Understanding constant current circuit

Question

I've been learning about different constant current circuit designs and just recently stumbled upon this one. All I know is that the highlighted resistor is the load resistor and when varied, like a potentiometer, the current running through the load resistor stays the same. Also, adjusting the other two resistors will change the current through the load.

I think this makes some sense, for creating a path of less resistance either through the diodes or transistor will cause more current to flow in that direction. Yet, I don't understand why the current through the load is unaffected by the loads resistance. Is it because the current above the transistor cannot sense the loads resistance below? Therefore, once passing through the transistor you will have the same amount of current due to the fact that the current only has one path to ground? Maybe thats total nonsense.

Lastly, not sure why the diodes either.

Answer 1

Zach, this circuit is pretty easy to understand if you understand the BJT first. (You will understand diodes, if you understand the BJT, so that's a given.) Everyone struggles with these things at some point, so it's fine you don't apprehend this well right now. Take it one step at a time.

There is plenty of information on diodes here (and elsewhere.) You are awash in information about them. I won't try and replicate any of that. It's enough for this circuit that you accept two things about diodes:

1. A forward-biased diode has a fixed voltage across it. For regular silicon diodes, this value is \$700\:\text{mV}\$. (For LEDs, which are also diodes, it varies with the color and type and you have to look at the datasheet for that.)
2. Everything I just said in point #1 is actually wrong. But for these purposes, you don't need to worry about that fact.

Now to the BJT. It also has a diode from base to emitter. So the rules above apply. But we add the following about the BJT:

3. When the BJT's base-emitter diode is forward-biased, the collector current is the same as the emitter current.
4. What I just said in point #3 is also wrong. But #3 is close enough for these purposes to not matter.

So. Now we can describe the circuit.

• The \$20\:\text{k}\$ resistor forward biases the two diodes by providing a path for the current to go to ground.
• The total voltage across the two diodes is therefore \$1.4\:\text{V}\$, with the rest left over for the resistor. Therefore, the base voltage for the BJT is \$10\:\text{V}-1.4\:\text{V}=8.6\:\text{V}\$.
• Therefore also the resistor current is \$\frac{10\:\text{V}-1.4\:\text{V}}{20\:\text{k}\Omega}\approx 430\:\mu\text{A}\$.
• The BJT's emitter is forward biased and therefore the emitter will be \$700\:\text{mV}\$ above the base or \$8.6\:\text{V}+700\:\text{mV}\approx 9.3\:\text{V}\$.
• So the voltage across the \$500\:\Omega\$ resistor is \$10\:\text{V}-9.3\:\text{V}=700\:\text{mV}\$ (one diode drop -- which if you look closely you should see why this will be the case in this circuit.) From this, we can compute that the current in that resistor is \$\frac{700\:\text{mV}}{500\:\Omega}\approx 1.4\:\text{mA}\$.
• Since by rule #3 above, the emitter current and collector currents are the same, it follows that the collector current is also \$1.4\:\text{mA}\$.

The collector current is always the same as the emitter current (within a reasonable approximation.) So, it doesn't matter what resistor you place between the collector and ground.

Except,

• The above conclusion isn't right if the collector current we just worked out causes a voltage drop across the collector resistor that exceeds the base voltage. So this means that the resistor cannot be larger than \$R=\frac{8.6\:\text{V}}{1.4\:\text{mA}}\approx 6100 \:\Omega\$. So it has limits.

Answer 2

The voltage across Vbe controls the emitter current. Since one of the base diodes matches the Vbe drop , that leaves the other diode to control the emitter resistor voltage drop.

Then for large variations of the 10V supply or collector load resistance , the emitter current is now regulated by Ie= Vf/Re for Vf=0.7V It is not perfectly constant since the Vf/If is not perfectly matched between the diode and transistor Vbe

Answer 3

There are a couple ways you could look at this circuit, but try this:

Assume diodes have a constant 0.6V drop across them, as long as they are forward biased. So the two diodes in series have a total of a 1.2V drop.

Since the base-emitter junction and 500 Ω resistor are in parallel with these two diodes, the voltage across them must be the same. And since the base-emitter junction of a transistor is also a diode, the voltage across that must also be 0.6V as long as it's forward biased, which it is here. So, the voltage across the 500 Ω resistor must be 0.6V.

By ohm's law then, the current through the 500 Ω resistor must be 0.6 V / 500Ω = 1.2 mA.

All the current through that resistor must go in the transistor's emitter. And an equal current must go out somewhere, and there are two options:

1. out the base
2. out the collector

Because a transistor has gain, ideally a lot of gain, the collector current will always be much higher than the base current as long as the transistor is appropriately biased. Let's just assume it's appropriately biased for now, and consider the conditions in which that isn't true later.

If we assume the transistor has infinite gain, then all the current entering the emitter of the transistor must exit the collector, because the base current is zero. Since the only path for current out the collector is through the yellow load, the voltage across that resistor must be whatever value required to make the current in the load equal to the current through the 500 Ω resistor.

Having grasped the basic operation of the circuit, consider:

The simulation shows 1.33 mA through the 500 Ω resistor, which is slightly different from the 1.2 mA calculated above. Why is this different? What's the voltage across the diodes in the simulation?

What happens if the load resistor is extremely large, say 100 MΩ? How much voltage would need to be across it to get the desired current, and can the circuit still work under those conditions?

Real transistors don't have infinite gain. How does that affect the circuit?

What can affect the accuracy of this circuit? Temperature? Device variation? How and why?