I was reading this document.


Suppose we implement a bidirectional power switch using two P-Mosfets in back to back configuration with common source tight together. (Figure 5, page 3 of the pdf).

We know that Bidirectional Power Switches are capable in the ON state of support bidirectional current flow and in the OFF state block positive or negative voltages.

I could not find in the document what happens in the OFF state with the current.

Current will be blocked bidirectionally as a fact of back to back mosfet configuration and the only current that will flow will be the leakage current of mosfets?

In the ON state what happens to negative voltages?

  • \$\begingroup\$ Thanks for the green check. Any follow-up questions? \$\endgroup\$
    – AnalogKid
    Commented Oct 11, 2023 at 16:09
  • \$\begingroup\$ Thanks for your clear answer, helps me so much understanding the topic. I've also simulate this circuit in my pc and in the osciloscope when circuit is off I see some very little current (in the order of uV) at the output also in the gate mosfets (tighted together with a resistor) so I can't figure if this is because leakage current of mosfet, or is because on state resistance as you mention. I know this is not easy to explain on simple words but maybe you know an example I can see in order to be clear about this. Thanks so much really for the excellent response! \$\endgroup\$
    – marcosbc
    Commented Oct 11, 2023 at 19:09

1 Answer 1


Without getting into leakage currents and on-state resistances, simplify things for a coarse analysis.

When the FETs are off, they are perfect open circuits. When the FETs and on, they are dead shorts. The diodes are perfect, with 0 V Vf.

Now, when both FETs are on, the circuit behaves like pieces of copper wire. They have zero resistance, and short out their own internal body diodes. Currents can move in both directions. No matter what the amplitude or polarity, the voltage at one end is the same as the voltage at the other end.

When both FETs are off, they are infinite impedances; cut wires. The two diodes are back-to-back, so there is zero current in either direction. Different voltages (positive, negative, AC, whatever) can be at the two ends, and they will not interact in any way.

Of course nothing behaves like this in the real world, but that is the intent of the circuit.


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