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I want to learn how to design a simple current source using a BJT.

I found this picture on another site (https://www.wellpcb.com/blog/pcb-projects/constant-current-source/?utm_source=chatgpt.com), but it doesn't seem to say how I'm supposed to choose the component values, and I haven't had much luck in finding other sources that do.

enter image description here

I assume R1 is chosen such that the Zener is in its breakdown region, but how do we choose Re?

What is Re's role here?

Does this circuit work for a larger range of input voltages?

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  • \$\begingroup\$ If you use a classic red LED (Vf ~=1.4 V) for the ZD1 position, you get a light that goes off when the load is disconnected (provided the voltage drop across Re is less than about 0.7 volts when the bias current through R1 flows through it. ) Also the temperature coefficient of a red LED is about the same as the Vbe temperature coefficient - using a high enough gain transistor makes for a reasonably stable with temperature current source. You can also find a specific zener diode voltage which has the same temperature characteristic as the Vbe, that also works. \$\endgroup\$ Commented 10 hours ago
  • \$\begingroup\$ cosmin, You may also want to reference this answer as an alternative approach. \$\endgroup\$ Commented 7 hours ago

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Transistors have approximately constant Vbe voltage, so the resistor Re defines what the constant current is that you want at the load, based on Vb that is defined by Zener.

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The trick to understanding this design is to understand the behaviour of an emitter-follower:

schematic

simulate this circuit – Schematic created using CircuitLab

The "input" here is the potential \$V_B\$, produced by source V1, and the "output" is emitter potential \$V_E\$. As I sweep input V1 from zero to +12V (blue), watch what happens to output \$V_E\$ (orange):

enter image description here

\$V_E\$ is always 0.7V lower than \$V_B\$, hence the name "emitter-follower":

$$ V_E = V_B - 0.7V $$

In other words, we can set whatever potential we like at emitter E, by applying a 0.7V higher potential at base B. We do that with resistor R1 and diode ZD1. Let's say that ZD1 is rated for 5.6V:

schematic

simulate this circuit

By doing this we know that:

$$ V_E = V_B - 0.7 = +4.9{\rm V} $$

By Ohm's law we can find emitter current through Re:

$$ I_E = \frac{V_{RE}}{R_E} = \frac{V_E - 0{\rm V}}{R_E} = \frac{4.9{\rm V}}{4.9{\rm k\Omega}} = 1{\rm mA} $$

Since collector current and emitter current are very similar (base current is assumed to be small in comparison), that means that most of this current must be coming from the collector, through whatever's connected up there; an ammeter in this case, showing this to be true.

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The circuit you’re looking at is a classic BJT-based constant current source.

R1’s job is to make sure the Zener diode is in its breakdown region, providing a stable reference voltage. When choosing R1, you need to ensure that enough current flows through the Zener so it stays in its stable operating range. You also want to consider the small base current of the BJT, though it’s usually minor compared to the Zener current. Essentially, R1 sets up a reliable reference voltage for the transistor.

Re determines the current through the transistor. The emitter current is almost equal to the collector current, so Re is the component that actually sets your constant current. The voltage across Re minus the base-emitter voltage of the BJT defines this current. One of the advantages of this setup is that the current becomes relatively independent of the load: as long as the voltage across the emitter stays roughly constant, the current will remain stable.

This circuit works as long as your supply voltage is high enough for the Zener to regulate and for the BJT to stay out of saturation. If the supply voltage drops too low, the Zener may not maintain its voltage, or the transistor may enter saturation, and the current will no longer be constant.

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\$R_1\$ needs to be sized so that \$I_{\small{Z}}\approx \frac{V_{\small{SRC}}-V_{\small{Z}}}{R_1}\$, where that current is specified in the datasheet for the zener. The load current will be approximately \$I_{\small{LOAD}}\approx\frac{V_{\small{Z}}-700\:\text{mV}}{R_{\small{E}}}\$. You will need to accept the fact that there will only be a compliance voltage for the load of \$\le V_{\small{SRC}}-V_{\small{Z}}\$. And \$I_{\small{LOAD}} \le 10\cdot I_{\small{Z}}\$.

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  • \$\begingroup\$ Your formulae render extremely badly for me. I could not correctly read your initial "VSRC-Vz" - it LOOKS like VsrcVz with a ash over the RC. The same expression after (for the load of" is only slightly corrupted. I don't recall that being the case in your prior material. I don't know if you are doing something differently. || A quick look at several of your prior answers shows these effects to be largely absent. It COULD be my system, but it seems not. See image here \$\endgroup\$ Commented 8 hours ago
  • \$\begingroup\$ Please advise if it's my system :-( . \$\endgroup\$ Commented 8 hours ago
  • \$\begingroup\$ @RussellMcMahon Renders fine for me. I'm using FireFox on Windows 10. But I thought this system uses an outside service to auto-generate the images and I don't see why it would render something different for you, than me. Is there a way to see what's being sent over to the 3rd party system so we can compare that part of it? \$\endgroup\$ Commented 7 hours ago
  • \$\begingroup\$ @RussellMcMahon What does your system show as the Latex? Does that Latex look wrong to you? \$\endgroup\$ Commented 7 hours ago

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