At first glance, this looks to me like the opposite of a Widlar mirror. If I understand correctly, the basic idea behind the Widlar circuit is to have a low output current by using exclusively low resistances. The resistor at the emitter of the output transistor causes the output current to be lower than the collector current in the input transistor.

The circuit considered here seems to be intended as a "concave mirror" of current. :-) The output current is about the tenfold of the input current. One advantage is certainly saving of energy, especially if the output current is relatively high.

As in other circuits as well, emitter resistors provide a negative feedback. While current mirrors without emitter resistors work well in integrated circuits because the parameter differences of the neighboring transistors are insignificant, and so is the difference in temperature, their performance is not satisfactory when built from discrete elements. I guess that the performance of mirrors, built from discrete elements, is far better with two emitter resistors than without, provided that the dependence of the resistances on temperature is low.

So, my opinion about the four configurations is:

1. Least complex, well suited for ICs.

2. Well suited for low output current in ICs.

3. I think it doesn't make sense ... but I might change my mind, depending on other contributions. It would yield larger output current than input current but I feel that it would be rather a temperature sensor than a current mirror.

4. Well suited for high output current and/or mirrors built from discrete elements.

 Dear Cyril,

Welcome!

Joerg gave an answer which covered some reason to add an emitter resistance in the input branch of the Widlar current source. I will try not to repeat a part of his answer in spite of the difficulty which i can face.

The emitter resistance in the output branch of the current mirror has two functions:

It reduces the output current of the current mirror since:

* IE2= (VB- VBE2)/ RE2 where, VB is the common base voltage, VBE2 is the base to emitter voltage of the output transistor and RE2 is its emitter resistance. 

* It increases the output resistance of the output current source which enhances the performance of the current source

The emitter resistance of the input transistor may have  input current thermal stabilization function in addition to other functions which can be explored. In this comment i will confine my self on the thermal stabilization function. The input circuit to the current mirror is to set the input current. That is the power supply voltage VCC, RB, the bias resistance, the transistor acting as a diode and the RE1 set a value for the bias current and consequently a value for VB. If the current of the transistor increases because of an increase in temperature The voltage drop on RE1 increases causing a decrease in VB1 since one can assume that VB is more or less constant so long VCC constant. This reduction in VBE1 will force IC to return back near to its initial value. 

This is the well known thermal stabilization effect of emitter resistance on the bis currents of the transistors.

If this resistance is removed the bias current will grow with temperature causing an equal increase in the output current in case of the conventional current mirror and also in case of the Widlar current source it will affect some increase of its current. So, the inclusion of RE1 has the function to stabilize the current mirror current against temperature changes. 

So, In the input side of the current mirror RE1 has its known function which is the thermal stabilization of the transistor bias currents.

Best wishes

Dear Cyril,

welcome!

What is the area ratio of the two transistors in the given circuit in the question? what is the current ratio required , that is what is the ratio IE2/ IE2? I intend to make some quantitative analysis of the circuit.

Best wishes

Dear Cyril,

Dear colleagues,

In this comment i will make some circuit analysis that may exploit the properties of this configuration brought by the respected colleague Cyril. He always bring us challenging problems.

From the simple analysis, one can write down:

VBE1 +IE1 RE1 = VBE2 + IE2 RE2 = VB,

with the symbols have their usual meaning except VB is the voltage of the common base a s hinted previously

Rearranging and substituting VBE= Vt Ln (IE/IS), where Vt is the thermal equivalent voltage= 25.6 mV at 300 kelvin, IE is the emitter current  of the transistor and IS is its saturation current and Ln is the natural logarithmic, IS= JS A, with JS is the saturation current density and A is the emitter area of the transistor, one gets:

Vt Ln ( IE1 IS2/IE2 IS1)=  IE2 RE2 - IE1 RE1

This relation can be used to study different cases of the current mirror:

The exact current mirror, where the left hand side is made zero by realizing the condition:( IE1 IS2/IE2 IS1) =1 which means that A2/A1= IS2/IS1= IE2/IE1, which is the condition of the conventional current mirror without the emitter resistors. 

