Talk:Circuit Idea/Simplest Transistor Current Source

Here is the building "scenario" that my students (group 65b) and I were using during Lab 3 to build the simplest transistor current source without feedback. Circuit-fantasist (talk) 07:58, 28 April 2008 (UTC)

Lab 3 (Group 65b): Building the simplest transistor current source
Tuesday, April 08, 2008, 10.30 h

What a current source is


We begin our lab with general questions: "What is a current source? What do we want it to do?" How does a current source behave? How do we figure out that some device is a current source? How do we make a current source?" Simply speaking, we have to form a notion about constant current source.

What have you known about the current source from the basic course of electricity? Here is the well-known symbol of a current source (see the picture). How to make a good guess that it is a current source? If I give to you a (black) box with two leads and I say that it is a current source, how do you convince yourselves that it is really a current source? Vladimir: It will begin warming... Why do you think so? And what will happen, if we connect the two leads (a short circuit - case 1 on the picture)? Is it dangerous for the current source (comparing with a voltage one)? Vladimir: Yes, it is... Atanas: No, it isn't... Indeed, there is nothing dangerous for a current source, if we short its two leads. As the current source is designed for a given current, it keeps up (limits) the current thus protecting itself.

Then, maybe the opposite condition - an open circuit (case 3 on the picture) - will be dangerous as the current source will try to pass the same current as before. In this situation, the current source is misled: it doesn't see the open circuit; it "thinks" there is something connected there. It just "sees" that no current flows and raises its internal voltage up with the "hope" to pass a current through the harmful "thing". Of course, the current source does not manage to pass any current through the open circuit and reaches the final (compliance) voltage; a saturation and even a breakdown occurs. Thus, the compliance voltage represents the maximum voltage that the current source can reach when strives to produce the desired current.



Passive current source


Atanas has said that, if we connect a resistor in series to the voltage source there will no problem with the short connection. Well, let's place a resistor; thus we obtain the simplest current source. Choose for concreteness the favorite pretty values for the electrical quantities: V = 10 V and R = 10 kΩ to obtain I = 1 mA.

Ideal load. First, make a short connection at the output of the circuit. Since there is no voltage drop across the load, all the input voltage V is applied across the resistor R. As a result, the current is exactly as Ohm's law says:

I = V/R = 10 V/10 kΩ = 1 mA

Keeping a constant current by using an ideal current load Passive voltage-to-current converter (Voltage causes current)



Imperfections. What are the problems of this humble circuit? Let's connect a load having some resistance to see what happens. As we can see, a voltage drop appears across the load and the current decreases.

The real load introduces some voltage drop VL, which affects the excitation voltage VIN. Now, not all the input voltage is applied across the current-setting resistor R but only the voltage difference VIN - VL. In other words, here the voltage difference VIN - VL determines the current IOUT instead the voltage VIN. As a result, the current decreases:

IOUT = (VIN - VL)/R = VIN/(R + RL)

This passive circuit is imperfect; it can't stand against the load "intervention" especially, if the load varies. It is a static, fixed, non-adaptable circuit... What do we do then?

Keeping a constant current by depreciating the load Passive voltage-to-current converter (voltage-controlled current source) Op-amp circuit builder (go to Stage 2 of this interactive flash movie)



Dynamic current sources
What a new idea might arise now? Let's begin reasoning... The problem pops up when we connect a load and, what is worse, if we begin varying its resistance. Note the load may be not only a steady or varying resistor; it might be a capacitor that is charging, a diode (ordinary, zener, LED, base-emitter junction, etc.) or even a voltage source (e.g., a rechargable battery)... It is not so important what exactly the load is; it is only important that a voltage drop appears across the load and this voltage drop confuses things.

Basic "dynamization" idea
If we look around, we can see that most things in this world are changeable, adaptable, dynamic... According to this observation, we have to make our current source react somehow to the load "intervention", to resist to its attemption to lower the current. That means something inside the current source has to change, in order to compensate the disturbing voltage drop across the load. What can change in this simple electric circuit?

Varying (dynamic) internal voltage


Vladimir: We might change the voltage... Yes, it is the one possible solution. Vladimir suggests, if the voltage drop VL across the load begins rising, the internal voltage VVAR of the current source to begin rising as well in the same direction with the same rate so that the difference VR between them to stay constant. This is the first idea: dynamic current source with following internal voltage. Till when will this magic continue? Of course, until VVAR finally reaches the positive supply rail. After that point, our wonderful "voltage-dynamic" current source becomes an ordinary one.

Have you heard the legend about Baron Munchhausen that was using his own boot straps to pull himself out of the sea? Our circuit is just one possible implementation of Munchhausen's bootstrapping idea in electronics.

