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I was wondering if there was any way of driving back-to-back MOSFETs (with their sources and gates tied together as shown in the following image) without using an optocoupler?

Typical opto-isolated SSR

The reason is that optocouplers are slow, I want to use a SSR to short out a ultrasound transducer after sending a transmit pulse-train to shorten the ring time (to get the receiver set-up quicker to measure shorter distances).

Assuming N-Channel MOSFETs, I can't think of a way of biasing the gate to say +5V relative to the source (which is changing all the time), without using an isolation technique like optocouplers or a transformer.

Is there a simple way to do this if you don't require isolation?

I intend to drive the transducer with something like shown below, where Cp and Rl models the piezo-electric transducer, and Cl/transformer is to make the load purely "resistive" and then match the impedance.

enter image description here

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  • \$\begingroup\$ There probably is a good way of achieving what you're trying to do. But it would depend on how you are driving your ultrasound transducer. Is it ground-referenced or floating? What is the drive voltage and frequency? If you can post the schematic of your existing transducer front end, it will make it much easier to answer your question. \$\endgroup\$ – Nick Alexeev Dec 5 '13 at 2:17
  • \$\begingroup\$ I have added a schematic of what I think the front-end may look like. It will probably be floating. The short MOSFETs could be located either on the left or right of the transformer, although if on the left-side I'm not sure how effective they will be. \$\endgroup\$ – gbmhunter Dec 5 '13 at 2:41
  • \$\begingroup\$ HP4800A is a vector impedance meter. I ought to be missing a piece somewhere. Will it have enough power to drive the transducer? Somehow it should allow you to turn the output on and off, right? \$\endgroup\$ – Nick Alexeev Dec 5 '13 at 3:28
  • \$\begingroup\$ Oh sorry, that schematic was taken from an example document, I should of mentioned the HP4800A will have nothing to do with it, the transformer will be driven with some kind of MOSFET switched square wave. \$\endgroup\$ – gbmhunter Dec 5 '13 at 4:12
  • \$\begingroup\$ Are you 100% sure that shorting the transducer achieves quickest time for it to lose its energy. I'd have thought shunting with a specific value load resistor is likely to dissipate the power/energy to the lowest value in the shortest time. I may be wrong, just a thought. \$\endgroup\$ – Andy aka Dec 5 '13 at 8:38

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