Electronic – Understanding constant current circuit

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Constant Current Circuit

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 that's total nonsense.

Lastly, not sure why the diodes either.

Best Answer

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:

  1. When the BJT's base-emitter diode is forward-biased, the collector current is the same as the emitter current.
  2. 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.
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