Control SystemsEnter Your Electronics & Design Project for a chance to win a \$200 shopping cart!Submit an EntrySubmit an Entry  Back to homepage Project14 Home Monthly Themes Monthly Theme Poll

This post is for the Project14 Control Systems theme. Probably the simplest possible entry: a Constant Current Driver with one (1) active element.

Control systems can be complex. With a number of analogue or digital components to keep a circuit "under control".

This one uses a single transistor. And it uses that transistor's inherent characteristics to provide a tight, closed control loop.

I'll do a theoretical analysis, and then check the reality on a breadboard.

If you're new to bipolar transistors, this post may help you "to think like a transistor".

This is one of the circuits from The Art of Electronics. In the 3rd edition, it's circuit 2.23 C.

If you are interested in another one of their practical examples: The Blinker in a Side-View Mirror - What's Inside?.

# The Circuit

This is a Current Source. A circuit that sources current into a load.

In essence, it's a common-emitter design: the emitter is common to both input and output.

image source: photo of my paper copy of The Art of Electronics, 3rd edition

What we want to achieve is that we can control that sourced current precisely.

Regardless of the resistance of the load, the circuit should push the "set current' through it.

A popular use for this circuit is to drive LEDs. They are components that are current-controlled, and this circuit is one option to drive them.

When I say always, precisely, exactly, equal, regardless, =, ... this is within a range. Good enough in this context, so that the details can be ignored.

That's part of engineering. Use what's needed, ignore what can be neglected.

I will touch some of the limits here, but this will not become a first-principles post.

#### Expected behaviour:

• Set the desired current through a load (LED in this circuit) .
• Define that current using Ohm's law: the current through the emitter resistor will be exactly the current we want to have running through the load..
• For that, the voltage over that resistor has to be known and constant.

#### Some things that will help us:

• in this common-emitter design, the emitter voltage is always one diode drop higher than the base voltage.
• and in this common-emitter design, the current through the collector is exactly the current through the emitter.

If you understand these two behaviours, and you understand that if we have a fixed voltage over a resistor, we also have a constant current through it, we are virtually done.

#### Set the Constant Emitter Voltage pt. 1

Our "current-set" resistor is in the emitter circuit.

So if we want to have a constant current through this resistor (and hence through the emitter), we need to have the emitter voltage at a known fixed point (Ohm's law).

The way to do that here, is by setting the base voltage.

In a PNP transistor that's conducting (and it should conduct - we want to run a current through it), the base voltage is 0.6 V below the emitter voltage. The voltage over the diode that is the basis-emitter junction.

If we bring the base to a known voltage, we know that the emiter will be 0.6 V higher. Also a known voltage.

#### Bias the Base

In this case, we use a voltage reference made out of 3 diodes. Normal 1n4148 ones.

Within a broad current range, the forward voltage drop over 1 diode is 0.6 V. 3 diodes will push the base 1.8 V lower than the supply.

Our supply is 10 V. The base will be at 10 V - 1.8 V = 8.2 V.

The 5K resistor will take care that there's a decent current flowing through the diodes. In our case, that current will be 8.2 V / 5 K = 1.64 mA.

#### Set the Constant Emitter Voltage pt. 2

We know that the base is at 8.2 V. And that the emitter will be one diode drop above that.

Then, the emitter voltage must be 8.8 V.

#### Calculate the Emitter Current

The 220 Ohm emitter resistor is between the supply and the emitter.

Supply is at 10 V. Emitter is at 8.8 V. So there's a voltage drop of 1.2 V over the resistor.

Ohm's low then dictates that there is 1.2 V / 220Ohm = 5.45 mA running through the resistor.

The resistor is in series with the emitter, so the emitter current is 5.45 mA too.

#### Calculate the Collector Current

The easiest calculation ever. In a common-emitter circuit, the collector current is equal to the emitter current.

The emitter current is 5.45 mA. That means that the collector current is 5.45 mA.

I explained the circuit, starting from a given schematic.

In reality, you will work the other way around.

You will know what constant current you want to source, and what the power supply value is.

You then find a working combination of components.

# Reality

I built this circuit and did the measurements. The results meet the expectations.

All voltages drawn on the circuit are relative to ground.

In theory, the circuit should source 5.45 mA. In reality I measured 5.4 - 5.5 mA.

# Practical Limits

This circuit is easy, and can even drive a range of currents by changing the base voltage.

There are limits in the operation mode that you have to deal with:

• emitter voltage should always be 0.6 V above the basis. It needs to have headroom for the basis-emitter voltage drop (should I say emitter-base voltage drop in a PNP?), or the circuit will stop conducting.
In a fixed circuit, this is fairly easy to calculate. In a circuit where you want to change the current by changing the base voltage, it limits the range.
• Check the saturation collector-emitter voltage. There has to be enough headroom in the circuit to avoid that that voltage drops under the saturation threshold.
• Check the transistor's limits, such as maximum current, maximum collector-emittor voltage, second breakdown.
• Check the transistor safe operating area. Collector-emitter voltage drop over the transistor * current through  the transistor has to stay within a safe envelope.

# Control without Negative Feedback?

No. There is intrinsic negative feedback in this design. The resistor in the emitter circuit causes emitter degeneration.

A change in the emitter voltage (caused by a change in a current change) will change the bias of the diode.

This is a self-correcting mechanism. If the current rises, the transistor will be driven less. And the current lowers again. This is a control mechanism with negative feedback.

You are not wiring the feedback. The common-emitter circuit with emitter degenerating resistor cunningly exploits the transistor's physical behaviour

(see The Art of Electronics, 3rd edition,  2.3.4 B:Emitter resistor as feedback).

# Summary

This is a smart and simple control mechanism.

Although you don't have a dedicated feedback circuit, it's inherently there.

A practical circuit that's ideal to drive devices that need a constant current, such as LEDs.

If you ever asked why a set of LEDs with series resistors have different brightness: the series resistor limits their current, but doesn't control it to a fixed value. This circuit can solve that.

Even with the big tolerance in LEDs from a single supplier, when you drive them with a constant current, they will give a similar brightness.