Is there a negative feedback here? No, it isn't. The varying voltage source follows blindly the load voltage; it does not "interested" in the magnitude of the current.


 * I am not sure if a coil can be treated as a way to "produce" dynamic voltages for keeping the current constant (at least temporarily). Or is there practical way of doing so in real circuits?
 * Thanks,
 * /bruin


 * Yes, an inductor behaves exactly as a dynamic voltage source that opposes the changes of the input (exciting) voltage source so that to keep the current constant. Both the time-dependent elements (inductors and capacitors) behave as opposing voltage sources; only, inductors oppose the input voltage in the beginning while capacitors - in the end of the transitional process. Circuit-fantasist (talk) 17:36, 4 June 2009 (UTC)





Varying (dynamic) internal resistance
Do you have another idea? What else can change in this humble circuit when the voltage drop VL rises and, as a result, the current lowers? Atanas: Let's vary the resistance R... Great idea! The resistance R is another attribute that can change to keep the current constant.

Actually, we have to make the current source change its internal resistance so that to keep a constant sum R + RL; as a result, the current I = VREF/(R + RL) will stay constant. Again, till when will this magic continue? If RL reises, R will lower until it becomes finally zero. After that point, our wonderful "resistance-dynamic" current source becomes an ordinary one.

Since the resistor R has the property to keep up a constant current, they name it current-stable resistor. Agin, is there a negative feedback here? No, it isn't. The varying resistor changes blindly its resistance; it does not "interested" in the magnitude of the current.

Since the resistor R has the property to keep up a constant current, they name it current-stable resistor. Let's realize this idea as it is simpler and more obvious than the idea of a dynamic voltage source considered above.



A transistor acting as a current-stable resistor


What electronic component has such a current-stable behavior? A student: The operational amplifier... No, no, it has to be a simpler element. Remember in this lab cicle we consider diode and transistor circuits; so, it has to be a diode or a transistor. A hint: the one has a behavior of a voltage source (more precisely, of a voltage-stable element); the other has a behavior of a current source (current-stable element). Well, which of the two behaves as a current-stable element? Of course, the transistor...

Well, let's then see why the transistor has such a behavior. You have already studied a subject about semiconductor elements; so, you have to know how it behaves. Do you have any reminiscence of it? What can a transistor do? A student: ...to amplify... No, it might sound paradoxical but the transistor does not amplify; it attenuates!?! The only thing that a transistor can do is to change its present resistance thus dissipating a power and, as a result, regulating the current through or the voltage across the load.

Here is an n-p-n transistor connected in series with a load RL and a power supply VCC. It is a three-terminal element (a triode) while a diode is a two-terminal element. A triode is too complicated for our reasoning:) so, we break down it into two two-terminal parts: an input base-emitter part and an output collector-emitter part. Each of them has an IV characteristic; now we are interested in the output characteristic of the collector-emitter part. Why they name it "IV characteristic"? Because we can do two things with it: to apply a voltage across it and to measure the current through it or to pass a current through it and to measure the voltage drop across it. Also, they name this curve output characteristic because it represents the output transistor's part. How does it look? Let's present the circuit operation graphically.

For this purpose, we have to superimpose two characteristics - one of the source and one of the load. First, let's draw the IV characteristic of the voltage source acting here as a power supply. What is it? It is a vertical line that is shifted with the magnitude of the voltage source VCC. Then, we have to show the presence of the load by inclining the line to the left according to the load resistance RL. The resulting curve represents an IV characteristic of a real voltage source having a voltage VCC and an internal resistance RL (they name it "load line"). After, we have to plot the transistor's output characteristic on the same coordinate system. The crossing point is known as an operating point; it represents the present magnitudes of the voltage and the current.







When we vary the load, the load line rotates and the operating point moves horizontally along the transistor's output characteristic. But why? How does it do this magic? What does the transistor actually do to achieve such an effect? Can someone explain? Let's try.

When we lower the load resistance the load line stands up; the transistor does the opposite - it increases its present collector-emitter resistance RCE (or just RT) so that to keep up a constant total circuit resistance. We may show RT by another line that begins from the coordinate system origin (the trick is to think of the present resistance RT as an ohmic one). So, this line will rotate around the origin and incline as well. As a result, the crossing point moves along a new almost horizontal line - the output transistor characteristic. This line is imaginary, unreal, artificial but we can see only this line; it represents the behavior of a current-stable resistor.



