# Converting 5V Square-Wave to +-5V Square-Wave

I am a student electronic engineer and was hoping to get some guidance on a project that I am working on. I am using the AD9850 module to output a 5V square-wave at 1.1MHz as can be seen in the figure below:

I was wondering if it is possible to somehow convert this square-wave into a +/-5V (10Vpp) square wave at 1.1MHz frequency. Is it possible to achieve this with just a 5V power supply or would an external power supply of 10V be required? The signal is to be used to excite an ultrasonic transducer.

• You could use an RF 1:2 step-up transformer or, you could use an amplifier with appropriate supply voltage(s). Aug 26, 2021 at 12:13
• The best design will have specs for all assumptions on frequency error, power output and impedance of load. Then decide how to drive a large capacitive load with a low impedance by appropriate push pull driver keep in mind/simulate Q, L, C, R with a differential output. SHow tolerances with all specs. Aug 26, 2021 at 12:53

## 1 Answer

simulate this circuit – Schematic created using CircuitLab

Seems that using AD9850 to generate your square wave could be done with a simpler numerically controlled oscillator.
Those two NOT gates would be driven with a 5V DC supply voltage. HCMOS gates like 74HC04 would have approximately 50 ohm drive impedance (each). Parallel connection of unused gates to NOT1 and to NOT2 increases drive available.
If you still need more drive, Justme's method of H-bridge drive would be more appropriate, and would also result in 10V p-p available.

• Thank you for your reply, would this method have the square wave swinging from +5V to -5V resulting in 10V peak to peak? Aug 26, 2021 at 13:51
• Actually in my answer I did have a H bridge made of inverters in mind, just like in your answer. But it's now deleted because apparently it did not answer the question. I guess it was too short. Aug 26, 2021 at 13:52
• Yes, the voltage available at the ultrasonic terminals would be 10V peak-to-peak. You haven't mentioned ultrasonic impedance @ 1.1MHz. It is possible that 1.1MHz current flowing through the transducer will load down those two NOTgate drivers, reducing the 10V p-p to something less. The 6 NOT gates inside 74HC04 could be paralleled into 2 ranks to help this potential problem (schematic reivsed). Note that a transducer with one terminal grounded cannot use this solution. Aug 26, 2021 at 14:27
• Brilliant solution. Instead of solving OP's question, you solved OP's problem. Engineering at its best! Dec 25, 2022 at 15:14