0
\$\begingroup\$

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.

\$\endgroup\$
  • 2
    \$\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
1
\$\begingroup\$

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 :

schematic

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).

\$\endgroup\$
  • \$\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
  • 1
    \$\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
  • 1
    \$\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
  • 1
    \$\begingroup\$ @Andyaka Ooops, that's very true... I edited the schematic in the answer. \$\endgroup\$ – Nicolas D Jan 29 '16 at 10:10

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.