constant-currentcurrentcurrent-sinkcurrent-source
I have designed a two-transistor current sink for my application. I need it to keep the current constant which flows through an IR-LED:
simulate this circuit – Schematic created using CircuitLab
The current should be constant at:
I = UBE1 / R1 = ~0.7 V / 120 Ohm = ~5.8 mA
Now the problem is that the current will not stay constant as desired when varying the voltage between 2.6 V and 3.3 V. It varies between 1.6 mA and 4 mA. What do I miss here?
This is a voltage controlled current sink that has a LED as its load but anything as a load would do: -
Like I say - ignore the LED and assume it's just a resistor up to some abitrary positive supply voltage that needn't be the same as the op-amp.
It works by ensuring the voltage across the emitter resistor is the same as the voltage on the potential divider on the +Vin of the op-amp - the op-amp controls the BJT to make sure that 1V is across the 10 ohm resistor thus 100mA will always flow through the LED/Load in the collector.
If you want a smaller current either make the potential divider produce a smaller voltage on the +Vin of the op-amp or increase the 10R emitter resistor. Here's a similar circuit that controls a current source referenced to the +V rail: -
This circuit uses FETs instead of a BJT just to demonstrate the possibilities. The op-amp and FET to the left control the current through a 1k resistor to +Vs. There's a cap across it to stabilize the voltage across the 1k to prevent FET leakage capacitance making it noisy.
The voltage across this 1k is therefore identical to Vin (use low Vos op-amps and it will be within a hundred uV. The 2nd part of the circuit take the +Vs referenced voltage and creates a constant current source for the load. In this circuit the current through the load is Vin/100.
Making a voltage controlled resistance is much much harder to achieve linearity but if that's the way to go good luck.
Building a constant current load with an opamp is a fairly straightforward way to do this. Take a look at the first diagram in this post, for instance:
An opamp adjusts its output to make the voltage on its - and + terminals identical. In this case, the negative input is the voltage across a current sense resistor (R1) and the positive output is the voltage across a potentiometer. Making the two voltages equal means that the current through R1 - and hence, your load, which should go where "V+" is on the schematic - equals the voltage on the + terminal divided by the resistance of R1.
In this case the output of the opamp is driving Q1, a FET, because the FET can sink far more current than the opamp.
To control the load digitally, replace R3 and R6 with the output of a DAC, or a suitably filtered PWM output. For a smaller load like 4mA, you'll want a much bigger resistor for R1; the value of R1 is a tradeoff between the accuracy you get from using larger set voltages, and the extra power dissipation and burden voltage across it.
Best Answer
You forgot to take into account the forward volt drop of the LED. It might be betweeen 0.8 volts and 3 volts depending on technology.
Thus, if your supply voltage is 2.6 volts, the "current control circuit" might only receive (maybe) 1.5 volts across it and, given that the two transistors might need at least 1.4 to 1.6 volts across their circuit to begin reasonable conduction, you are on the edge of it just starting to conduct.
If your supply voltage did rise higher I'm sure it would begin current limiting around the 5 or 6 mA mark. Here is a slightly improved version of your circuit: -
Note that Rb connects directly to the positive rail - it will help but only a little bit because to get conduction from both transistors you still need 1.4 volts to 1.6 volts from Q2's collector to ground (your IO port).
Pictures from here.