Talk:Circuit Idea/Can a "diode" current mirror exist?

Lab 3: Transistor circuits (trying to create a "diode" current mirror)
Tuesday, April 08, 2008, 13.45 h (extracted from the lab records made by a solid-state recorder) Circuit-fantasist 18:56, 29 May 2008 (UTC)



The idea of this lab is to transform the simplest transistor current source (that your colleagues from group 65b made an hour ago) into a current mirror circuit. You can see what they have discussed on the whiteboard and what they have made on the prototyping PCB. Well, let's first examine and discuss their creation and then try to metamorphose it to the desired circuit.

Look at the circuit on the whiteboard that is left from your colleagues. What do you see there? What is it? What is the idea of this circuit? What does the transistor do there? What is its behavior? What does the potentiometer do? What are these steady resistors? Why have they connected them in series with the varying ones? What does every component do in this circuit?

Well, let's begin reasoning to reveal the secret of this typical transistor circuit...



Understanding the transistor
We have to begin from somewhere; we need some idea to start the discussion... So, let's first consider the ubiquitous transistor. It is a three-terminal device... How have they connected it here? How do we connect three-terminal elements? They have connected one of its leads to the ground; so, the three-lead transistor is connected as a four-lead device. Also, in order to understand the more complex three-terminal transistor, we can break down it into simpler two-terminal parts. For example, we can first separate the output collector-emitter part as a two-terminal "element" and the input base-emitter part as another two-terminal "element"; then, to begin reasoning what does the transistor do.

Varying the input current
The simplest way to drive the transistor is by current. For this purpose, we have to connect only a base resistor Rb between the power supply and the transistor's base; for simplicity let's connect the same load resistor RL = Rb. Only, as a result, we get a current amplifier (Ic = βIb) instead the desired current mirror (Ic = Ib). How do we equalize the currents?

Varying the input voltage
We can connect a second resistor (actually, a potentiometer) in parallel to the base-emitter junction to divert the excessive current from the base to the ground. Where can we drive a transistor from? What is the input? The voltage between the base and the emitter drives the transistor. We can recognize in the left part of the circuit the ubiquitiuos potentiometer and a steady resistor connected in series. So, obviously the idea is to drive the transistor by a voltage. What is the mode of the transistor operation (saturated, active or cut off)? There are also measuring devices connected - two voltmeters and a (milli)ammeter.



What can we do with this circuit? We can vary its attributes one after another and observe its reaction to our intervention. Let's begin varying the input voltage. What happens, if we move the P's slider downwards to the end position? What is the input voltage then? Zero. Yes, there is no any input voltage. The output collector-emitter part of the transistor is in cut-off condition. The resistance between the collector and the emitter is infinite; no current flows in the output circuit. So, you can think of a transistor just as a varying resistor that has infinite resistance at these conditions (zero input voltage). Why the voltmeter V1 shows maximum voltage (equal to the supply voltage)? It is not so simple to explain why... Well, we have already discussed this phenomenon. We have a resistor whose upper end is connected to the positive rail of the power supply. What is the potential (the voltage regarding to the ground) of its lower end?

One possible viewpoint is to think of this circuit as a voltage divider with R1 = RL and R2 = ∞ (R2 >> R1; so, K ≈ 1. Another viewpoint is to see that the resistor R1 does not resist since there is nothing to resist (no current flows). Actually, there is no transistor connected in the circuit; we may erase it from the whiteboard...

The transistor is just a varying resistor
Now begin moving the potentiometer's slider up. At given moment the output voltage V2 begins lower; stop when it becomes somewhere between the supply rails (0 - 10 V), for instance in the middle (5 V). What is the mode of the transistor (saturated, active or cut off)? It is a "semi-opened"; so, it is in active mode. How does the transistor do this magic? Does it do something special? No, the transistor does not do anything supernatural in this region, it does the same what a tube, an FET, a rheostat, etc. would do, if they were at its place - it just changes its present resistance. So, the transistor is a varying resistor; only, it is an electrically controlled resistor. Well, if we have set a total load resistance RL = 10 kΩ and VCE = 5 V, what has the transistor done? It has set its present resistance RCE = 10 kΩ as well. It is just the same resistor as the collector one. Think of this circuit as a voltage divider with two identical resistors; its transfer ratio is K = 0.5. This is the notion about a transistor that we need to understand how transistor circuits operate. All the active elements do the same; they are electrically controlled resistors. So, all the transitor circuits contain both ordinary resistors and electrically controlled resistors (active elements). They form the well-known voltage divider where transistors are put on the place of R1, R2 or the both.