When the left hand side becomes zero it follows that IE2/IE1= RE2/RE1

And therefore the current ration is realized twice, one by adjusting the area of the transistors and the other by emitter resistance ratio.

The other realization of the current mirror by an approximate method where, the write hand side which is the difference in the base to emitter voltage of the two transistors even with using equal area transistors can be made much smaller the voltage drops on the emitter resistors. This can be made VB appreciable larger than VBE. 

It follows that The right hand side is approximately equal zero and IE1 RE1=IE2RE2,

which gives that the current ratio is the inverse of the emitter resistance.

In both cases the current ratio of the mirror will be fixed by the resistance ratios.

Best wishes

Hello everybody here! Excuse me for the late reaction, but I was in the countryside where I had to do some physical (not so pleasant) work to please my wife:) But now I am free to engage in my favorite RG mental work:)

Joerg, thanks for the strong initial thoughts that started an interesting discussion... and provoke Abdelhalim (thanks!) to give us the brilliant detailed analysis above... Aparna and Josef, thanks for the recommendations!

I will try to enrich your explanations with a more general, philosophical look at these four variants of a current mirror (you know this is the strong side of my thinking:) Let's begin with a simple current-mirror classification with respect to the transfer ratio:

1. In the classical current mirror (no emitter resistors), the output current is equal to the input current (IOUT = IIN); so it acts as a current follower.

2. In the current mirror with an emitter resistor in the input part, the output current is bigger than the input current (IOUT > IIN) so it acts as a current amplifier.

3. In the current mirror with an emitter resistor in the output part, the output current is lower than the input current (IOUT < IIN) so it acts as a current attenuator.

4. In the current mirror with emitter resistors in both parts, the output current can be: lower than the input current (IOUT < IIN at RE1 < RE2) so it will act as a current attenuator; equal to the input current (IOUT = IIN at RE1 = RE2) so it will act as a current follower; bigger than the input current (IOUT > IIN at RE1 > RE2) so it will act as a current amplifier.

It turns out the last 2-emitter-resistor circuit is the most universal because, depending on the ratio of the emitter resistances, it can work either as an attenuator, follower or amplifier?

Now let's explore the presence of the negative feedback in the four current-mirror versions:

1. In the classical current mirror (no emitter resistors), the input base-emitter junction is directly connected in parallel to the output collector-base junction of the transistor. So there is a 100 % voltage type (shunt-shunt) negative feedback. The transistor is as though "inverted" - it adjusts its input quantity (VBE) to match its output quantity (IC) to the external input quantity (IIN) thus fuctioning as a non-linear (logarithmic) current-to-voltage converter.

Roughly speaking, the transistor acts as a diode voltage stabilizer; hence the name "active diode". This provides almost ideal operating conditions of the input current source implemented with only one resistor (the current depends only on the input voltage and resistance).

2. In the current mirror with an emitter resistor in the input part, additional voltage drop appears across the emitter resistor. It is proportional to the input current; so the emitter resistor acts as a linear current-to-voltage converter. But note this voltage is outside of the negatiwe feedback loop. So there is no current type (series-shunt) negative feedback (emitter degeneration); there is only the initial voltage type (shunt-shunt) negative feedback. The stage properties are not modified.

3. In the current mirror with an emitter resistor in the output part, there is a voltage drop across the second emitter resistor. It is proportional to the output current; so it acts as the same as above linear current-to-voltage converter. But note that now its voltage is subtracted from the input base voltage. So there is a current type (series-shunt) negative feedback (emitter degeneration) that increases the output resistance of the second stage. In total, there are two negative feedbacks - parallel (in the input part) and serial (in the output part).

4. In the current mirror with emitter resistors in both parts, as above, there are two negative feedbacks - parallel (in the input part) and serial (in the output part).

Then let's explore the four current-mirror versions regarding the linearity...

But here I will interrupt for a while this interesting discussion to share something interesting that happened to me this morning... and it really excited me...

I decided to see if there are any new interesting materials about the current mirror on the web. And so I came across a very interesting question (the first link below), put in a little naive but heartfelt way in the StackExchange forum. The reader was trying to understand how the input part of the current mirror works by reading a material posted in the Analog Devices Wiki educational section (the second link below).

