Introduction: Zener Diode Shunt Regulator

Picture of Zener Diode Shunt Regulator

Every circuit needs a power source, but different circuits have different power requirements. For most small scale electronics, the power consumption is (is or at least could be with better design) very low. For those times when you don't need a lot of power, just a precise voltage level, the simplest means of regulating the supply voltage is by using a Zener diode.

In this Instructable, I will compare a typical voltage regulation circuit with the Zener diode shunt regulator circuit and show you the proper way of selecting the perfect components for your circuit needs. Correct usage of this circuit can save time and money, but it is not well suited for all designs. So grab those breadboards and let's gown to business!

Step 1: Zener Diode Basics

Picture of Zener Diode Basics

In case you don't know, a diode is a special type of electrical component that lets current flow freely in one direction, but blocks it from flowing in reverse - in the water analogy, the diode would be a one-way valve, only allowing water to flow in a specific direction. A typical diode will have a voltage drop of around 0.7V when forward biased. 

All diodes have a "reverse breakdown voltage" which if applied to the diode in reverse will cause current to flow backwards through the component, typically destroying it in the process. This value is typically in the hundreds to thousands of volts. A Zener diode (similar to an "Avalanche Diode") is a special sub-class of diodes that will allow current to flow in the reverse direction if the applied voltage is above a certain level without damaging the component. Of course, there are limitations to the voltage level and/or current flow, but those are things that the design engineer must take into consideration. 

Some common Zener diode breakdown voltages are: 1.8, 3.3, 5.1, 7.5, and 12.6, making them ideal for use in many small circuits.

Step 2: Linear Voltage Regulator Overview

Picture of Linear Voltage Regulator Overview
For 95% of small circuits, the voltage regulation is performed by a linear drop out regulator. The common 78xx series is very popular due to the broad range of regulator values represented by the "xx." For example, a 7805 regulator will output 5V. A 7812 regulator will output 12V. These are all DC values. There are two main problems when it comes to using these regulators:

  1. They are not very efficient - that excess voltage is wasted as heat.
  2. They require a specific range input voltage for operation. 

There are some regulators called LDO for "low drop-out," meaning the input doesn't have to be that much higher than the output. These are nice, because a standard 7805 regulator has a drop out voltage of 2V, meaning you need at least 7V at the input for proper operation, but they can cost a lot more. The power inefficiency is a bit more of a problem. For example, if you have a supply of 12V dc regulated down to 3.3V dc with a load drawing 1A, The power in is about 12W while the power out is 3.3W, meaning there is a loss of 8.7W. That is only 27.5% efficient!  It's for this reason that DC-DC switching regulators are used for applications when larger amounts of power are needed and power-line noise (from the switching frequency) is not an issue. These are more complicated to design and also cost a lot more.

In addition to these concerns, most regulators will need a capacitor at both the input and output to stabilize the voltage for proper operation. The good thing about using this type of regulator is that it should work perfectly (without additional consideration) just so long as the supply voltage stays within the operating range and the output current does exceed the maximum.

Step 3: The Zener Diode Shunt Regulator

Picture of The Zener Diode Shunt Regulator

When you know the level of the supply voltage and you have a small load, using a Zener diode as a regulator can be a great option; however, without proper components, this circuit can be far more inefficient than a linear regulator. 

Since the Zener diode is placed in the circuit under a reverse bias, it will allow current to flow through it as long as the supply voltage is above the breakdown voltage of the diode. The series resistor is in place to burn off the excess voltage. Again, this energy is wasted as heat in the resistor. The reason this circuit can be more inefficient is the fact that current will always be flowing through the resistor so long as the supply voltage is above the diode breakdown voltage, even without an attached load.

The value of the resistor determines the current. For example, using our previous numbers of a 12V supply and a 3.3V Zener diode, 8.7V will be dropped across the resistor. The correct value resistor will allow just enough current to pass as is needed to power the load circuitry plus a tiny bit consumed by the Zener diode. If no load is attached, then the entire current will be consumed by the diode.

It is for this reason that knowing the maximum power requirements of the load are very beneficial. Consider a microcontroller circuit that flashes an LED at 20mA. The maximum current consumption of the microcontroller will depend on how fast it is running, among other things, but could easily be less than 100uA. Just for safety's sake, we will say we need 30mA of current supplied to the entire circuit.

