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I am trying to get my head around this (current sink) circuit. I have seen this circuit with two NPN transistors and connected base - but to swap around one of them seems a little strange to me.

Does this arrangement have a special name? Can someone please explain it to me?

strange current sink

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    \$\begingroup\$ I think the circuit is in error. Where did you get it? Link? I suppose it could be operating Q2 in a reverse-avalanche mode, but that is unusual. \$\endgroup\$
    – AnalogKid
    Commented Jan 6 at 21:26
  • \$\begingroup\$ github.com/riktw/LabPowerSupplyE361xA/blob/main/Kicad/… \$\endgroup\$
    – huababua
    Commented Jan 6 at 21:35
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    \$\begingroup\$ I just think the linked schematic is incorrectly drawn, It will work of course but it just causes confusion when it needn't have if drawn the more conventional way \$\endgroup\$
    – Andy aka
    Commented Jan 6 at 21:55
  • \$\begingroup\$ The schematic has several problems. Among other things, D12 is backwards so the relays will not pull in and Q6 will go boom. \$\endgroup\$
    – AnalogKid
    Commented Jan 6 at 23:00
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    \$\begingroup\$ Q2 is flipped from the normal way, so it is operating in reverse, which is not conventional and has no advantage in this case that I can see. The B-E breakdown voltage is not exceeded and the reverse beta (typically about 8 as opposed to about 300 when used normally) is still more than sufficient so the circuit will function. \$\endgroup\$ Commented Jan 6 at 23:12

3 Answers 3

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It looks like an erroneous version of a standard constant current circuit.

enter image description here

Figure 1. Image source (mine) from LEDnique.

How it works:

  • On power-up Q1 and Q2 are off. There is no collector current so L1 is off.
  • If the digital control input on the left is brought high (5 V) Q1 will turn on. Current will flow through L1, Q1 and R2.
  • As the voltage drop across R2 increases to about 0.6 V Q2 will start to turn on and shunt some of the base current away from Q1.
  • The result is that the circuit will settle at whatever Q1 emitter current will drop 0.6 V across R2.
  • When the input signal drops to zero Q1 and L1 turn off.
  • If an unswitched version is required then just connect R1 to Vbb.
  • Note that the usual series resistor is not present with the LED. Q1 acts as a variable resistor in this case adjusting to maintain the required current.
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  • \$\begingroup\$ There is usually a lot of opposition to claims like "Q1 acts as a variable resistor" (ie vs. comparing the transistor to a variable resistor). \$\endgroup\$ Commented Jan 8 at 12:39
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    \$\begingroup\$ @Circuitfantasist, let them oppose. I could call it a variable conductance which would tie in better with "semiconductor". ""All models are false. Some are useful!" \$\endgroup\$
    – Transistor
    Commented Jan 8 at 12:54
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    \$\begingroup\$ Note that the word "transistor" comes from "transfer-resistor", so I think the variable resistor concept was baked in from the start. \$\endgroup\$
    – AnalogKid
    Commented Jan 9 at 1:14
  • \$\begingroup\$ About the circuit - seems like a lot of work to go through for less than 1 mA. Given where this circuit is in the overall schematic, I'm thinking someone got the R1 value incorrect. \$\endgroup\$
    – AnalogKid
    Commented Jan 9 at 1:15
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Looking for the idea

I first thought seriously about this famous circuit solution of a constant current source three years ago when I laid out my insights in one of my questions and answers. Now, looking again at this circuit configuration with two transistors facing each other, I struggled to find some even simpler and more logical explanation of this interlocking connection. Thus I came to the conclusion that first I need to unfold this "ball" of links and elements to distinguish some familiar functional blocks and then connect them in cascade one after the other as we usually draw schematics.

Showing the circuit evolution

Drawn in such a compact form, the circuit diagram takes up little space... but the idea remains hidden. So I realized that the best way to explain it was to show the circuit evolution step by step through CircuitLab experiments.

I have used an ammeter as a current load, changing its internal resistance manually (in the parameters window) or through the DC Sweep Simulation. Initially, its resistance is zero, hence the designation RL=0.

For convenience, in these conceptual circuits, I have tried to use "decimal" (multiples of 10) values of electrical quantities whenever possible.