The text below is written by a student from this group:

The BJT transistor as a three-legged creater (??? Circuit-fantasist (talk) 12:28, 27 May 2008 (UTC)) looks much like a light switch (2 legs for IN and OUT, and one for the ON/OFF button), and basicly that's what it is - a switch. But this one here is like those fancy light switches with a rotating knob or even the more fancy ones - with a metal "touch-plate", which can be used to set a different light intensity in a continuous range by rotating the knob, or keeping your finger on the metal plate for a short time. But unlike these mechanically controlled switches, the transistor is an electrically controled one - on this scheme the output collector-emitter(C-E) current is controlled by the input base current or the base voltage, whichever suits us more.

So here comes the simple idea - if we apply a constant base voltage, we will set a constant C-E current. When the C-E voltage increases (e.g. due to decrease in voltage on the load resistor Rl) the C-E resistance will increase too (remember we set the C-E current to constant using a constant base voltage). So what do we get - a dynamic varying resistor which keeps a constant current flow. In short - a current-stable resistor. Which is exactly what we need in order to make a constant current source.

This behaviour is shown on the IV diagrams below. They can be explained as a mechanism with the I line acting as a "fixed rail", on which the crossing of the two "pivots" - Rl and Rt lines - point A slides. As we decrease the load (Rl that is) point A slides down the Rl line, up the Rt line, and to the right on the I line. This means that the transistor resistance increases, so the whole resistance of the circuit remains the same, and therefore the current flow remains the same. As we do the opposite - guess what - exactly the opposite things happen ;)

--V.konushliev (talk) 08:42, 19 May 2008 (UTC)



Building a simplest transistor current source on the whiteboard
''Please, read carefully this story. IMO, this is the fanciest and the most incredible story about a transistor circuit that I have ever seen on the web! What do you think about this assertion?'' Circuit-fantasist (talk) 05:55, 20 May 2008 (UTC)

Looks scary at first sight :) But none of the elements involved bites, and everyone has it's role.


 * Maybe, it looks scary because we have placed here the final, complete and perfect circuit solution? Maybe, we would build it step-by-step snapping every drawing on the whiteboard to show the circuit evolution? Maybe... Circuit-fantasist (talk) 05:37, 20 May 2008 (UTC)



In the middle, surrounded by a blue line is the "boss" of the circuit - the varying load resistor Rl. It's he who leads the concert, and the whole curcuit is built for him - to satisfy his needs and desires for constant current. The nameless resistor just beneath the boss is his "bodyguard" - just in case someone curious moves the slider of the "boss resistor" to its very end and try to short-curcuit the power source through the transistor.

Everything outside the blue line is the BJT current source itself - a whole team of elements working for the "boss". There's the "hero" of the curciut - the workhorse - the mighty bipolar junction transistor (BJT), who does all the dirty job (donkey work:) of keeping the constant current. Connected to its base is his "manager" (modern heros need to have managers), the varying resistor (potentiometer) P, who is basicly a varying voltage divider, and by the means of base voltage Vb tells the transistor what to do. In this case - using a constant base voltage the manager tells him to keep the current flow at a constant rate. The "manager" also has a "bodyguard" - Rb keeping the power source safe from short-curcuit's caused by someone playing with the potentiometer P.

Then there are our watchfull eyes - the voltmeter V2, used to keep an eye on the voltage drop on the load resistor (the boss), another voltmeter V1, watching over the voltage drop on the transistor's collector-emitter junction, and an ammeter, which gives us a sneak-peak on the current flow.

Last, but not least - the power source VCC, who brings the whole curcuit to life :) -- V.konushliev (talk) 13:22, 19 May 2008 (UTC) 

Mounting the circuit on a PCB
It was easy so far but now we have to materialize the circuit on a prototyping printed circuit board. Can someone solder? If yes, please! Here are soldering iron and solder. How do we solder? Look at this links: 1, 2. You may use also various tools...

The transistor is n-p-n type packed in metallic body. The collector is connected to the body; so, you might measure the collector voltage by touching the body. You may supply the circuit by the computer or by an external power supply. What do you prefer? If you use the PC supply, there are two fixed voltages (+12 V and -12 V). Maybe, it is better to use the two varying lab supplies because we might provoke our transistor current source by varying the supply voltage. If you prefer, you might use a single power supply...







For our educational purposes it is preferable to arrange the elements in a similar way to the circuit diagram drawn on the whiteboard. For example, place the negative rail at the bottom and the positive one at the top; drive the transistor from the left and take the output from the right, etc. Now, let's make some photos..





What do we connect as a load? Let's it to be something varying - this 10 kΩ rotating "potentiometer" will suit us. Connect it as a rheostat (a two-terminal varying resistor). By the way, why the slider is connected to the one end? What is the idea of this connection? Also, as it is very likely someone to burn out it, connect a protecting resistor in series with the potentiometer.