And because he had not found the exact answer to his question, he had sought  a help here. Unfortunately, here he is answered formally by all sorts of mathematical analyses... but not with the simple truth spoken in a few words. And I decided to answer him simply first taking a look at this ADI material.

But reading the material, I gradually began to feel that it was very familiar to me... until, at some point, I realized that I was actually reading something I wrote. It turned out that the author has allowed him to skillfully compile my original explanations of the underlying ideas in the current mirror circuit solution, even using the structure of my "building scenario". Obviously, he had used my material in Wikibooks (the third link below)...

Now I feel conflicting emotions. On the one hand, by putting my ideas on the web (especially in Wikibooks and Wikipedia), I have provided them for free. On the other hand, I think it is not moral to introduce ideas like yours without even mentioning where did you get them from.

Because there is some absurdity in this situation. On the one hand, circuit designers (including ADI) constantly complain to me that it does not make sense to present basic circuit ideas in this profound way... and ignore my explanations. For example, more than 10 years ago, I offered to the respected Dan Sheingold, an editor of Analog Dialogue, to write a series of short materials revealing basic circuit ideas in such an intuitive manner... but this endeavor failed.

On the other hand, it turns out that now ADI explains the basic current mirror ideas through my explanations. What is the logic of all this? Where is the ethics here?

Am I right? If so, what do I do?

https://electronics.stackexchange.com/questions/280270/input-stage-of-a-current-mirror-circuit?rq=1

https://wiki.analog.com/university/courses/electronics/text/chapter-11

https://en.wikibooks.org/wiki/Circuit_Idea/How_to_Reverse_Current_Direction

Obviously in such embarrassing situations everyone has to deal with his/her problems (forums create an illusion of empathy but in practice we are alone in this world)...

I think I made the right decision by first registering in Analog Devices Wiki thus becoming a wiki editor. Then I edited the source of the page and quoted the Wikibooks material at the beginning.

Now I only have to wait for the editor's reaction...

https://wiki.analog.com/university/courses/electronics/text/chapter-11?force_rev=1

 Dear Very dear Cyril,

Not at all. I have to thank you for your very interesting and expressive questions and comments on the basic analog circuits. I try to do my best to share in the discussion and make my contribution understandable and useful. I will not forget your words to me when started my activities on the researchgate. Your words were and still are very encouraging to me. Cyril you have your own unique style for describing the basic circuit operations. Your explanations are scenario stories for very descriptive operations of the electronic circuits.

You are a very faithful and respected professor and friend.

What do you wrote and edited will remain lightening the way for many young scientists.

For me, you will remain one of my best friends whom i met in my whole life.

My best wishes to you and your family and many thanks for the researchgate that gave us the chance to meet in science.

Abdelhalim

Thanks Abdelhalim!

You made the impossible possible - you regain my faith in the NETWORK! You made me feel you and all my "virtual friends" here as real friends... even as more real than the real ones around me...

Thanks for the superlatives ... but I just do the complicated - simple, and the simple - even simpler:)

Inspired by the courageous thoughts above, I continue looking for the most ingenious idea in the simple BJT current mirror...

• To cascade a current-to-voltage and voltage-to-current converter is the first "topological" idea  thanks to which the direction of the current is reversed.
• At first glance, it seems that it would be best these two converters be linear. But in this case, the voltage drops across them will significantly increase when the input current increases. This will cause serious problems: the voltage drop across the input current-to-voltage converter will influence the input current source (especially, if it is a simple resistor-type current source); the voltage drop across the output voltage-to-current converter will decrease the operating range (compliance voltage) of the output current source.
• And here, in my opinion, the most ingenious idea arises - to cascade a compressing (logarithmic) current-to-voltage converter with an expanding (antilogarithmic) voltage-to-current converter. So, in this arrangement, the voltage is a undesirable quantity... and therefore it should be suppressed... it should stay low down, near the ground. This idea seems absurd but in practice, it is well known from the analogue signal processing, and widely used, e.g. in audio engineering.