To figure out the necessary series resistance, subtract the output voltage from the supply voltage and divide it by the desired current: (12V - 3.3V) / 30mA = 290Ω. The other thing to consider here is power dissipation. The resistor will drop 8.7V at 30mA, dissipating 0.261W of power. A 0.5W resistor should be used. If no load is attached, the Zener diode will consume the entire 30mA dissipating 0.099W of power. A 0.2W or greater diode should be used. Even with our example load attached, the Zener diode will be consuming most of the current when the LED is not being lit. This is why this circuit can be very inefficient.

Step 4: Summary and Thoughts

So, to figure out what parts to use in this circuit:

  1. Decide what voltage level you need for the circuit.
  2. Calculate what current requirements the circuit will have.
  3. Calculate the necessary series resistance to provide the desired current.
  4. Calculate the worst case power consumption by the resistor and diode.
  5. Pick out parts accordingly.
R = ( VS - VDIODE ) / ITOTAL
PR
( VS - VDIODE ) * ITOTAL   =   ITOTAL2 * R   =   (VS - VDIODE)2 / R
PDIODE = VDIODE * ITOTAL

 
I have attached a datasheet with some power values for a line of Zener diodes from General Semiconductor. The information is a bit dated, but it provides for a good example. The table shows the Zener part number, breakdown voltage, current consumption, and maximum regulation current. Remember, a safe bet is to use a part rated for at least twice its potential power dissipation. 

Practical Considerations
You might wonder, given the fact that this circuit can be so inefficient, when (and why) should we use it? The answer is pretty simple. This circuit can be used when the load is a fairly continuous value (not drastic changes like powering a blinking LED) and is not too large since the components cannot handle very much current. It can be used to power a few chips that need a regulated voltage level lower than the supply. These chips can then control larger loads (like motors or LEDs) connected directly to the source via transistors (BJT or MOSFET).

Another great time to use this regulation circuit is when the supply voltage may be too low for a regular regulator to work. For example, I recently designed a Quiz Game system for a friend. He wanted it to be powered by 4 AA batteries. Here is the problem: 4 rechargeable batteries will produce about 4.8V (4* 1.2V cells) while 4 regular alkaline AA batteries will produce 6V (4 * 1.5V cells). The latter voltage is too high to supply many chips such as a microcontroller, so it needs to be regulated. I really wanted to use 5V to power the µC, so the best solution with the parts I had on hand was to use a Zener diode to regulate the voltage. 

If rechargeable batteries were used, the 4.8V supply would pass through the resistor into the circuit, unchanged by the Zener diode. The current draw by the µC would determine the voltage drop across the resistor, but it is a negligible amount. If regular batteries were used, the 6V supply would be regulated down to 5.1V by the Zener diode. There are other obvious solutions, but I didn't feel like ordering any new parts since I had a few Zener diodes and plenty of resistors on hand.

Lastly, this circuit can be used if space or cost is a huge factor in the design. A resistor and Zener diode will cost a lot less than a regulator and a couple of capacitors, and they will take up a lot less space on a circuit board. Hopefully, this information will point you in the right direction when it comes to designing your circuit's power supply!

 

Comments

theb42 (author)2016-11-22

Say you are charging a super capacitor where "VBat" is located in your circuit and the voltage across the capacitor is just 2V currently and you have a 3.3V zener diode and resistor attached to the capacitor as shown in your circuit and want to start dumping the excess voltage when the capacitor reaches the 3.3V zener voltage.... My question is there a small current that flows through the resistor and the zener diode even when the zener voltage has not been reached? What happens when you hook up a 2V supple to the 3.3V zener diode as shown? How much power is lost?

Kurt E. Clothier (author)theb422016-11-25

If I understand what you're asking, you are saying that VCC would be charging a super cap, so the current through R1 is in the opposite direction of my schematic.

Like my circuit, there will always be current flowing through R1, but in your circuit it would often be very minimal. Similarly, there will always be current flowing through the Diode, even when it's value is not reached - it's called leakage current, and you'd have to consult a datasheet for an actual value, but I've seen it range from pico to micro amps.

As for actually charging the cap, the diode would serve no purpose. Capacitors will store as much potential as they can (and explode if you go above their rated value). If the cap has 2V, and you attach it to a 3.3V line, it will immediately draw current until it is also 3.3V. You can calculate how long this takes if you know the voltage differential (3.3 -2), the capacity (Farads), line/load resistance, and possibly the maximum discharge rate of your source VCC.

If you then connect the cap to a 5V source, it will draw more current until it reaches 5V. It will NOT go above 5V, because that is all the potential available to it. However, the circuit I have shown is for providing a zero-current reference or regulating source voltages, and it is a good trick when your source could vary between an acceptable potential and something a bit too high (like 4 rechargeable batteries at 1.2V each vs 4 alkaline batteries at 1.5V each.). This circuit is not meant for charging purposes of any kind.

