Circuit Idea/How to Reverse Current Direction

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Revealing the Secret of Current Mirror

Circuit idea: Connect a reversed and a direct voltage-to-current converter one after another.

A current mirror consists of two consecutively connected inverse converters

BackgroundEdit

There are many professional publications about the famous current mirror circuit and its versions[1]. Only, as a rule, they look formal, specialized and "sterile". These sources give a lot of details about current mirror versions but they do not show the basic ideas behind them.

Our mission. We, circuit ideators, have a different "mission" in Wikibooks - to reveal the philosophy behind circuits. We know that before showing in detail how to calculate an electronic circuit we have first to show what the very basic idea is behind this circuit. In Circuit Idea, we try to do that relying only on our human intuition, imagination and common sense.

Fig. 1 What does the transistor Q1 do in this odd circuit?

The simple but complex current mirror. At first sight, the current mirror looks a simple bare circuit (Fig. 1); but actually, it is an odd, strange and exotic circuit that is never explained (there are only some guessworks). It is a great paradox that everyone knows what a current mirror is but nobody knows how it operates.

As a rule, the classical current mirror explanations begin with the assertion that the current-setting transistor Q1 acts as a diode. But it is very primitive and confusing to say "the input transistor Q1is a diode". Actually, it is not a diode; it is exactly a transistor operating in the active mode. It would be a diode, if its collector was disconnected. Then all the current IREF = VCC/R (R is omitted on the picture) would flow through the base-emitter junction acting really as a diode. But here the output collector-emitter part of the transistor is connected in parallel to the base-emitter "diode". In this way, it serves as a shunting regulating element that diverts the great amount (β/(1 + β)) of the current. But how and why the transistor does this magic?

There is something strange and confusing in this arrangement... As everybody knows, the base-emitter voltage VBE is the input quantity of the bipolar transistor and the collector current IREF is the output quantity. But here all is on the contrary - the collector current is the input and the base-emitter voltage is an output!?!? What an idiocy?

Questions to be answered. In order to understand this odd circuit, we need to answer dozens of questions that are never answered. Here are some of them.

What does the transistor Q1 do in this circuit? What is its function there? What is the difference between it and a diode? Can we replace it by a bare diode or a base-emitter junction (leaving the collector disconnected) as it is shown in another current mirror story [2]? Why and how the collector current IREF serves as an input quantity and the base-emitter voltage as an output one (we thought the base-emitter voltage VBE was the input quantity of the bipolar transistor and the collector current IREF was the output quantity)? What does the transistor Q2 do in this circuit? What is its function there? How do collector voltages vary when we change the input current IREF (the input voltage VREF , the "programming" resistor RREF and the load resistance RL )?

Since there are no satisfactory answers to these questions, let's forget temporarily all kinds "cut-and-dried" citations and to try disclosing the mystery of the famous circuit by ourselves. Just forget them and begin thinking by yourself!

The general idea behind the whole circuitEdit

It seems our story will be not based on "verifiable sources" just because there are not such sources; so, one can say it is an original research. Only, we will endeavor to explain the famous circuit so simply, clearly and evidently that there is not any need to verify these assertions. By the way, the very NOR stipulates such a possibility - to use sources, "the accuracy of which is easily verifiable by any reasonable adult without specialist knowledge".

The problem: current direction inverterEdit

Fig. 2. A current mirror is a 3-terminal device

In circuitry, especially in the area of microelectronics (e.g., inside op-amps), sometimes we need to invert the direction of a current (to make the flowing in current a flowing out one and v.v. - the flowing out current a flowing in one) without changing its magnitude. In this way, the output current just follows, "copies" the input one but this copy is inverted, "mirror" one. In other words, the input current "programs" the output one and the whole circuit serves as a programmed current source (current-controlled or dependent current source).

Obviously, in order to do that, such a current direction inverter (having a more popular name - current mirror) has to be at least a 3-terminal device (Fig. 2). The reason of that is because the two entering currents have to "go out" from somewhere (Fig. 2a) and v.v., the two exiting currents have to "go in" somewhere (Fig. 2b). In this arrangement, the one lead serves as an input, the other - as an output and the third - as a common terminal (usually connected to the positive or negative rail and more rarely - to the ground).

How to create a current mirrorEdit

Basic structure. In electronics, we have two kinds of circuits producing or, more frequently, just controlling current - current sources and current sinks. The difference between them is that, regarding to some circuit point, sources "inject" current while sinks "absorb" current. Obviously, in order to implement the general arrangement above (Fig. 2) into a concrete current mirror, we need both the circuits. As a rule, these circuits are voltage-controlled; so, in order to control them by current, we need to connect current-to-voltage converters before them. Let's now draw the block diagram of the two possible current mirror arrangements.