Resistor current source

In principle, setting a constant current is done by connecting a resistance element in series with the load. In the simplest case, this is an ordinary constant (ohmic) resistor R. As you can see, when the load resistance is zero, the current exactly follows Ohm's law - I = Vcc/R = 10 [V]/10 [kΩ] = 1 [mA].

schematic

simulate this circuit – Schematic created using CircuitLab

But when the load resistance starts to vary (0 ÷ 10 kΩ), we see that the current does not remain constant. The greater the resistance R, the more constant the current will be, but this also requires a higher supply voltage. The problem is that the resistor is "static" (with constant resistance).

STEP 1

Transistor current source

We can solve this problem with a clever trick - changing the resistance R in the opposite direction to RL so that their sum remains constant. As a result, the current will stay constant, I = Vcc/(R + RL) = const. The role of such a dynamic resistor can be played by a transistor. Its "resistance" (Vce/Ic) complements the load resistance RL so that their sum is approximately equal to 10 kΩ.

schematic

simulate this circuit

As a result, its IV curve is close to the horizontal curve of a good current source.

STEP 1

Note that ultimately the transistor is controlled by the (differential) base-emitter voltage which can be a difference of two (single-ended) voltages (Vbe = Vb - Ve). So if we raise both voltages by 1 V...

schematic

simulate this circuit

... nothing will change.

STEP 2

Negative-feedback current source

So far, we have exploited the transistor's property to behave as a current-stabilizing nonlinear element at a constant base-emitter voltage. It did this by changing its "resistance". But cannot we simultaneously make it change its input voltage Vbe (by changing Ve)?

They thought of doing it a century ago and called it emitter degeneration. For this purpose, it is enough to replace the constant voltage source Ve with a "proportional source" with voltage Ve = I.Re, i.e. resistor Re (a voltmeter with 1 kΩ internal resistance). Now when RL increases, the current will try to decrease. Ve will decrease; that will increase Vbe and the current will increase (restore its previous magnitude).

schematic

simulate this circuit

The IV curve is apparantly horizontal. At RL = 9 kΩ, the transistor exhausts its supply of "resistance" (saturates), and the current decreases.

STEP 3

2-transistor current source

Finally we arrived at our circuit idea. Above, the transistor Q1 was "self-controlling" by changing its emitter voltage. Now we give it a chance to control its base voltage at the same time. For this purpose, we amplify Ve by another transistor Q2 and apply it to the Q1 base. Something interesting happens here - Q1 controls both Vе (direct) and Vb (amplified) at the same time. For example, if Ve tries to decrease, Vb will increase many times over and overall the input voltage Vbe will increase.

schematic

simulate this circuit

As a result of the added amplification, the IV curve becomes absolutely horizontal.

STEP 4

Conventional circuit diagram

Well, let's finally flip Q1 against Q2 to get the conventional compact schematic.

schematic

simulate this circuit

Voltage source

In the circuit above with a current-type negative feedback, we had to invert Ve before applying it to the base of the transistor because the stage is not-inverting. For comparison, in a transistor diode circuit there is no need for such inverting because the amplifier stage itself is inverting.

schematic

simulate this circuit

As you can see, the current-stabilizing transistor is made to act as a voltage stabilizer with an almost vertical IV curve (here it is horizontal since the voltage is plotted along the ordinate).

STEP 5

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The schematic is likely in error, but the idea is workable, with slight changes.

A reverse-connected B-E junction acts like a 5-7V "Zener" diode. So we could use it to establish a reference voltage on the base.

schematic

simulate this circuit – Schematic created using CircuitLab

This constant current source would draw roughly about 5mA through the load.

Would I put it in a product? Probably not without further qualification, including aging tests (expensive!), and assuring that the product is traceable at least back to the fab.

Will it work on the breadboard? Sure.

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  • \$\begingroup\$ Interesting.... You might want connect an ideal diode with VF = 5÷7 V in parallel to simulate the breakdown voltage. \$\endgroup\$ Commented Jan 9 at 7:25
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    \$\begingroup\$ @Circuitfantasist Indeed, but unfortunately in CircuitLab such "helpers" can't be hidden and they pollute the core idea :/ \$\endgroup\$ Commented Jan 9 at 17:07

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