying to keep its base at 2V when it tries to stay at 0.7 V above common (what you mean by ground in your circuit).

In the OP circuit, the base of Q1 is held at +2.0 volt. The emitter voltage will be about 0.7 volt below that, which is 1.3 volts.

But the OP circuit shows Is is a conventional current source, not a sink. If we invert that source to make a sink, it will set the bias current through the emitters to Is.
How much of Is comes from the +2 volt source will depend on the possible saturation of Q1. That will be determined by the quite unnecessary 1k collector resistor, the supply voltage and Is.

You are correct that the LHS transistor Q1 will be on. The base emitter junction of Q2 will be reverse biased, so it will be off. The voltage on Q2 base will be zero, only because of resistor R. You are correct that the op-amp output voltage will therefore be twice zero.

 

[alan123hk]

I have mentioned already one restriction: Due to active feedback (feedback factor can be above unity) the differential gain of Baluncore`s circuit (post#22) must not be too low (a gain of two leads to instability).

This circuit is basically a very good design. But just like a general negative feedback circuit, it does enter an unstable state in some special cases. But this does not necessarily mean that there is a problem with the basic structure of the circuit itself, usually the problem can be solved by simply modifying the parameters

For example, as shown in the figure below
Just change the GBW of the op amp from 10Meg to 1Meg or reduce the negative feedback factor (R5 from 1K to 300) to eliminate self-oscillation and resume normal operation without reducing the voltage gain.

 

[alan123hk]

With a slight modification again, we can make this circuit have the following nice properties (At least it looks like this in the simulation).

The input impedance is close to infinity, and the output impedance of the op amp is very low
Operating frequency from DC to 10MHz. In this frequency range, the input and output are almost completely in phase. Absolutely stable, no self-oscillation occurs.
The voltage gain is set to 3db, which can be lower than 2. In fact, the voltage gain can be set arbitrarily, etc.

 

[Baluncore]

What is U1 op-amp, generic ?

Beware SPICE AC analysis. Instability is always possible, as AC gain plot does not show phase to infinity, you must check the time domain.

Why do you not show to beyond 1 GHz? You have a high pass filter in the loop. The rising graph looks to me like it might oscillate in the UHF band, before the gain falls off.

 

[alan123hk]

What is U1 op-amp, generic ?

I just took this op amp from the parts library inside the LTSpice.
Name: UniversalOpamp
Description : A linear, single pole op amp with no internal nodes or input voltage range limits.

Beware SPICE AC analysis. Instability is always possible, as AC gain plot does not show phase to infinity, you must check the time domain.

I checked the time domain and found no parasitic oscillations. But I'll still try to dig deeper into this.

Why do you not show to beyond 1 GHz? You have a high pass filter in the loop. The rising graph looks to me like it might oscillate in the UHF band, before the gain falls off.

Agree, this is a matter of concern.
The following is the simulation results from 1Hz to 1GHz.

 

[Baluncore]

I get a similar result.

 

[Baluncore]

@alan123hk
It seems that the 39 dB gain at 50 MHz is an artefact of the universal op-amp model. The circuit will be better behaved if you substitute something real, like an OP27 or OP37. The OP07 is too slow and accurate.

 

[alan123hk]

It seems that the 39 dB gain at 50 MHz is an artefact of the universal op-amp model. The circuit will be better behaved if you substitute something real, like an OP27 or OP37. The OP07 is too slow and accurate.

I found that while using the OP07 or OP27 reduces the overshoot of the frequency response curve, there is some parasitic oscillations in the output.
I believe these unwanted oscillations can also be removed with some modifications, but it seems the problem is more difficult than last time and may take longer to find a solution.

 

[LvW]

I found that while using the OP07 or OP27 reduces the overshoot of the frequency response curve, there is some parasitic oscillations in the output.
I believe these unwanted oscillations can also be removed with some modifications, but it seems the problem is more difficult than last time and may take longer to find a solution.

