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I have a square wave that is 5v peak-to-peak. I need a circuit that can amplify this square wave up to any given supply voltage between 9 and 50v. I don't need much current, nothing more than 100ma certainly, 20-50ma would work. My preference is to use components as cheap as I can get. How can I amplify this square wave for minimal component cost? Is there a way to achieve a square wave with a high peak-to peak voltage without in input pwm?

Edit:

The 0-Vsupply square wave will be fed into the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Vout will always float 10v above the supply, minus a few diode drops. The part I am stuck on is how to get a square wave at the supply voltage.

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  • \$\begingroup\$ Use a transformer. \$\endgroup\$
    – Andy aka
    Dec 22, 2019 at 0:16
  • \$\begingroup\$ @Andyaka I will already have a supply voltage available between 9 and 50v. I see all kinds of level shifters out there but they only work for lower voltages. \$\endgroup\$
    – Hackstaar
    Dec 22, 2019 at 0:20
  • \$\begingroup\$ "... for minimal component cost?" - that depends. How much signal current is available at 5V, what is the frequency range, and what are low/high voltages (eg. 0-5V)? How fast does the output have to transition, and how close does it have to get to the supply rails at 50mA? What load is it driving? \$\endgroup\$
    – Bruce Abbott
    Dec 22, 2019 at 0:33
  • \$\begingroup\$ @BruceAbbott none of these parameters are critical. Let me add a schematic. \$\endgroup\$
    – Hackstaar
    Dec 22, 2019 at 0:39
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    \$\begingroup\$ I'm going to answer your question, but -- is this an XY problem? It looks like maybe your real question should be "how do I get a low-current supply that floats 10V above my 50V supply rail?" \$\endgroup\$
    – TimWescott
    Dec 22, 2019 at 1:16

2 Answers 2

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If you don't mind burning up power, but want simple. You need to size your resistors to match whatever you choose for C1 (so if C1 = 1uF, choose R = 100 ohms, which means that Q1 needs to be a power transistor, not a cheezy little 3904).

schematic

simulate this circuit – Schematic created using CircuitLab

If you want to be more efficient, but can stand more drop (the output will pull down to about 0.9V, and up to 50V - 0.7V).

schematic

simulate this circuit

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I found a simple circuit that works for me. Just tested on my oscilloscope. I had originally overlooked this one because power dissipation goes up with supply voltage, but in my application a very small on-time duty cycle will suffice. This will decrease the average power dissipation through the collector resistor to within an acceptable range.

enter image description here

Schematic by Herbert Weidner

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  • \$\begingroup\$ You'll get better turn-off of the transistor if you put a 1k-ohm resistor from base to ground. I'm pretty sure that a 5k-ohm resistor won't give you enough oomph with your chosen 1uF cap, unless your entire circuit is fairly lightly loaded. \$\endgroup\$
    – TimWescott
    Dec 22, 2019 at 1:39
  • \$\begingroup\$ What transistor are you using, and what are the on and off times? \$\endgroup\$
    – Bruce Abbott
    Dec 22, 2019 at 2:38

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