3
\$\begingroup\$

I created a voltage follower/buffer for a first stage transimpedance amplifier and want to limit its maximum output voltage to roughly 3.3 V to protect the following ADC from voltage spikes.

Simulating the circuit on Falstad, I get 3.3 V of output voltage when inputting 4-5 V. But the real circuit on my breadboard gives 2 V when the input is at 5 V. I'm using 1N4728A Zener diodes for the clamping.

Is there anything I got wrong in my circuit? What else could be the reason for my real output differing from Falstad?

enter image description here

\$\endgroup\$
1
  • 2
    \$\begingroup\$ I think you're making a common mistake about Zener diodes: seeing them as 'magic voltage drops' that produce a fixed and exact voltage drop across themselves, whatever the weather. Actually, they have quite dynamic behaviour and the drop across them varies with current, temperature etc. \$\endgroup\$
    – TonyM
    Jan 28 at 0:41

3 Answers 3

6
\$\begingroup\$

The 1N4728A specifies its zener voltage at a test current of 76mA. With 5V output on your follower, you'll have well under 5mA in the diode, and your zener voltage may be lower. The tolerance of the part is +/-5%, but only at the test current. (Still, 2V seems very low, so double check your P/N maybe.)

Your simulation model is probably not modelling the I-V characteristics accurately, so you don't see the issue there.

To verify, use a current-limited bench supply to test your zener and record the voltage at your approximate current (3V/1K = 3mA)

\$\endgroup\$
3
  • \$\begingroup\$ tested it. that was the issue... thank you, totally a case of read-the datasheet better from my side... do you have an alternative idea to use instead of a zener diode? I can't really supply the required 75mA through my op-amp-output \$\endgroup\$
    – Miromed
    Jan 27 at 21:27
  • \$\begingroup\$ Beat me to the answer. Classic trap for young players. To use a Zener diode as a clamp at the rated voltage, you have to ensure a quiescent current equal to the test current published in the datasheet. Manufacturers often don't publish the I-V characteristic curve of their devices, so the test operating point is our only reference \$\endgroup\$ Jan 27 at 21:28
  • \$\begingroup\$ @Miromed There are a lot of options- If you Google "clamp circuits" you can check them out. One way might be a PNP BJT with a 3.3 V-0.7 V = 2.6 V reference on the base and the emitter connected where you have the zener. (Collector to ground). \$\endgroup\$
    – John D
    Jan 27 at 21:46
2
\$\begingroup\$

Try this, move the feedback point to the cathod of Zener-diod. I haven't tried this in real. It's an idea I just got.

  • When the output voltage is not higher than Zener voltage, the output voltage will be exactly same as input.
  • if the input voltage goes over zener voltage, the output of OP amp will be ~ Vcc, current to the zener diode will be (Vcc - Vz) / R.

Note that TL431's minimum cathod current for normal operation is ~0.5mA. Thus, be sure that do not use too large resistor on the out of OP amp. 1 ~ 2KOhm would be enough.

Limited Buffer

\$\endgroup\$
2
  • 1
    \$\begingroup\$ Replace the low-voltage Zener with a TL431 and two resistors and you'll have a winner. \$\endgroup\$ Jan 28 at 3:22
  • 1
    \$\begingroup\$ @SpehroPefhany Thanks, I've just updated as your recommendation. \$\endgroup\$ Jan 28 at 4:00
0
\$\begingroup\$

Note, that besides the well-pointed subjects about Zeners, you didn't mention the OpAmp you're using. Many parts do have a limited voltage output range, that's easily 1.5-2V below the positive supply voltage: check on the datasheet.

Funny enough, this could also help you limit the output range. A humble LM358 still can be supplied by 5V and output a maximum of around 3.5V, yet providing a close to zero Volt range on the other side. There's another downside though, as its common-mode input voltage range is also limited to 3.5V. This is another important issue, as some OA, when exceeding input range, exhibit phase reversal...

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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