Alan123hk, I agree to all of your conclusions (post#39, 40 42, and 45). The circuit works when we carefully take the stabiliity criterion into consideration. There is no fundamental problem.
However, we must remember what happens when we have ACTIVE feedback (feedback factor>1).
In this case, the "critical point" with unity loop gain is located at a frequency which is LARGER than the opamps transit frequency.
Therefore, it is of great importance how the opamps open-loop gain response looks like for such high frequencies (above the transit frequency). That is the reason the various opamp models behave so differently. This applies in particular for simple opamp models with only one single pole - these models cannot reveal the real behaviour of Baluncore`s circut.
There are two basic methods to overcome the mentioned stability problems:
(1) Application of one of the known methods for external frequency compensation
(2) Restriction to larger closed-loop gains (certainly larger than "2") which means: (a) Reduction of the opamps resistive feedback factor or (b) gain reduction of the transistor amplifier.

The degree of the necessary actions (case a or b) depends on the opamp used (and, in particular, on its phase shift function for frequencies above the transit frequency).

 

[LvW]

Beware SPICE AC analysis. Instability is always possible, as AC gain plot does not show phase to infinity, you must check the time domain.

Partly true - I think.
(1) An AC analysis can reveal instability problems when we do a LOOP Gain analysis (instead of a closed-loop gain analysis). In this case, we can apply the well-known stability criteria (Bode, Nyquist).
(2) Even a time-domain analysis cannot always reveal instability problems "per se".
To see if the system ist stable or not it is best to switch-on (at least one of) the power supplies at t=0.
Otherwise, the program can eventually find a fixed and stable bias point and "sees" no instability - this is, in particular, true if we use a simplified opamp model. However, a "real" model with internal delay-elements (time constants) will show (in most cases) such instability properties.

 

[DaveE]

Therefore, it is of great importance how the opamps open-loop gain response looks like for such high frequencies (above the transit frequency). That is the reason the various opamp models behave so differently. This applies in particular for simple opamp models with only one single pole - these models cannot reveal the real behaviour of Baluncore`s circut.

Yes, this! If you are working with high frequencies (relative to your amp, of course) you need to know about that 2nd pole. It's virtually always around where the dominant pole compensation crosses 0dB. Here's one of the most important parts of the datasheets:

 

[alan123hk]

The circuit works when we carefully take the stabiliity criterion into consideration. There is no fundamental problem.
However, we must remember what happens when we have ACTIVE feedback (feedback factor>1).
In this case, the "critical point" with unity loop gain is located at a frequency which is LARGER than the opamps transit frequency.
Therefore, it is of great importance how the opamps open-loop gain response looks like for such high frequencies (above the transit frequency). That is the reason the various opamp models behave so differently. This applies in particular for simple opamp models with only one single pole - these models cannot reveal the real behaviour of Baluncore`s circut.
There are two basic methods to overcome the mentioned stability problems:
(1) Application of one of the known methods for external frequency compensation
(2) Restriction to larger closed-loop gains (certainly larger than "2") which means: (a) Reduction of the opamps resistive feedback factor or (b) gain reduction of the transistor amplifier.
The degree of the necessary actions (case a or b) depends on the opamp used (and, in particular, on its phase shift function for frequencies above the transit frequency).

I have a lot of respect for your comments on the circuit and totally agree with you. Your description of the operation of the circuit and what analysis methods the circuit should use is simple and clear, nothing deliberately hidden. This method is especially useful for beginners to learn.

 

[alan123hk]

I've tried making an op amp with a few resistors, diodes and transistors, of course such a simple op amp would have poor specs, but would be fun for hobbyists.

Preliminary Specifications :
Frequency Range DC - 1MHz
Maintain an input impedance of not less than 20K ohms in the frequency range of DC-1MHz
Output Voltage Swing at least 2V
Other specifications are still unknown...