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I am buildind a circuit like this, but I am confused about D1, is it necessary? enter image description here

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  • \$\begingroup\$ Can you please provide more information about the use of the circuit. As it stands is not useful for other people browsing your question. Adding missing part numbers and values for the components would be nice too (U1, U2, Q1, D1 especially). \$\endgroup\$ – Lorenzo Donati supports Monica Jul 19 '16 at 10:31
  • \$\begingroup\$ U2 appears to be placed backwards. \$\endgroup\$ – brhans Jul 19 '16 at 11:32
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The circled diode provides a -0.7Vdc negative rail or reference. Note that the anode of the diode is connected to circuit ground (mostly hidden by your red circle - but visible at high magnification).

This negative rail allows the (+) input of the 2nd op-amp to operate near or below ground.

Note that I would connect pin 4 of the op-amp to the cathode of the circled diode - this will improve operation when the (+) input of the 2nd op-amp is near zero Vdc.

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  • \$\begingroup\$ can you please explain your note about connecting pin 4 to the cathode of diode? \$\endgroup\$ – Haris778 Jul 19 '16 at 8:40
  • \$\begingroup\$ Pin 4 is the negative power supply connection for the op-amp. Moving pin 4 to the -0.7V rail improves operation when the (+) pin of U1b is at ground. \$\endgroup\$ – Dwayne Reid Jul 20 '16 at 0:14
  • \$\begingroup\$ I can expand on that further in my answer if you need. \$\endgroup\$ – Dwayne Reid Jul 20 '16 at 0:14
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I'm not certain what your circuit is attempting to do, but when you see a diode in parallel with a base-emitter of a transistor (Q1), then this is usually to add a small voltage drop. The base-emitter is a similar diode junction, so the circuit is usually matching voltage drops.

This technique can reduce effects such as crossover distortion in some types of opamp-driven circuits that require a transistor, making the circuit overall more 'ideal' in behavior.

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  • \$\begingroup\$ The D1 diode is not in parallel with the b-e junction of the Q1 transistor. In fact neither diode anode or cathode even connects to any terminal of the transistor. \$\endgroup\$ – Michael Karas Jul 19 '16 at 4:06
  • \$\begingroup\$ I see that i missed the junction. The diode raises the 'gnd' for U2, the output of which is going through the transistor, so I suspect that the diode is still to reduce distortion caused by the transistor. \$\endgroup\$ – slightlynybbled Jul 19 '16 at 4:13
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    \$\begingroup\$ -1 - It is kind of silly to be guessing at the function of the circuit. Saying you "suspect" and "I'm not certain what your circuit is attempting to do" are not ways to go about answering questions here. \$\endgroup\$ – Michael Karas Jul 19 '16 at 4:38

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