How to set the input voltage? How to make a lower voltage by a higher one? Of course, by another potentiometer supplied by the same power supply (here it is connected as a real potentiometer).



Investigating the circuit...
In the beginning, we may draw and superimpose the current loops over the circuit diagram. Then, in order to measure the currents and voltages, we have to connect ammeters and voltmeters.

I have an idea for a syncrinised presentation - I will "move" the slider of the input potentiometer on the whiteboard; at the same time, you will move the real slider on the PCB. In the beginning, the input voltage is 0 V; so, no any base current flows. The transistor is "closed". Increase carefully the input voltage and look at the input transistor characteristic; imagine we walk along it:) At some moment, a current will begin flowing through the base-emitter junction and β times bigger collector current will flows through the load.

What happens, if you go too far? The transistor will saturate. So, we have to stay in the middle where the transistor operates in the active region.

What elements constitute our current source? These are all the elements excluding the very load, of course... Well, now we have to see, if it is a constant current source. How? We have to disturb, "provoke" it and to observe its reaction to our "intervention". But let's afford an opportunity to Vladimir to tell what has happened in the laboratory (he can do it very well). Circuit-fantasist (talk) 19:41, 27 May 2008 (UTC)

The text below is written by a student from this group:

Everything here is based on the simple law of Ohm $$R=V/I$$



...by varying the load resistance RL...
... we observe a change in V1 and V2, but no change in the ammeter. We're happy: the current flow is constant :) How did that happen?

When we increase the load resistance Rl so does its voltage drop, which can be seen on voltmeter V2 (logically simple - more resistance generates more pressure). As V2 shows an increment, V1 respectively shows a decrement (their sum represents the voltage of the power sourse Vcc). Since we made the transistor act as a current-stable resistor, it "senses" the decrement of the C-E voltage drop and obediently reduces its resistance, so the sum of resistances in the circuit branch remains the same. Therefore the current flow remains the same too.

When we decrease the load resistance, exactly the opposite happens (V2 shows a decrement, V1 shows an increment, the transistor rises its resistance, the sum of resistance remains the same, and so does the current flow).

...by varying the supply voltage VCC
...we observe the same pleasing facts: no change in the ammeter, although V1 and V2 show changes. We're happy again: the current flow is still constant :)

This time both the voltmeters V1 and V2 showed increments of voltage drops, with the increment of the supply voltage Vcc. The transistor again "senses" the change in voltage drop and rises its resistance. But if the transistor reacts the same way as in the case above, how does the current remain the same, if there is no change in Rl? The difference comes with the fact, that the "manager" of the transistor tells him to rise it's resistance even more, by applying a larger base-emitter voltage. Of course this is due to the fact, that the "manager", as a simple voltage divider, divides a larger voltage (Vcc), and therefore gives a larger result (Vbe). At the end, the sum resistance of the curcuit branch rises enough to compensate the increment in supply voltage Vcc and keep the current flow the same.

And again, with the opposite input (decrement of Vcc) the opposite phenomena is observed, but the result stays the same - constant current flow. A job well done by the BJT and it's "team" :) --V.konushliev (talk) 15:11, 19 May 2008 (UTC) 


 * Vladimir, now I am realizing that I have misled you to some extent in this experiment. The circuit is not as correct as it seems. It would be more consistent, if we have made all the possible to make the input "manager" keep a constant voltage. Here we rely on the constant power supply; so, we shouldn't change its voltage. We should supply the potentiometer by another voltage source or replace the input voltage-divider circuit by a voltage-stabilizing one (e.g., by connecting a diode into the lower leg). This would be a prelude to the famous current mirror circuit (see group 66b and group 67b). Please, edit your explanation according to these circumstances.


 * Vladimir, I have a suggestion to. Would you contribute the Circuit idea story about the simplest transistor constant current source (and not only it)? Your writing style is unique; so, you might tell your lively story in a separate page as a different viewpoint at this famous circuit. Circuit-fantasist (talk) 06:46, 20 May 2008 (UTC)

Resources
Reinventing constant current source reveals the philosophy behind constant current source Understanding the simple BJT current source

Simplest?
I was surprised to see a transistor with 4 resistors, when the simplest transistorised current source is a tr with one resistor from Vcc to base. Maybe a rename is in order? 82.31.66.207 (discuss) 16:34, 24 July 2013 (UTC)
 * Yes, you are right.--Circuit-fantasist (discuss • contribs) 19:27, 10 August 2013 (UTC)