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I have built the circuit shown below and been testing it on a breadboard.

Everything works except the positive half clipping circuitry.

I need it to clip after 3.3V. It currently just clips below 0V, but a signal greater than 3.3V can still be seen. If I use 3.3V from a power bench instead of the inverting opamp circuit I used, which gets 3.3V from a 12V supply, then it works fine. So something is wrong with my inverting opamp circuitry, and I assume it has to do with not enough current to switch the diode on. How would I go about fixing this?

Basic receiver circuitry

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  • \$\begingroup\$ I would use an actual voltage regulator, not an op-amp, to generate your 3.3V rail. Also, don't see a capacitor on the 3.3V rail to ground. Is there one somewhere??? \$\endgroup\$
    – Kyle B
    Sep 13, 2021 at 16:10
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    \$\begingroup\$ @KyleB most voltage regulators don’t sink current very well. The short-circuit current of the LT1214 is in the 50m A range. Adding a cap of more than perhaps 100pF to the 3.3V will likely make U4 oscillate. \$\endgroup\$
    – Spehro Pefhany
    Sep 13, 2021 at 16:23
  • \$\begingroup\$ @SpehroPefhany Right, that's why I don't think he should use an opamp as a regulator. I don't see this as sinking current so much as "absorbing a fast spike". That's why a voltage regulator and a BIGGER cap is what I'd go with. Like 100uF or 220 maybe paralleled with 10uF ceramic. All the current that needs to be 'sunk' is AC and comes through that 100nF cap, so it can't be much. 100uF >> 100nF Should easily be able to deal with the spikes. A proper regulator IC costs what... $0.25??? No brainer IMO \$\endgroup\$
    – Kyle B
    Sep 13, 2021 at 16:32
  • \$\begingroup\$ SHould also mention to OP that this won't 'clip at 3.3V'. It will clip at 3.3V PLUS the diode drop (around another 0.3V). So this would clip at 3.6V. If he looks closely at the other side, he'll probably find it clips at -0.3V, not 0V \$\endgroup\$
    – Kyle B
    Sep 13, 2021 at 16:38
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    \$\begingroup\$ @KyleB it has to sink 50mA plus, which is much worse than no-load. This is a commonly seen issue with connecting diodes to supply rails as clamps, so I’m emphasizing it, no intention to pick on you. For example, the clamp may work in normal operation when other chips are drawing sufficient current, and fail when the processor is in sleep mode. That one cost a client company a great deal of money and reputation. \$\endgroup\$
    – Spehro Pefhany
    Sep 13, 2021 at 16:52

1 Answer 1

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Your clipping depends on U4 being able to sink enough current to overcome what U2 can source. So, one output is fighting another.

You can add some resistance before the clamp to limit the current to a few mA, maybe 5-10mA maximum. That would also prevent U4 from going into current-limiting on the negative side. Forcing the output into current limiting causes a lot of heating, possible to the point of damage without a heat sink, according to the LT1214 datasheet.

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    \$\begingroup\$ Thank you very much! Worked like a charm today. \$\endgroup\$
    – Ian Edwards
    Sep 14, 2021 at 21:07

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