Part of a PCB I designed looks like this:


Simple phototransistor with a resistor to measure the voltage over. More light => more voltage over the resistor.

However, the circuit behind the output doesn't like it when the impedance of it's input is too high. Lacking a better idea, I decided to try what effect an opamp as isolation buffer would have:


The opamp (LM358N) got hot and didn't perform as expected.

VDD is 3.3V, not even close to the maximum Vin the LM358 is rated for.

Did I cut a corner I shouldn't have cut or did I make a seriously stupid mistake here?

To avoid any confusion: the circuit without opamp works exactly as intended. Just not in combination with the next stage of the design.

  • \$\begingroup\$ How did you connect the other op-amp in the LM358 package? If you left it's inputs unconnected, it's possible that it's oscillating and that's what's heating up. \$\endgroup\$ – The Photon Mar 26 '16 at 21:43
  • \$\begingroup\$ @ThePhoton All unused inputs are directly connected to GND. \$\endgroup\$ – Mast Mar 26 '16 at 21:44
  • 1
    \$\begingroup\$ That's still a potential problem. Configure the other op amp as a follower with the input grounded. \$\endgroup\$ – WhatRoughBeast Mar 26 '16 at 22:02
  • 3
    \$\begingroup\$ When T6 conducts it will pull the +ve input to the op-amp up towards Vdd. But an LM358 is not a rail-rail op-amp, and it doesn't work well when the +ve input gets within 1.5V of Vdd. You need to use a rail-rail op-amp. \$\endgroup\$ – Steve G Mar 26 '16 at 22:24
  • \$\begingroup\$ @SteveG I could've sworn I picked a rail-to-rail, wow, that's a major mistake there. \$\endgroup\$ – Mast Mar 26 '16 at 22:40

Your Answer

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

Browse other questions tagged or ask your own question.