If you continue moving up the slider, you will probably reach the saturation where V2 ∞ 0 V and the transistor is opened totally. What is the current now? It has a maximum value according to Ohm's law: I = VCC/RL.

Varying the load resistance




How have we arrived at a conclusion that the output part of the transistor behaves as a constant current source? We have somehow to check this specualation. For this purpose, we have to provoke the circuit and to see how it will react to our intervention. Well, what will we see, if we vary the load resistance RL? The output transistor characteristic can give us the answer. Do you remember the subjest of semiconductor devices? These characteristics are favorite for them. Only, they do not show how to use these transistor behavior. A student: the characteristic is horizontal; so, the current will not change, if the load varies. Yes, it is right. But what does the transistor do, what is its behavior to obtain such a characteristic?

Does the current change, if we vary the load resistance? A stident: Yes, it does... But how do you explain it? Because the answer is exactly opposite - the current does not vary... Obviously, there are things that contradict our intuition and this is one of them... Well, look at the picture on the right (from Group 68b); it shows graphically how the transistor does this magic. Two IV characteristics are superimposed on a common coordinate system. The first curve is an IV characteristic of a real voltage source having a voltage VCC and an internal resistance RL (they name it "load line"). The slope of this curve depends on RL. If RL = 0, it is a vertical line; if RL > 0 it inclines proportionally to the resistance. The second curve represents the present output transistor resistance (RCE or shortly, RT).

The transistor does the following trick: if we try to decrease/increase the load resistance (to set upright the load line), it does the opposite - increases/decreases its present resistance (RCE or shortly, RT) so that the crossing (operating) point moves along the almost horizontal output characteristic of the transistor. As a result, the current stays almost constant. You can see how simple is the idea of this dynamic constant current source. It consists only of two elements: a voltage source and a dynamic resistor connected in series with the resistive load. The resistance of the dynamic resistor supplements the load resistance to a constant value (RL + RT = const); so, the current is constant - I = VCC/(RL + RT). Shortly, the dynamic resistor keeps up a constant total sum of the resistances.

A conclusion: ''If you connect in series a transistor and a resistor and begin varying the resistance of the resistor, the transistor will also begin changing its present resistance so that the sum of resistances to stay constant; as a result, the current stays constant. The transistor acts as a dynamic resistor that varies, as a current-stable resistor (an element that has the property to keep up a constant current)''.

And what about the voltage drop across the resistor? Does it vary? Move the slider and look at V2. If we lower RL, the voltage drop VL across the load lowers as well. But the transistor increases its present resistance RT and the voltage drop V2 across it increases as well. And vice versa; if we increase RL, the voltage drop VL across the load increases as well. Now, the transistor lowers its present resistance RT and the voltage drop V2 across it lowers as well.

A conclusion: If we vary the load resistance, the current stays constant but the voltages vary in wide limits.

However, at given moment the transistor depletes its reserve of resistance; it saturates and the voltage V2 across it becomes zero. The magic of dynamization ceases and if you continue changing the load resistance, the current will change according to Ohm's law.

Varying the supply voltage
Besides the load resistance, we can vary the supply voltage as well. Look at the picture on the whiteboard again and answer the question, "Will the current stay constant, if we vary the supply voltage VCC?" Well, try it on the lab set up. As we can see, the current varies. Why? What is the problem?

When we vary the supply voltage, the transistor would react to this new "intervention" as before: if we incrase VCC, it would increase its present resistance RT and v.v. The transistor would keep up a constant ratio VCC/(RT + RL). But still the current varies... What is the reason? A student: The input voltage varies. Yes, this is the reason... The tragedy of this circuit is that they have supplied the input voltage divider by the same power supply. But we have not another power supply. What do we do then so that the current to stay constant when we vary the common supply voltage?

