# Simple Square Wave Amplification Technique?

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: 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.

• Use a transformer. Dec 22, 2019 at 0:16
• @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. Dec 22, 2019 at 0:20
• "... 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? Dec 22, 2019 at 0:33
• @BruceAbbott none of these parameters are critical. Let me add a schematic. Dec 22, 2019 at 0:39
• 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?" Dec 22, 2019 at 1:16

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). 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). simulate this circuit

• is 2N3904 ok at 50v? Dec 22, 2019 at 8:05

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. Schematic by Herbert Weidner

• 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. Dec 22, 2019 at 1:39
• What transistor are you using, and what are the on and off times? Dec 22, 2019 at 2:38