Lastly, connecting a 2V source to a 3.3V zener diode would have essentially no effect, except for the tiny leakage current I talked about earlier.

HảiK1 (author)2015-06-07

Thanks for your article

stevenarango (author)2014-04-12

great instruct able. lots of great info. thanks

benjamin712 (author)2014-03-27

Thanks for the article. I am building a wind mill generator and have a charge controller rated for a maximum input of 15V. I want to protect the controller in the event of voltage spikes from the gerneator motor above 15V. Attached is a possible design I came up with using the information above. I was hoping you could tell me if it would work as desired. (Would all voltage above 15V short to ground?).

Note* I have not determined the total resistance equivalent of the load or measured current values. Hopefully above at least 1A. Thanks!

Your use of this type of diode is generall called an Avalance diode instead of Zener, but it's basically the same thing. Have a look at this page for more info:

http://en.wikipedia.org/wiki/Avalanche_diode


It looks to me like you have the diode "pointing" to ground. That means ALL current would short directly to ground. To use the diode as a clipper, it needs to "point" at the voltage. Remember, a diode only lets current flow in the direction it is pointing unless the voltage level exceeds its "breakdown voltage." We can exploit that by using Zeners with low breakdown levels to only allow higher voltages to pass.

I would recommend you post your circuit and questions on the following website.

http://electronics.stackexchange.com/


It's a community of Electrical Engineers and they will gladly offer you advice and direction. Just be warned, if you don't word it very well, they might seem a bit rude... its just to protect the integrity of the sight. Take a look at a few other questions (or read the about page) to see what I mean.

http://electronics.stackexchange.com/about

lam (author)2014-03-07

Dear Sir Kurt E.

Please help me to know in 2 circuit (photo) is same working (function) or not ??? Thank you so much

lam

Kurt E. Clothier (author)lam2014-03-07

Iam,
No, the diodes in your circuit do not do the same thing as in mine. What you have is a type of "clipper circuit" that limits the amplitude of the output wave form:

http://en.wikipedia.org/wiki/Clipper_(electronics)

lam (author)Kurt E. Clothier2014-03-07

Dear Sir Kurt E.

Thank you so much for your reply.

Kind regards

lam,

jhorapalok (author)2014-01-30

Thanks for the write-up.... very informative indeed.

No problem, glad it helped!

franssoa (author)2013-06-21

Thank you. Clear instructable, very understandable even for my bad english.

vicdex (author)2013-06-05

You are a very good material, go on with your good instruction,

Phil B (author)2013-02-07

Thank you for this. It is very well done and helpful. I was familiar with the idea of using a zener diode as a shunt to clamp the voltage in a circuit at a desired level, but had given no thought to disapating surplus input power with a dropping resistor.

Once we had a wireless telephone that gave us a lot of crosstalk in the earpiece. I happened to check the output of the 12 volt wall transformer and found it was 13.4 volts. I clamped the voltage of the transformer with a 12 volt zener diode and the crosstalk problem disappeared. (Actually, to be certain the zener diode could handle the power, I placed to zeners in parallel. I know that is not the way to do it, and the zeners may not be closely enough matched so that one bears the larger part of the work, but it worked out fine for me.)

Kurt E. Clothier (author)Phil B2013-02-07

Thank you for the comments. I've found that pretty much anything in electronics that has a specified voltage level on it is really +-10% of that voltage. I've mostly always used Zener diodes for clamping purposes as well, but thought it might be helpful for some people to see how they could serve as an actual, stable power supply for small loads. The burn off resistor is very important, because that extra voltage is dropping somewhere, and the diode typically can't handle that kind of current flow.

Phil B (author)Kurt E. Clothier2013-02-07

Thank you. I have also seen regulator circuits that use a power transistor clamped at its base with a Zener diode, but I never attempted building one.

Phil B (author)2013-02-07

Thank you for this. It is very well done and helpful. I was familiar with the idea of using a zener diode as a shunt to clamp the voltage in a circuit at a desired level, but had given no thought to disapating surplus input power with a dropping resistor.

Once we had a wireless telephone that gave us a lot of crosstalk in the earpiece. I happened to check the output of the 12 volt wall transformer and found it was 13.4 volts. I clamped the voltage of the transformer with a 12 volt zener diode and the crosstalk problem disappeared. (Actually, to be certain the zener diode could handle the power, I placed two 12 volt zeners in parallel. I know that is not the way to do it, and the zeners may not be closely enough matched so that one bears the larger part of the work, but it worked out fine for me.)

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