Fig. 3a. A current mirror sinking the output current
Fig. 3b. A current mirror sourcing the output current

If we have a current source, we convert the input current ("entering" the current mirror) into a voltage and then use this voltage to control a current sink; as a result, we obtain a current sink (Fig. 3a). Conversely, if we have a current sink, we convert the input current ("exiting" the current mirror) into a voltage and then use this voltage to control a current source (Fig. 3b); as a result, now we obtain a current source. We can already generalize this basic current mirror structure in a conclusion:

A current mirror consists of two consecutively connected current-to-voltage and voltage-to-current converters.

Fig. 4. The current mirror consists of two consecutively connected inverse converters

Characteristics. It is interesting that the two converters may be linear (then IOUT = V/R = IIN.R/R = IIN) but it is not obligatory. They might be non-linear devices having whatever transfer or IV characteristics that may even depend on other quantity (e.g., temperature); the only requirement is their characteristics to be reverse. In this way, if the one converter implements a function y = f(x) and the other represents the reverse function x = f -1(y) the whole function is y = f(x) = f(f -1(y)) . Note that the two converters might be connected in any order, including the direct after the inversed one (Fig. 4). At first sight, this result looks strange and nonsensical but in this way the main problem - reversing the current direction, is solved. So, we can formulate the next conclusion:

A current mirror consists of two consecutively connected converters having reverse characteristics.

Simple reversing. Usually, there is not a couple of direct and reverse converter and we have only one kind of converter. If this is a reversible converter, we can use it as a direct converter and then connect another such converter in an opposite direction to get a reverse converter.

Reversing by negative feedback. The problem is when the converter is not reversible and it is a bare one-way device. In this case, we cannot swap the circuit input and output ports; we cannot apply an input quantity to the circuit output and to get an output quantity from the circuit input (unfortunately, this is our case). Then, the only way to "reverse" the circuit is applying the ubiquitous negative feedback. Only systems with negative feedback have the unique property to reverse the cause and effect relations between the input and output quantities of the objects. They adjust their internal input quantity so that their internal output quantity to become equal to the "true" external input quantity. As a result, the internal input quantity follows, depends on the external input quantity; actually, the internal input quantity serves as an external output quantity. In this way, they can "reverse" the objects.

Implementing the general idea into a BJT electronic circuitEdit

Once revealed the general idea we can create as many as we want current mirror circuits (this is the power of this heuristic approach). In all these versions, only the electronic components (BJT, FET, op-amps, etc.) will be different; the general idea behind them will be the same. Well, let's begin with the most popular of them - the basic BJT current mirror.

An output partEdit

We can drive a bipolar transistor by voltage or by current. If we change the base-emitter voltage as an input and use the collector current as an output (Fig. 5), we can think of a BJ transistor as a non-linear voltage-to-current converter having an exponential characteristic. So, we can use it directly as an output part of our simple BJT current mirror:

The output part of the simple BJT current mirror is just a bipolar transistor acting as an exponential voltage-to-current converter.

Fig. 5. Implementing the output part of a BJT current mirror
Fig. 6. Superimposed IV curves of the output part

Operation. How does the transistor T2 behave in this arrangement? In order to know, let's carry out an experiment - set a constant input voltage VREF = 0.5 ÷ 0.7V and then vary the load resistance RL (with the same success, you can vary the supply voltage V2 or even both the resistance RL and the voltage V2). The result is surprising - the transistor changes its present resistance RT2 between the collector and the emitter so that to keep up a constant total resistance Rtot = RL + RT2 = const (Fig. 6). As a result, the output current remains constant IOUT = V2/Rtot. In this way, the output collector-emitter part of the transistor T2 acts as a current-stable non-linear resistor. They usually say the transistor T2 acts as a simple current sink.

An input partEdit

Now, we need to make the BJ transistor serve as the needed opposite current-to-voltage converter. Only, we cannot "reverse" it directly since the transistor is a one-way device, whose base-emitter junction can control the collector current; the opposite is just impossible. What do we do then?

We have already known the remedy - its name is negative feedback. In our case that means to make the transistor adjust its base-emitter voltage "VOUT" so that the collector current to be IIN = V1/R. For this purpose, we have just to connect its collector to its base, in order to apply a "100% parallel negative feedback" (Fig. 7). As a result, although it seems strange, in this "reversed" transistor the collector current serves as an input quantity while the base-emitter voltage serves as an output quantity!

The input part of the simple BJT current mirror is just a bipolar transistor with 100% parallel negative feedback.

