Building a voltage source for small circuits with a zener diode? Let's explore how this is possible, let's build a shunt regulator.
Zener Diodes as Shunt Regulator
You have developed a small circuit, but the voltage it is designed for is not available in the target environment? This not an uncommon situation. The 5 V or 3.3 V that are used in many microcontroller circuits are not used everywhere.
What do you do in such a case? You need a voltage regulator. A voltage regulator is a circuit that provides your circuit with the constant voltage you require. Today, we are going to look at a very simple voltage regulator circuit: a shunt regulator. We can build it using just a zener diode and a resistor. Let's have a look!
What is a Shunt Regulator?
When talking about voltage regulators there are two main types to differentiate between: linear regulators and switch mode regulators. Switch mode regulators are efficient, but complicated to build. They use a switching element like a transistor or MOSFET to regulate the voltage and can both reduce the input voltage, this is called a buck converter, or increase it, which is called a boost converter. Linear regulators can only reduce the input voltage.
There are two types of linear regulators, a shunt regulator is one them and at the same time one of the simplest voltage regulator circuits. Shunt regulators as opposed to in series linear regulators, regulate the voltage by providing a current path in parallel to the load through which they sink the surplus current.
For a shunt regulator to work the overall current has to be limited. This is typically achieved by connecting a resistor in series to the circuit. The core component of the regulator is, however, the actual regulating element, also called control element. This is the component that ensures the constant output voltage. For that, it not only provides the alternative current path, but more importantly, controls how much current needs flows through it so that the voltage remains constant. In our circuit, this is achieved using a zener diode. It limits the output voltage to its zener voltage and sinks the remaining current.
Does the circuit look familiar to you?
It's practically identical to the one used when using a zener diode as voltage reference.
Voltage references are one of the most common applications of shunt regulators. However, if we want to power a small circuit the current limiting resistor has to be chosen much smaller than for a voltage reference.
An Example Circuit
Let's have a look at an example circuit. The shunt regulator in the example powers a small microcontroller, an ATtiny85. The ATtiny85 normally operates at 5V. The shunt regulator circuit allows us to use it with 9 V or 12 V batteries. It achieves this with the help of a simple 5.1V zener diode.
The ATtiny85 runs a small program, that lets the connected LED blink. However, this is not the focus of this article. There are plenty of tutorials on how to program the ATtiny85 out there. If it interests you, have a look at the LED candle project which I made about a year ago.
Above, you can see the circuit in action. The LED lights up and blinks. The ATtiny85 works and is not damaged by the high input voltage. The shunt regulator does its job.
Building a Shunt Regulator
How was this circuit designed? Let's cover the theory that you need to build such a simple regulator.
Defining the Requirements
The first step is to collect our requirements. What is the input voltage range? What voltage and current do we need for the circuit?
The example circuit uses an ATtiny85 which is typically operated at 5 V. With its internal clock at 8 MHz it requires 5 mA of current. Its acceptable voltage range is 2.7 V to 5.5 V with an absolute maximum rating of 6 V. Additionally we need to power the LED. With its 1 kΩ current limiting resistor it requires an additional 5 mA at 5 V. In total, we require 5 mA if the LED is off and 10 mA if it is on. To play safe, let's add a safety margin to the power budget and plan with 15 mA. If reliability is important, you might want to use a bigger safety margin and plan with 20 mA. However, as we are going to see later, choosing the power budget too generous also has its disadvantages.
As power supply I would like to use are 9 V block battery or a 12 V lead acid battery. The input voltage range is thus roughly 9 V to 15 V.
Picking a Zener Diode
With our requirements set, let's select a zener diode. For 5 V we could either use a 4.7 V or an 5.1 V zener diode. In the example circuit I picked a BZX79-C5V1 5.1 V zener diode from my part box. It has a tolerance of 5 % and will produce an output voltage somewhere between 4.8 V and 5.4 V at a zener current of \(I_Z = 5 mA\). The zener current is the current that should at least flow through the zener diode to guarantee that output voltage stays between the specified values. The current through the diode should not fall below this value.
Calculating the Resistor Value
Last, but not least, we need to determine the value of the current limiting resistor. It needs to be selected in a way that ensures that enough current for both our circuit and the zener diode flows. This assumption should hold true, even at the lower end of our voltage range. We can calculate the resistor value using the following formula:
\(R_1 = {{V_{in} - V_Z} \over {I_Z + I_L}}\)
With 9 V as our lower voltage limit it gives the following result for R1:
\(R_1 = {{9 V - 5.1 V} \over {5 mA + 15 mA}} = 195 Ω\)
With standard resistor values we can use either 180 Ω or 220 Ω. As you can see I chose 220 Ω for the example circuit. This choice reduces the power budget further to:
\(I_L = {{V_{in} - V_Z} \over {R_1}} - I_Z = {{9 V - 5.1 V} \over {220 Ω}} - 5 mA = 12.7 mA\)
This is quite a tight power budget. It's your choice whether that is acceptable for you. The blink circuit in this example is by no means a critical application and only serves as a short time example. There are, however, valid reasons for a deciding not too generous power budget when using a shunt regulator. To understand why, let's have a look at the power dissipation at the upper end of our voltage range.
Power Dissipation
At 15 V the power dissipation of the resistor R1 is:
\(P_{R_1} = {V_{R_1}^2 \over R_1} = {{15 V - 5.1 V}^2 \over {220 Ω}} \approx 450 mW\)
For this we already need to pick a 1/2 W resistor instead of an 1/4 W one. A 1/4 W resistor only works up to 12.5 V. This needs to be considered in our component choice for R1.
The overall current at 15 V is:
\(I = {{V_{in} - V_Z} \over {R_1}} = {{15 V - 5.1 V} \over {220 Ω}} = 45 mA\)
This current only depends on the voltage difference and the value of the current limiting resistor. If the circuit consumes less power, then more power has to be dissipated by the zener diode. If the LED is off the ATtiny85 requires only 5 mA leaving 40 mA that flow through the diode. In this case the power dissipation is:
\(P_Z = V_Z \cdot {I - I_L} = 5.1 V \cdot 40 mA \approx 200 mW\)
Conclusion
These calculations show, that we can build the circuit with a 1/2 W 220 Ω resistor and the BZX79-C5V1 zener diode. The 200 mW power dissipation is within the 400 mW limit of the diode.
However, while the circuit works, the calculation shows that it wastes a lot of energy. 650 mW of power dissipation in the regulator alone are certainly not optimal, especially for a battery operated circuit. But, there is only so much we can too about that.
Our options are:
- Reducing the voltage difference (prefer a 9V battery over a 12 V battery)
- Selecting a higher resistance value which limits our power budget further
- Choosing a more efficient voltage regulator type
More Efficient Alternatives
In most cases a series regulator is a suitable alternative. It is more efficient and similarly easy to use. In case you need a high efficiency or high currents, it's best to use a switching regulator.
Summing Up
As we have seen, a zener diode based shunt regulator is a very easy, but highly inefficient voltage regulator. The huge drawback of shunt regulators is, that their power consumption is solely dependent on the input voltage and its current limiting resistor. It can be used to power circuits that only require few milliamperes of current. The condition for this is, that the difference between input and output voltage is small.
