Voltage Regulator using Transistor

A voltage regulator with a transistor usually consists of a bipolar junction transistor (bjt) with high current handling capability in an emitter follower configuration, driven by zener diode and resistor potential divider (PD) network. We first use a Zener diode and resistor across the input rail to make a PD that provides a regulated output. This output from the PD then drives the base junction of the transistor so its output is regulated as well. An advantage of using an emitter follower transistor is that it allows for greater power handling, than a Zener diode could alone.

These types of circuits are usually used in cassette tape recorders to provide regulated voltage to the DC motor. This way, as the batteries drain, the motor receives the same voltage, thereby keeping the speed constant.

For the transistor type, you need to pay attention to its current handling capability and h_FE figure. Many general-purpose transistors can handle currents up to 1 A and they are useful if all you need is 500 mA or less.

If the transistor h_FE was 50 and you needed it to provide 500 mA output current to a device, then we should remember the following formula from transistor theory class.

h_FE = I_OUT / I_B

Therefore, with rearrangement we get the following,

I_B = I_OUT / h_FE

I_B = 0.5 / 50

I_B = 0.010 A

The current passing through resistor R splits between the Zener diode and the base junction, therefore the following identity applies.

I_R = I_Z + I_B

We find from the Zener diode documentation that we need to have at least 10 mA (or 0.010 A) passing through the Zener diode to maintain it in the breakdown region, therefore we can calculate I_R as follows.

I_R = 0.01 + 0.01

I_R = 0.02 A

If we used a Zener diode with a voltage of 6.8 V, and the supply voltage were to be 10 V, then the voltage drop across the resistor R is 3.2 V, because we subtract the two voltages, as it is a PD.

Now that we have the voltage across the resistor R and current going through it, it is simply a matter of using Ohm’s Law to calculate its value.

R = V / I

R = 3.2 / 0.02

R = 160 Ω

Power Requirements

When the output current from the circuit is zero (when we remove the load), all the current (0.02 A) will pass through the resistor R and the Zener diode. Since we know the voltages across both elements in the PD and the current flowing through it, we can calculate their power requirements.

Power = current × Voltage

Zener diode Power Requirement = 0.02 × 6.8

Zener diode Power Requirement = 0.136 watts

Resistor Power Requirement = 0.02 × 3.2

Resistor Power Requirement = 0.064 watts

As you can see, the Zener diode power requirement is very small because we need a very small current from the PD going to the base of the bjt, and the bjt handles the power across the load.