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I'm trying to build a switch for a high frequency, high-voltage AC signal required for a research project. I have these two circuits and I don't understand why they behave differently. Note that these are not my final circuit diagrams, I plan on having an LED and phototransistor control the gate voltage eventually. I get that I need to apply a voltage difference between the source and the gate pins in order for both MOSFETs to conduct. However, both of these circuits behave differently under simulation. enter image description hereenter image description here

This circuit (above) switches the AC signal on and off fine, giving expected behavior.

enter image description hereenter image description here

However, this circuit (above) clips the negative half cycle of my AC signal. I don't understand what the difference is between these circuits. I want to build the second circuit because it makes the actual construction a bit easier in real life, I just built the first circuit first to make sure that I understood how to open/close these MOSFETs. In reality, I would also want to have separate grounds for my AC circuit and my 5V logic circuit to help with isolation; is that the correct direction?

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3 Answers 3

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M1 has a drain to source short in the second circuit, so it's doing nothing.

That's because of the grounds.

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  • \$\begingroup\$ I see that now, thank you! If those two grounds were separate, however (like the ground connected to the source of M1 is a ground for my 5V voltage control circuit while the ground connected to the drain of M1 is a ground for my AC circuit), would that solve the issue? \$\endgroup\$ Commented Jul 28, 2023 at 20:55
  • \$\begingroup\$ If they are totally isolated from each other (as in the first circuit) it would. If there's some capacitance then the signal could be affected, depending on the circuit impedance, frequency and capacitance value. \$\endgroup\$ Commented Jul 28, 2023 at 21:15
  • \$\begingroup\$ I think I know what you think you're saying. To be clear, you might post a 3rd schematic using two different GND symbols to show the isolation. \$\endgroup\$
    – AnalogKid
    Commented Jul 28, 2023 at 22:51
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Start with your second circuit, both drain and source of MOSFET M1 are connected to ground. That's a dead short, and this transistor can be physically replaced by a wire, leaving you with this:

schematic

simulate this circuit – Schematic created using CircuitLab

With the source permanently at 0V, and the gate permanently at +5V, then \$V_{GS}=+5V\$. The MOSFET is switched on, always and regardless of the potential difference across V2. Therefore when \$V_5 = +5V\$ node OUT is effectively connected to ground via M3, and \$V_{OUT}=0V\$. That's the reason for your flat "zero" orange trace.

Notice the body diode present inside the MOSFET. If the MOSFET is switched fully off (\$V_{GS}=0V\$), this diode is the only element inside the MOSFET that remains, and the circuit becomes:

schematic

simulate this circuit

That's a basic half-wave rectifier, where the diode has 0.7V across it during the negative half of the waveform of V2, which is why you see -0.7V "flats" in the blue trace, corresponding to the times when the diode is forward biased and conducting to ground.


You first circuit only works because gate potential is specified relative to source potential, and not to ground. It is the potential difference between gate and source \$V_{GS}\$ that determines the state of the MOSFET, not any absolute potential \$V_G\$ (relative to zero volts ground) at the gate.

In your first circuit, the sources of the two MOSFETS are permitted to adopt whichever potential they wish, and V5 is connected in such a way that ensures that the gates are always 5V above the source, regardless of changing source potential. Therefore \$V_{GS}\$ is not dependent upon V2 in any way, and the MOSFETs' states are similarly independent of V2. This independence from V2 is lost the instant you connect either source or drain to a fixed potential.

The need for two MOSFETs is due to the body diode, which causes a MOSFET to conduct from source to drain even if the channel is switched fully off. By placing two MOSFETs in series, with the body diodes in opposite directions, you eliminate this parasitic, inconvenient conduction path when the MOSFETs are supposed to be off, regardless of polarity of voltage across the pair. Conductivity is therefore controlled exclusively by the channel, but the price you pay is that the MOSFET sources cannot be tied to any fixed potential, including ground. For this reason, and as you remarked, it is challenging to maintain the condition \$V_{GS}=+5V\$ or \$V_{GS}=0V\$ while the source is flapping up and down in potential.

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The first circuit is necessary. Typically an isolated photovoltaic gate driver such as TLP3905 is used. If greater speed is required, a DC-DC isolated power supply will be needed, with any switching type gate driver isolator.

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