A student: ...to add another transistor... I don't know, tell where you will place this additional transistor. You have to have some idea to place it. What does a tranistor stabilize? Do all colleagues understand that a transistor stabilizes current? A transistor is a current stabilizer while a diode is a voltage stabilizer. The characteristic of such a voltage-stable element is a vertical line. All the voltage-stable elements (diodes, zeners, LEDs, etc.) have such a characteristic; they behave as a voltage source.

A true diode sets the input current




Another student:...to place a diode... Yes, this is the right answer, to place a diode but now you have to say where. Please, write it on the whiteboard; here are whiteboard markers. Do not are confused that it looks not so pretty because we "invent" the circuit scribbling on the whiteboard. The other alternative is to show "cut-and-dried" pretty circuits on slides or Power point presentations...

Well, your colleague begins drawing a diode; is she drawing it in the right direction? By the way, I know there is diodes that are usually connected in a reverse direction. What are they? Students: ...zener diodes... Right but let's use an ordinary diode. Only, let's connect it correctly... Is the circuit written correctly?

Another student: Let's erase the potentiometer... Go and draw what you mean. Maybe, it is better not to remove the potentiometer; instead, draw it higher up and connect it properly, as a rheostat (do not forget to connect the slider). Leave it as an element that allows us to regulate the input current. It is time to make some shots. This room is not so suitable to make photos as a Hollywood studio but your colleagues do something great; so, they deserve to be perpetuated:)

Don't you think it's time to discover the web for similar solutions? For this purpose, write "current mirror" in the Google window. As usual, the corresponding Wikipedia page (too formal and boring) stays at the first position.

Let's then see Tony Kuphald's page about current mirrors. In the beginning, he reasoning about Ebers-Moll diode equation. We can see the load and the transistor and now we exclaim with surprise - here is the same diode! Tony Kuphaldt has proposed what our colleagues has propsed as well - to improve the circuit by connecting a forward biased diode in parallel to the base-emitter junction. Then, the author begin telling what the profit is in connecting an ordinary diode.

Let's scrutinize the Tony Kuphaldt's circuit since I have the following idea. As he is a very good teacher and an excellent web writer, I intend to write an email to him where to describe our Wikibooks "open student project". More concretely, I will tell him how you and I, all together, have arrived at his idea; then, we have checked it and have found out that it is right (or not). That is why is so important to judge his circuit.

Tony claims that the output current flowing through the load is equal to the input current flowing through the diode: "Current through resistor Rload is therefore a function of current set by the bias resistor, the two being nearly equal." But is it true? Let's try it.



A base-emitter transistor "diode" sets the input current
Who can solder? Choose a diode, solder it on the prototyping PCB and measure the currents. As I can remember, your colleague has told that a transistor consists of two diodes. Here we need a diode whoose IV curve is identical to the IV curve of the transistor driving the load. What an idea can arise then? Imagine we have not an ordinary diode; we have only transistors...

Replacing the true diode with a base-emitter transistor junction




Can we use the transistor as a diode? Yes, we can. In this case we can connect the base-emitter junction of an identical transistor (leaving the collector no connected) as a diode. A student: Yes, this was the lower circuit of Tony Kuphaldt...

They frequently use this "trick"... Even the transistor can be "semi-damaged"; we can use only one of its junctions (in this case wee nedd exactly the base-emitter junction). Who will draw the new circuit? Let's the colleague to draw it. Leave the collector non-connected. Try to draw the circuit more symmetrically and turn it with the base to the other base; thus we will arrive easier to the great current mirror circuit. Finally, draw where currents flow by green current loops.

Then look for another but the same transistor and solder it.



A problem appears


Only, a problem appears - the output transistor is saturated! So, let's begin discovering why...

We can find out the current-setting diode diverts only "one" base current Ib = V/2Rbias and the corresponding collector current of T2 is β/2 times bigger - Ic = IL = β.V/2Rbias.

We can conclude that the "diode" current mirror acts as a current amplifier (IOUT > IIN) but not as a true current mirror (IOUT = IIN)! So, we should notify Tony Kuphaldt about our find and invite him to join our discussion, in order to defend his speculation:)

Well, this was a copy of Group 65b's work. Let's continue their work by improving the "diode" current mirror.