So, the transfer characteristics of the converters may not only be non-linear (as I have been thinking so far), but they must be such. The good news here is that although the particular transfer characteristics are nonlinear, the whole transfer characteristic is linear.

Dear Respected Colleague Cyril,

I quoted your answer  because i want to extend it: In the classical current mirror (no emitter resistors), the output current is equal to the input current (IOUT = IIN); so it acts as a current follower.

What is the idea of the current mirror with two emitter.... Available from: https://www.researchgate.net/post/What_is_the_idea_of_the_current_mirror_with_two_emitter_resistors/1 [accessed Aug 8, 2017].

In the classical current mirror, the ratio of the output current to the input current is normally controlled by the transistor emitter area A of the two transistors. This is realized by the physical lay out of the transistors and the emitter area definition of the transistors by the fabrication masks. It is so that IC2/IC1= A2/A1.

So, for AI=A2 the current mirror current ratio will be one like a flat mirror, it can be greater than one with amplification like concave mirror or the ratio can be made smaller than one like a convex  mirror.

So, in the classical current mirror, the current ratio is normally controlled by area ratio of the constituting transistors.

Best wishes

Abdelhalim,

Thanks for the clever "geometrical" idea. The interesting thing here is that the problem of current scaling is solved without worsening the circuit characteristics (e.g., voltage compliance)...

And I continue thinking about the most fundamental idea behind the current mirror...

Two days ago, I decided to help the StackExchange reader (with a user name Sumanth). Here is our dialogue in the beginning of the page (my user name there is Circuit fantasist):

"Sumanth, do you still want to really understand how the current mirror input stage works? – Circuit fantasist 2 days ago   

@Circuit fantasist, I think jonk's answer has done an amazing job in clearing my doubts. Thank You! – Sumanth yesterday

 I mean a simple and clear explanation that shows the fundamental idea behind the current mirror... – Circuit fantasist 6 secs ago"

An interesting situation - the reader believes in the lengthy jonk's analysis below... which is really an "amazing job"... and yet it does not reveal the fundamental idea...

https://electronics.stackexchange.com/questions/280270/input-stage-of-a-current-mirror-circuit?rq=1

I continue thinking how to present in the most striking way the ubiquitious current mirror circuit...

And here is another powerful idea - to think of the current mirror as of an inverting current buffer (booster).

• The voltage buffer (e.g., an emitter follower) has extremely high input resistance (to not influence the imperfect input voltage source) and extremely low output resistance (to be not influenced by the imperfect voltage load).
• Oppositely, a dual current buffer should have extremely low input resistance (to not influence the imperfect input current source) and extremely high output resistance (to be not influenced by the imperfect current load).

So, both buffers perform consequtively two actions and accordingly, have two parts - input and output. But it is interesting that in the case of voltage buffering they can be implemented only by one transistor while the current buffering requires two transistors.

Indeed, a single transistor can act as an amplifying current mirror where the base current is the input current and the collector current is the output current. To make it a following current mirror, we should divert the excessive input current from the base. We can do it by connecting a shunt active diode in parallel to the second base-emitter junction... and this would be another striking way to build the classic 2-transistor current mirror...

We already have three power views at the bare 2-transistor circuit:

• "compressor-expander"
• "current buffer"
• "shunted amplifying current mirror"

 An interesting association, dear Aparna...

Cyril - only by accident I stumbled upon your question regarding the Widlar source.

In the following, I quote Horowitz/Hill (Art of Electronics, Fig. 2.47 with two emitter resistors, 1k each):

"One problem with the simple current mirror is that the output current varies with changes in output voltage ....This is because of the...Early effect. .....One solution is the circuit shown in Fig. 2.47.  The emitter resistors are chosen to have at least a few tenth of a volt drop. This makes the circuit ...better, since the small variaitions of VBE with VCE are now negligible."

A discussion of this circuit can be found in he following link:

https://electronics.stackexchange.com/questions/202549/current-mirror-improvement-by-adding-resistors

Hi Lutz,

I think we are about to uncover the secret of the current mirror with two emitter resistors. We only have to build the circuit in the correct logical sequence and show its advantages and disadvantages compared to the classical current mirror... and of course, to think a bit about the philosophy behind these circuits.

