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Several times here on the EE Stack Exchange, I've seen people say not to connect a load between the emitter of an NPN transistor and ground, but instead connect it between the power source and the collector. In that case the transistor is a current sink, not a current source.

Why shouldn't you draw current from the emitter, like this below?


simulate this circuit – Schematic created using CircuitLab

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If you are saying that you've seen a certain pattern of opinion here on EE.SE, then please provide the links. –  Nick Alexeev Aug 19 '14 at 0:03
What will you connect to the base in your proposed circuit? –  The Photon Aug 19 '14 at 0:19

3 Answers 3

up vote 2 down vote accepted

Your question says "current source" but I think you're really asking about switching voltage to a load.

With a emitter-follower you're stuck with around 600mV of voltage drop (plus whatever the driving circuit drops when supplying the base current) unless you have a higher voltage source available. So if you are driving an LED and have a 3.3V power supply and a GPIO that has 2.9V on it when driving base current, you're going to get about 2.3V to your LED, which will be rather limiting, plus the transistor dissipates power because of the current and voltage lost.

If you use a saturated transistor (PNP or NPN) then your voltage drop across the transistor might go down to a few tens of millivolts, and almost the full supply voltage is available for the load, and the transistor stays cool.

On the other hand, there are circumstances favorable to the use of an emitter follower, for example if you wish to draw current from a higher voltage (perhaps unregulated) supply and provide a relatively constant voltage (when 'on') to a load from that.

Most of the simple questions here don't require that kind of circuit, but I've certainly used it to benefit in commercial products where saving a bit of money can make a big difference.

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A current source should have high impedance.

Looking in to the emitter of a BJT in forward active mode you see a low impedance.

This means the base acts more like a constant voltage source (with a value of Vbase - 0.7 V) than like a constant current source.

This circuit is commonly used when a voltage source is wanted. It is called an emitter follower. The controlling signal is applied to the base, not the collector.

Another similar configuration is the common base amplfier. For common base, the controlling signal is applied to the emitter, and an output current is produced at the collector. This acts as a near-unity gain current buffer, much like the emitter follower is a unity-gain voltage buffer.

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So to put it another way, an emitter follower lets you regulate the output voltage? And wiring the load between the supply and the collector lets you instead regulate current? –  Duncan C Aug 19 '14 at 0:16

The main reason is because bjt current through the the emitter is proportional to the Base-emitter voltage difference in an exponential relation.

With the base set at a certain voltage, and ground not moving because it's ground you will always get the same amount of current flow through the BJT regardless of what the load resistance is.

In your case though, with a set base voltage, your emitter voltage is no longer pinned. This results in exponentially more current if your load resistance drops. There's no current control at all at that point because you have virtually no control over the emitter voltage.

From Ebers-Moll model on wikipedia:


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But if you want your transistor to serve as a low-impedance switch (like to light a lightbulb) then wouldn't an emitter-follower be the way to go? Then the resistance of the load would be what limits current. –  Duncan C Aug 19 '14 at 0:22
@DuncanC Generally no. If you want to switch something you want gain. Generally you want a small voltage change to result in a high voltage change elsewhere. In emitter follower configuration, you have to go from 0 to basically full supply voltage to perform a switch because your emitter follows your base voltage. In common emitter topology, only a small voltage change of 0 to 0.7 volts results in a switching action on the output. –  horta Aug 19 '14 at 14:01

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