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Q2: How does a Constant Current Sink Work ?

Q2: How does a Constant Current Sink Work ?

2. How does a Constant Current Sink Work ?


CCS – A constant current, no matter “what” …

Actually, what is depicted on the drawing looks more like a sink, a drain to me, like as sinking, draining, or “flushing to Ground”, rather than “sourcing from the High Voltage supply”. For the sake of a brief discussion, assume R1 = 70R. R2=200K. So, to solve this riddle, you only need one piece of information: the voltage drop on a conducting base-emitter diode is ~ 0,7Volts (more or less).

Let us assume, that both of the transistors are conducting slightly. The Collector of T2 is the stable “current drain”. R2 is a huge resistance, that supplies a polarization voltage for the base of T2, from a high voltage source.

a). Have a close look at R1. it is connected to a conducting base-emitter diode of T1. So there is a voltage of ~0,7V. It just so happens, that this voltage is ALSO on R1.

b). We now know, that the current flowing through R1 is 0,7V / 70 R = 10mA.

c). The current “for” R1 is originating from T2, which is conducting. The majority of the 10mA “originates” from the collector of T2, since the base current of this transistor is fairly small.

d). The collector of T2 is the the point that we use as a constant current sink (drain). But WHY does it sink a constant current ? I will prove to you that it is constant, because … it is impossible for this current to change.

e). Assume: for some unknown reason, the T2 collector current increases. This would imply that the R1 current increases. This would imply that the voltage on R1 increases. BUT …. this is not possible, as even a tiny voltage increase on the base of T1 shall cause it to massively conduct. This would imply that the voltage of it’s collector is massively pulled DOWN, due to the low resistance path via T1 to ground, in comparison to a huge value of R2. But … this voltage will be pulled down only a little tiny bit. Why? Because the collector of T1 is connected to the base of T2. Even a small decrease of the voltage on the base of T2 shall result i T2 conducting LESS. The collector current of T2 returns to the point of equilibrium, i.e. to our 10mA. Then, T1 also returns to it’s state of equilibrium. The huge gain of both T1 and T2 is used solely for this purpose, of “returning to equilibrium”.

f). Contrary to the above, lets assume: for some unknown reason, the T2 collector current decreases. This would imply that the R1 current decreases. This would imply that the voltage on R1 decreases. BUT …. this is not possible, as even a tiny voltage decrease on the base of T1 shall cause it to decrease it’s conduction. This would imply that the voltage on it’s collector rockets sky high, due to the high voltage supplied via R2. But this will not happen. Why ? Since the collector of T1 is connected to the base of T2, and a minimal rise of the voltage on the base of T2 shall imply that it conducts MORE. The base emitter diode of T2 will conduct stronger, and T2 will open up, allowing a greater flow of current from the collector of T2 through R1. The current flowing through R1 returns to the point of equilibrium, i.e. to our 10mA. This also implies that T1 also returns to it’s point of equilibrium. The huge gain of both T1 and T2 is used solely for this purpose, of “returning to equilibrium”.

The two transistors are connected, as if each one is eating the others “tail”. This is a very strong feedback loop, established between two transistors, where each has a fairly high gain. The overall gain of the open loop would be the product of the two gains. For example: if beta_T1 = 200 and beta_T2 = 200, then the open loop gain would be, my best guess, in the order of 40 000. But we close the loop and use all that horrendous gain for only one purpose: to control the stability of the voltage over R1.

But since this voltage is highly stable, and the resistor is (hopefully) stable, then so is the current, and specifically the one flowing through R1 and through the collector of T2.

The current flowing through the collector T2 is STABLE and constant. This circuit will FORCE all the other components above it to change their parameters in such a manner, so that they “provide” the needed 10mA of drain current. If it is a simple resistor, the forced change will imply a change of voltage upon that resistor, so as to MEET the demand for the 10 mA of current. Not more. Not less. Exactly what is needed, i.e. 10mA = 0,7V / 70 ohms. Use the R1 resistor to “program” the current of your current drain.

For examples sake: Assume a voltage supply of 30 Volts. Assume a resistor of a value of 2kohm, connected to that supply from its top, whereas its bottom end is connected to the CCS (collector of T2).

Since the current MUST be 10mA, that means a voltage drop on this resistor is equal 20V. The collector of the transistor T2 is elevated 30-20 =10V above ground.

Now, a different example: Assume all as above, but a different resistor, say 0,5kohm. Since the current MUST be 10mA, that means a voltage drop on this resistor is equal 5V. The collector of the transistor T2 is elevated 30-5 =25V above ground.

Obviously, If the resistor is ZERO ohm, the current will STILL be 10mA. But the potential of the collector of T2 will be equal to the maximum 30V (power supply voltage).

But what happens if the resistor is 10Kohms ? The circuit does not function properly. Why ? 10k multiplied by 10mA would imply a voltage drop of 100V. This will obviously not happen, as our power supply delivers only 30v. With such a high value of resistance – the circuit falls out of regulation

One more thing needs to be explained. Why do we use a “huge” value for R2 and connect it to a fairly high voltage ? The high value of R2 limits the current flowing into the base of T2 to one comparable to a trickle. If Beta_T2 = 100, that would be around 10mA/200 = 0,05 mA. Some of the current flowing through R2 gets routed via collector to emitter to ground, via the T1 transistor, if the voltage rises on R1. You could say that T1 is an over-voltage / over-current shunt, that limits the voltage on R2. T1 constitutes, together with R2, a gain stage of very high gain. The slightest variation ( = super super super tiny) of voltage/current on R1 immediately translates to a 200x bigger, but “still” tiny corrective variation on the collector of T1 (= base of T2). A huge value of R2 allows to “maximize” this effect. Then, finally, the “tiny” change of voltage on the base of T2 has a x200 leverage in terms of “correcting” the transistor’s behavior, so as to preserve the “preset” current flowing through R2.

Since 200 x 200 is 40000, I presume that the stability of such a current source should be comparable to the “stability” of the resistance of R1, “bettered” by a factor of 40000. Unfortunately, in real life – not so optimistic. There are things such as thermal drift, and this applies both to the transistors, as to the resistors. But none-the-less, You will still be very happy with the final performance of this cute circuit.

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