I suggest to do this with such an attractive (mental) experiment:

In this way, we will be able to investigate the behavior of all these famous circuit solutions and touch ourselves on what their inventors were thinking when inventing them...

Cyril - I think, you forgot to mention that all these experiments should be performed with a variable current load (causing a varying voltage at the output node.

Lutz, thanks for the quick response!

One of the four variable resistors performs precisely this role... Let's specify these resistors:

• RIN (RC1) - the input resistance (of the input current source)
• RL (RC2) - the varying load ("causing a varying voltage at the output node")
• RE1 - the resistor in the emitter of the input transistor (acting as an input current-to-voltage converter)
• RE2 - the resistor in the emitter of the output transistor (acting as an output current-to-voltage converter - emitter degeneration)

We can think of this arrangement as of a kind of balanced bridge:)

Of course, we should connect voltmeters in parallel to each element (resistor, transistor), and ammeters in the two legs.

Here are two more interesting questions in StackExchange related to the topic... Apparently, there is still something unexplained in these circuits, which makes readers continue asking questions...

https://electronics.stackexchange.com/questions/256433/how-does-this-circuit-given-with-diode-works-different-from-the-standard-current?rq=1

https://electronics.stackexchange.com/questions/84950/replacing-diode-connected-bjts-by-simple-diodes-in-a-current-mirror-circuit

Today, I took a look at the current mirror material in the ADI Wiki... and I was surprised to find that there was no reaction to my insertion at the top of the page - it has been staying already a second week as a "draft" (i.e., it is not officially approved)...

I just do not know how to characterize such a behavior... Perhaps plagiarism is something already well accepted in the research community... and I am behind the times in my understanding:)?

https://wiki.analog.com/university/courses/electronics/text/chapter-11?do=#chapter_11the_current_mirror

Oh, I forgot this network was not Facebook, where everyone immediately reacts to injustices in this world... Anyway, I will swallow this insult too... and will go on...

Dear Aparna,

You have reminded me to look at what is going on with the ADI Wiki page...

A moment ago, I threw an eye on it and what to see - it was already "approved"...  that is, мy insertion is removed?!? and new op-amp circuits are inserted before the transistor ones...

But let's put aside this story... and start with our interesting imaginary experiments with current mirrors... In fact, my idea is not just to experiment, but rather to reveal the philosophy of the current mirror through its evolution... Notice that I have no ready philosophy... but only some "ingenious" guesses:) that I would like to verify here, in this discussion... and to build a complete philosophy...

1. RE1 = 0; RE2 = 0; RC1 and VIN = max; RC2 = 0, VCC = max. This means that initially we have a classic BJT current mirror without emitter resistances. It is at the best input conditions (maximum input voltage and resistance) and best output conditions (maximum supply voltage and zero load resistance).

Remember that: VIN is the input voltage (the internal voltage of the input current source); RIN is the input resistance (the internal resistance of the input current source); RL is the varying load; VCC is the supply voltage of the output part (to which the load is connected).

Now we can vary the input current IIN in three possible ways - by changing the input voltage VIN or the input resistance RC1 or both (we can think of the combination VIN + RIN as a simple "resistor-type" current source).

To see the clever idea behind the simple current mirror, imagine for a moment that we pass the input current directly through the load RL (i.e., there are no any transistors and VIN, RIN and RL are simply connected in series). In this possibly simplest current-sourcing arrangement, a "harmful" voltage drop VL will appear across the load... and the current will be IIN = (VIN - VL)/RIN instead simply IIN = VIN/RIN as we would want. To decrease the error, we should decrease the load resistance up to zero... which is not possible since the load is an external component... and we have no control over it.

The problem here is that the input resistor and the load are connected in the same circuit (loop) and they share the same voltage VIN. So, if we can remove the load... move it to another place... the problem will be solved...

In the simple BJT current mirror, this problem is solved in an exquisite way - in the input circuit, the load is replaced by a "voltage-compressing" non-linear element (diode). The voltage variations across it are "crushed" and it behaves almost as a short connection with the ground.