OperationEdit

But how does the transistor T1 behave in this arrangement? In order to know, let's carry out another experiment - set a constant input (supply) voltage V1 and then vary the input current-setting resistance R (again, with the same success, you can vary the input voltage V1 or even both the resistance R and the voltage V1).

Fig. 7. Implementing the input part of a BJT current mirror
Fig. 8. Superimposed IV curves of the input part

The result is still more surprising than before - now, the transistor changes its present resistance RT1 between its collector and emitter so that to keep up an almost constant resistance ratio K = RT1/(RT1 + R) = const. As a result, the output voltage remains almost constant VOUT = VCE1 = VBE = const (Fig. 8). In this way, the output collector-emitter part of the transistor T1 acts as a voltage-stable non-linear resistor. But this is the recipe of making various active diodes ("ordinary", "zener" or "rubber", adjustable...)! It is just a wonder! The parallel negative feedback has made a current-stable resistor (the output part of T1) behave as a voltage-stable one! This is the same transistor but in the first case it serves as a current-stable element while in the second case it serves as a voltage-stable element.

A "reversed" transistorEdit

At the same time, the input voltage source V1 and the current-setting resistor R form a composed current-source that "want" to produce a current IIN = V1/R through a voltage-stable element (the transistor T1). By the way, some mystic cascode circuits are based on the same arrangement (a current source supplies a voltage-stable element and v.v.). It is interesting that, in this situation, the voltage-stable element changes its present resistance, in order to "help" the current source to establish the desired current. For example, if we decrease the resistance R to increase the current, the transistor T1 will also decrease its present resistance thus helping us to increase the current and v.v. Doing that, the transistor T1 adjusts its base-emitter voltage so that the collector current to be always IIN = V1/R. As a result, although it seems strange, the collector current serves as an input quantity while the base-emitter voltage serves as an output quantity! The negative feedback has reversed the one-way transistor!

An equilibrium pointEdit

Fig. 9. Discovering the equilibrium point of the current-setting transistor

It is interesting to discover how the transistor T1 manages to reach the equilibrium point (a system with negative feedback always reaches the equilibrium). For this purpose, let's change firstly the magnitude of the input voltage V1 or the resistance R. The transistor responses to this "intervention" by changing its present resistance RT1... but till when? In order to understand, let's carry out an interesting experiment - break the feedback loop and drive the "true" base-emitter transistor input by a separate voltage source "VOUT" (Fig. 9). Then, let's begin increasing/decreasing the "true" transistor's input voltage "VOUT"; in return, the transistor will begin dropping/raising its collector voltage VC1 by decreasing/increasing its present resistance RT1. Figuratively speaking, the two voltages "move" against each other. In order to imitate the negative feedback behavior, we have to stop changing "VOUT" when the two voltages become equal (a zero indicator connected between the collector and the base can show this moment); this is the equilibrium point. Now, if we short the zero indicator (connect the collector to the base or, as they say, close the feedback loop), the system will remain at rest as it is at the point of the equilibrium.

Another example: a transdiodeEdit

By the way, there is another paradoxical circuit that does the same but it is an almost perfect - the logarithmic converter based on the so called transdiode (a BJT transistor connected in the op-amp feedback loop[3]). As here, in this odd and also never explained circuit configuration, the collector current serves as an input quantity while the base-emitter voltage - as an output quantity?!? The only difference is that there an additional op-amp adjusts the transistor's base-emitter voltage so that its collector current to be exactly equal to the input one. The op-amp does this magic by observing the virtual ground point and keeping it (almost) equal to zero.

Assembling the whole circuitEdit

Finally, we have only to connect the output of the input part (the base-emitter junction of T1) to the input of the output part (the base-emitter junction of T2) to build the famous BJT current mirror circuit!

+ =

Fig. 10. Assembling the whole BJT current mirror.

The only problem is that the transistor T2 "sucks" another base current IB from the input current. As a result, the output current is smaller than the input one:

IOUT = IIN - 2IB = β.IB - 2IB = (β - 2).IB.

ConclusionEdit

What is the use of applying this stiff heuristic approach - step-by-step building instead giving the "cut-and-dry" classical circuit? The benefit is that we can build now various current mirror circuits knowing only one powerful general idea! We know what the transistors T1 and T2 do in this circuit; we know the truth about the basic current mirror circuit!

ReferencesEdit

1. Current mirror is a comprehensive exposition of the topic.
2. Current mirrors is a popular exposition of the current mirror topic written by the well-known web author Tony Kuphaldt.
3. Integrated DC Logarithmic Amplifiers