Design and performance of the 1-transistor regulator

7th February 2016

Cheap off-the-shelf voltage regulators such as the DE7805 provide very good regulation, are easy to use and very robust.

The 5 in the part number tells you that this is a 5 volt regulator. You put more than 5 volts in and 5 volts comes out.

However, with an integrated circuit, containing dozens of transistors, you never know what it's thinking. Maybe it's not actually a voltage regulator but in fact a Terminator brain, awaiting it's chance to destroy humanity.

Therefore, I consider it fun and interesting to study the performance of the "1-transistor" voltage regulator.

The resistor and zener form a shunt regulator. If the voltage across the zener tries to exceed it's breakdown voltage, it will conduct as much current as is necessary to keep the voltage at that level.

So we keep the base of the transistor at a constant voltage, and use it as an emitter-follower.

We choose a zener diode 0.6V higher than the output voltage we want due to the transistor's 0.6volt switch-on voltage.

We choose the value of the resistor to provide enough current for the zener. It's usually best to run zeners between 10-50% of their max power rating (go above this and it will heat up significantly).

To find the zener's max current; divide the zener's power rating by it's voltage. Here's the numbers I am using, for my 500mW zener.

OK so: 0.5/5.6=0.09. 90mA. We want 10-50% of that so between 9mA and 45mA (0.009 and 0.045 Amps).

The transistor needs some current from the resistor too. Find out how much by dividing the output current you want by the minimum current gain (hFE) of your transistor. Here's the numbers I'm using, with my BD135 transistor.

OK so: 0.3/40=0.0075. 7.5mA. So add that to the minimum zener current calculated above to work out the minimum total current. For me 16.5mA (0.0165 Amps).

Now to work out the resistor value we subtract the zener voltage from the supply voltage and divide it by how much current we want. Here's the numbers I am using.

Let's work out the maximum resistor value first: (9-5.6)/0.0165=378. 206Ω. This is the maximum resistor value we can use, else we won't get enough current when our supply is at 9V.

Now we work out the minimum resistor value: (13-5.6)/0.045=164. 164Ω. This is the minimum resistor value we can use, else we will put too much current when our supply is at 13V.

I'll use 180Ω since it fits nicely between my min and max values.

If you find your calculations don't work out because your maximum is lower than your minimum then you either need a more powerful zener or a transistor with more gain.

Finally we need to work out how much cooling we need. Find out the maximum rise in temperature the transistor can withstand. Find out the maximum power the transistor will dissipate in normal use. Divide the former by the latter. Subtract the thermal resistance of the junction-case if known. Here are the numbers I am using:

OK so: ((150-50)/(0.3*(13-5)))-10=35. We need a 35°C/Watt heat-sink. Remember when choosing heat-sinks that they may need airflow to perform as specified.

I want to keep my transistor as cool as possible to make it perform better so I will go overkill with a 3.3°C/W heat-sink, and a fan.

OK now let's make it, using a BD135 transistor, a 180Ω resistor and an Aavid Thermalloy 6399B heat-sink.

the 1-transistor circuit, with a couple 120Ω resistors as load. our input power is an adjustable-voltage switch-mode power supply. note very powerful fan behind the transistor's heat-sink


I will test both the DE7805 and the 1-transistor regulator to compare their performance in different ways. Unless stated otherwise the tests are done with a 12V power supply and an 80mA load.

Supply rejection

I vary the input voltage between 9.5V to 12V and see how much it affects the output.In mV/V. Lower is better.

Output impedance

I vary the load current between 80mA-160mA and see how much it affects the output voltage. Convert to a resistance value by V/I. In Ohms. Lower is better.

Unloaded accuracy

Due to leakage current the 1-transistor circuit will have a slight voltage increase when unloaded. I measure the output voltage with a 20mA load and no load to find out how much. In mV. Lower is better.

Quiescent current

What is the current draw with no load? In mA. Lower is better.


How resistant is it to short-circuit?

Cutoff voltage

The minimum difference between the input and output voltage before the output falters.


Well as you can see the 7805 performs better in all of the tests. This is due to it's complex circuitry, including temperature compensation and op-amp-like error correction.

However that doesn't mean the 1-transistor design is bad; it does it's job just fine.

Possible enhancements

Here I present 4 simple enhancements you can make to the 1-transistor circuit. You can use multiple of them together if you wish!

Cap across zener

Putting a capacitor across the zener helps improve filtering of high-frequency ripple. I think this is a great idea.

Note that it's sometimes useful to put a little cap in parallel with the bigger one.

Short-circuit protection

Having such easily blow-up-able circuits can be a bad idea. Do not try to limit the current by limiting the current at the transistor's base - it will probably go into thermal runaway. Measure the total current draw and limit that. Like this:

The current-sense transistor can be any general purpose BJT transistor, just make sure it can handle the power dissipation and current. Estimate the max power dissipation by: zener_voltage*((max_supply_voltage - zener_voltage)/resistor_value). So for my circuit 5.6*((13-5.6)/180)=0.23. 230 milliwatts. The resistor value should be calculated by dividing the transistor's switch-on voltage (0.6) by the current limit you want. So for a 400mA limit: 0.6/0.4=1.5. So 1.5Ω.

Note that if you want the circuit to withstand short-circuit indefinately then your output transistor will need to handle a power dissipation of max_supply_voltage * short_current.

This current limiter doesn't affect performance much but you have to take into account it will reduce the supply voltage by up-to 0.6V. So when you do your calculations reduce your minimum supply voltage by 0.6.

Note that, if you put a large capacitor across the zener diode, it can discharge rapidly when the output is shorted, irrespective of short protection. So if you're using short protection and you want a large cap for smoothing, put it across the output, not the zener.

Darlington or Sziklai pair

If you want a lot of output current you may find yourself needing a huge zener diode, and huge power wastage. A higher-gain transistor can help solve this.

Darlingtons and Sziklai pairs are two ways of making high-gain transistors. I prefer the Sziklai pair because it has a lower switch-on voltage thus allowing the regulator to have a lower dropout voltage.

The 1K resistor helps improve the output transistor's turn-off time.

Building this design with the exact same resistor and zener as before, but with a sziklai pair made from a BC337 and a TIP32C, I am able to draw over 2 Amps! Also the output impedance reduced to about 0.1Ω!

Improve unloaded accuracy

Putting a resistor across the output will constantly draw a small load, thus stopping the voltage increase when unloaded:

For my circuit the 1K resistor reduces the voltage rise from 300mV to 60mV. However it increases quiescent current from 37mA to 42mA.

Further entertainment


Please remember that the 1 transistor regulator only works with BJT transistors! FET transistors do not have a constant Gate to Source voltage, rather a ratio of voltage to current called Transconductance. I tried a MOSFET for fun and my output impedance was 6Ω!

Test setup

You may find it interesting to know some extra details about my test setup:

Achieving 78-like performance with discrete components

If you look on the web you can find some higher performance discrete designs. This and this both look good.