Then, in the output part (the "other place"), the voltage variations are enlarged back by a "voltage-expanding" non-linear element (transistor).

So the input transistor Q1, connected in this odd "collector-to-base" way, serves as a current-sensing non-linear resistor absorbing minimum voltage from the input voltage source VIN. It only samples the input current without interfering the input voltage source.

This technique requires a supply voltage VCC with the same as the input voltage VIN polarity. As a result, the two voltage drops (VR1 and VL) are located in voltage regions with the same polarity.

Note the current flows in opposite directions through both elements. So, this is a "current inverter".

As a comparison, in the op-amp inverting configuration, the same problem is solved in another no less ingenious way - the load is left in the input circuit... and the "harmful" voltage drop across it too... but it is compensated by adding an equivalent voltage. As a result, the input current source is shorted by a virtual ground.

This technique requires a supply voltage VCC with an opposite polarity. As a result, the two voltage drops (VR1 and VL) are located in two opposite voltage regions... and the whole operating range is VIN + VL.

Also, the current flows in the same direction through both elements. So, this is a "current follower".

This comparison give me another idea how to improve an imperfect ammeter...

In the RG question below, I have showed how to make it almost ideal by inserting it in the negative feedback loop of an op-amp.

It seems, with a similar success, we can make it more perfect, if we connect it as a load in the output part of a BJT current mirror...

Is there any sense in this idea? Or it can only serve to provoke the desire for discussion of the participants here:)?

I argue that it has one very significant advantage and one very significant disadvantage compared to the op-amp version...

https://www.researchgate.net/post/How_do_we_improve_the_real_ammeter_How_do_we_create_an_almost_ideal_ammeter_What_does_the_op-amp_really_do_in_the_circuit_of_an_op-amp_ammeter2

Aparna, thank you for your responsiveness!

RE is the resistance of the external emitter resistor (a rheostat in this experiment). So, when saying RE = 0, this simply means that, at this initial step, there is no emitter resistor connected... and we investigate a simple current mirror.

My idea here is to extract the most ingenious ideas in this circuit by following its evolution beginning with this simple solution and finishing with the brilliant Wilson circuit. I think this approach is more worthy of this audience than the widespread instructive explanations...

But... where is the audience:)?

Well Aparna,

If you break away for a moment from your beloved program, you will get to know what is the disadvantage of the simple BJT current mirror compared to its op-amp analog:) Such simple (but powerful) truths are not written even in such electronics bestsellers as H&H... and ADI Wiki as well:) 

IMO the input transistor can be easily damaged if a voltage source is directly connected to its collector. BTW there is the same problem with any open-collector output stage. I remember that once a colleague said, "if you bring an open collector out, always some fool will connect it to the power supply":)

But note that while the open-collector transistor can "survive" the heavy input shock (if the base current is small), the current-mirror input transistor will not... since its base-emitter junction (connected in parallel to the collector-base one) will surely be damaged. So the input part of the BJT current mirror is fully unprotected...

So, keep in mind this fact when improving an imperfect ammeter by connecting a BJT current mirror at its input:)

The FET current mirror can be also damaged if a voltage source is directly connected to its drain (gate)...

Once investigated the input part of the current mirror... and seen that it behaves as a kind of virtual ground... let's do the same with the output part.

Here, we can try to change the output current IL through the load again in the three possible ways above - by changing the supply voltage VS or the load resistance RL or both.

But in contrast to the input part, the load current (almost) does not change since the transistor Q2 resists (by varying its collector-emitter "resistance") our efforts to change the current... and the output behaves as a kind of "virtual infinity":) The current can be changed (controlled) only by the input base voltage; all other quantity variations are considered as a disturbance and are compensated...

It is interesting to see that when varying VCC, the voltage drop across the load (almost) does not change... and the load behaves as a kind of "zener diode" (hmmm... interesting - a resistor acting as a zener?!?) As a result, the collector voltage VC2 follows the supply voltage variations (as far as I can remember, this trick was used to shift voltage variations in integrated circuits?)...

We can see that, when varying (increasing) RL, the voltage drop across the load can vary up to VS (i.e., the circuit has the possibly maximum compliance voltage).