I have a LM471 op-amp and a single DC 5V source. Is it possible to double the source voltage using this op-amp? Can the op-amp be powered by the very signal to be amplified? I tried simulating such situation in Multisim and it seems power/signal have to be from different sources. Why is that?

A segment on my 7 segment display requires voltage higher than 5V. I want to boost my 5V power (double it, for example) and use the boosted voltage to lit the display. I don't want to use a separate power source just to power the op-amp. I understand the out current will be lower, but it doesn't matter in my situation.

Of course different elements might be necessary in this circuit, but I don't want to use things like 555 timers, just analog parts.

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    \$\begingroup\$ POWER out MUST be less than POWER in. There are ways to produce voltages > Vin and > Vsupply - usually using more than 1 opamp or an opamp + other parts BUT you need to say what you really are trying to do as your question could mean several things. \$\endgroup\$ – Russell McMahon Jan 6 '15 at 23:44
  • \$\begingroup\$ I'll assume that's a lm741. Not the best opamp to power from only 5V. So I think you are asking for a circuit that will somehow bootstrap 5V into a higher voltage.. that then power's that same circuit to a higher voltage. That's kinda fun...but I think you'll need at least one more pass element. You've got to feed more power to the power supply than just goes through the opamp. \$\endgroup\$ – George Herold Jan 6 '15 at 23:51
  • \$\begingroup\$ What exactly is an "LM471"? This? \$\endgroup\$ – Spehro Pefhany Jan 6 '15 at 23:52
  • \$\begingroup\$ How to unscrewing large bolts with swiss army knife? This question (in its original version) fits that profile. \$\endgroup\$ – Nick Alexeev Jan 7 '15 at 0:16
  • \$\begingroup\$ @NickAlexeev That would be an odd car (not impossible, just really odd) that required unscrewing bolts rather than (lug) nuts to change a tire. \$\endgroup\$ – Spehro Pefhany Jan 7 '15 at 0:26

If you have a microcontroller (as I suspect) driving your display, you could probably use a spare pin with pwm to drive a simple capacitive voltage doubler. Thus you'd need only common parts (caps, diode, a bjt/mosfet) - no need for 555, opamp or whatever and no inductor so little added noise.

Edit : here is the classic voltage doubler circuit :


simulate this circuit – Schematic created using CircuitLab

C1 should be much smaller than C2 (lower C1 gives lower ripple, but lower curtent capacity and slower voltage ramp-up), and critically diodes should have lowest possible Vf. You won't get exactly doubling from this circuit as it is, but for what I guess your application is that may not matter much.

Drive PWM pin by setting it to 127, and choose frequency matching C1 (or C1 matching the PWM frequency).

  • \$\begingroup\$ Can you post a link to a sample circuit of such doubler? \$\endgroup\$ – pedro Jan 10 '15 at 10:24
  • \$\begingroup\$ I added a schematic of the classic voltage doubler. Plenty to be found on the net (google "capacitive voltage multiplier" or "charge pump") \$\endgroup\$ – Nicolas D Jan 12 '15 at 13:09
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    \$\begingroup\$ True; the current at input is at least 2 times the current at output. Also if you try to pull much current from a mcu pin, the voltage at that pin will decrease. So obviously this won't provide 50mA at 10V. However if you only need 1 or 2mA, if 8V and some ripple is fine, then this approach may be the simplest (lowest component count, only common parts) \$\endgroup\$ – Nicolas D Jan 12 '15 at 13:58
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    \$\begingroup\$ The circuit in this answer is wrong - it will produce approximately 4V dc and misses the point. D1 should not connect to ground it should connect to +5V. \$\endgroup\$ – Andy aka Dec 11 '15 at 18:26
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    \$\begingroup\$ @Andyaka Ooops, that's very true... I edited the schematic in the answer. \$\endgroup\$ – Nicolas D Jan 29 '16 at 10:10

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