enter image description here

I'm trying to build a PSU based on a design which has it's maximum input voltage 15v. However mine is rectified from 200VA 24Vrms centre tapped transformer. I need to use a zener diode to ensure the voltage doesn't go above 12V for the TLC272 precision opamp. The original design only had 15V so it wasn't a problem. I was wondering if connected it up as in the schematic, would I get a maximum of 12V but full 34V for regulator. I chose 12V because it's easy to obtain and it was the as low as Vin could be on the original design.


2 Answers 2


The way you've drawn it the zener diode clamps the power supply reducing the 34V to 12V. What's even worse is that there's no current limiting and the diode will be destroyed.

The zener diode needs a series resistor to take care of the voltage difference between the 34V and the 12V.

enter image description here

To calculate the resistor value you need to know how much current the zener diode needs and how much current should go into the load (the opamp in your drawing). Zener diodes are often specified at the rather high current of 50mA, though more modern types will work properly at current less than 1mA. (This one only needs 50\$\mu\$A.) Let's suppose it needs 10mA and that the opamp will also consume 10mA, that's a total of 20mA. That's the current through the resistor. Since we already know the voltage across the resistor Ohm's Law will give us the value:

\$ R = \dfrac{\Delta V}{I} = \dfrac{34V - 12V}{20mA} = 1100\Omega \$

If you check the power dissipated in the resistor (something you always should do) you'll see that this is 0.44W, so use a 0.5W power resistor.

My original answer talked about 50mA zener current. That's very high, and I also realized that at 50mA a 12V/400mA zener would operate out of spec, as also jippie pointed out. It appears that the high current is only used for lower voltages (76mA for the 3.3V 1N4728A), and at 12V the datasheet talks about a safer value of 20mA. It's OK to use a lower current, though load and line regulation will be worse.

  • 1
    \$\begingroup\$ Normal zeners have a maximum power rating of 400mW. At 12V this means the max. current through the zener diode equals 0.4W / 12V = 33mA. If there is no load (no opamp connected) this means the resistor equals (34V-12V)/(33mA) = 660 ohm minimum, @733mW. The next question is if 33mA is sufficient to power the opamp (my feeling says yes). The important question to be answered is: How much current does the opamp draw? If you know the max current the opamp draws, you can calculate the maximum resistor value. (34-12)/I(opamp,max). You should add a decoupling capacitor in parallel with the zener. \$\endgroup\$
    – jippie
    Mar 29, 2012 at 19:52
  • \$\begingroup\$ @jippie - I also realized that at 50mA a 12V/400mW zener would operate out of spec, I changed my answer. \$\endgroup\$
    – stevenvh
    Mar 30, 2012 at 6:12

As shown the zener reduces the voltage in about the same way as you would slow down a moving car by standing in front of it. A zener diode is used to dissipate excess energy when energy supply is limited. If you used a big enough zener to clamp the supply to 12V it would damage he supply and the 34V supply would become a 12V supply.

Placing a resistor between supply and zener cathode (top end here with cross bar) will limit the current and isolate the zener from the supply.

Rzener <= V/I = (Vsupply - Vzener - headroom)/I_opamp_max.
If Iopamp max = say 10 mA.
Rzener <= (34V-12V - 2V) / 0.010 <= 2000 ohms. Say 1800ohms.
2V headroom allows the zener to "regulate" at all loads.

Resistor power rating > Izener x Vdropped = 0.010 x (34-12) = 0.22 Watt. Use at least a 0.25 w resistor and preferably a 0.5W one.

  • \$\begingroup\$ OK so I do need that power resistor then. \$\endgroup\$
    – Ageis
    Mar 29, 2012 at 21:25
  • \$\begingroup\$ @Ageis - only if you don't stand in front of moving cars as a preferred means of slowing them down :-). \$\endgroup\$
    – Russell McMahon
    Mar 29, 2012 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.