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Gate driver circuit

This picture shows some part of a high frequency (about 250kHz) MOSFET gate driver circuit.I searched for various types of gate drivers in google and didn't encounter anything like this. Can someone explain this configuration.

What is the purpose of the diodes?. why mosfet Source isn't connected directly to the gate driver supply ground?. what is the purpose of two 2k resisters and C12 capacitor connected between +12V and the ground?.

In here +12V and the GND are the gate driver power supply terminals.

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  • \$\begingroup\$ Electric components don't need a direct connection to ground (except maybe lightning rods). It is lifted about 3*0.6V (three diodes drops) using R21, R22? and three diodes in series. \$\endgroup\$
    – Oldfart
    Jun 21, 2019 at 8:42
  • \$\begingroup\$ Please show the whole circuit. The power mosfet in the picture has a dangling wire at the drain, wire goes that wire? To +12V as well? If the shown picture is a mosfet gate driver, how is this driver connected to the mosfet that has to be driven? \$\endgroup\$
    – Huisman
    Jun 21, 2019 at 10:52
  • \$\begingroup\$ DC drops (I * R) or inductive kicks ( L * dT/dT) make "grounding" a challenge. How high a current, switched how fast, is in your circuit? \$\endgroup\$
    – analogsystemsrf
    Jun 21, 2019 at 16:34
  • \$\begingroup\$ @Oldfart sorry i cannot understand your comment. what did you exactly meant by "ground"? is it earth or what. In here i meant by "ground" is the gate driver power supply negative terminal. \$\endgroup\$
    – Damith Pavithra
    Jun 22, 2019 at 9:33
  • \$\begingroup\$ @Huisman power mosfet drain and source is connected to a different power circuit with a separate power rails. Gate driver output is just directly connected to the " gate driver signal" wire. Gate driver power supply and the above shown +12V/GND is same. \$\endgroup\$
    – Damith Pavithra
    Jun 22, 2019 at 9:34

2 Answers 2

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In such cases, it is useful to redraw the circuit: as @Oldfart says in their comment, we have the following situation

schematic

simulate this circuit – Schematic created using CircuitLab

The source of the power MOSFET is connected to the ground via three of the four diodes \$D_1, D_2, D_3, D_4\$ packaged as a couple of double diodes: the resistors \$R_{21}\$ and \$R_{22}\$ bias this series connections of diodes with a approximately ten milliamperes, in order to let the source see few ohms of differential resistance (at least in the small signal regime) through ground and (perhaps this is the real reason of their presence) set a threshold of \$V_\mathrm{th} = 3V_\gamma+V_{GS_\mathrm{ON}}\$ for the gate driver signal voltage, where \$V_\gamma\$ is the threshold voltage of the diodes and \$V_{GS_\mathrm{ON}}\$ is the gate source turn-on voltage.

Edit
What happens if the source of the power MOSFET is connected directly to the ground, bypassing all the diodes and the capacitor \$C_{12}\$ with a short circuit? From my point of view, it is unlikely that this stage is used in some analog circuit, thus the only consequence I see is that, in this case, you simply have \$V_\mathrm{th} \simeq V_{GS_\mathrm{ON}}\$: this means that you obviously cannot set this threshold by changing the bias of \$D_1, D_2, D_3\$. Without knowing more details of the circuit, this is the only hypothesis I can do about how the circuit would work.

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    \$\begingroup\$ Your D4 looks to be in the wrong direction. \$\endgroup\$
    – mmize
    Jun 21, 2019 at 13:49
  • \$\begingroup\$ @mmize thank you very much! Corrected schematics. \$\endgroup\$
    – Daniele Tampieri
    Jun 21, 2019 at 13:54
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    \$\begingroup\$ @DanieleTampieri thank you for the answer. when the gate is off, it shows about negative 1.7V across gate and source. What did you mean by "differential resistance" can you explain why they have used this arrangement. what happen if i remove diodes and directly connected the source with gate driver ground \$\endgroup\$
    – Damith Pavithra
    Jun 22, 2019 at 9:37
  • \$\begingroup\$ @DamithPavithra the differential resistance is the small signal resistance: for the diodes, since \$I_A=I_s[\exp(V_A/V_T)-1]\$, we have $$ r_D=\frac{\mathrm{d}V_A}{\mathrm{d}I_A}\simeq \frac{V_T}{I_A}$$ where $$V_T=\frac{kT}{q}\simeq 25\mathrm{mV}\text{ at }T=21^\circ\mathrm{C}$$ is the thermal voltage. For the the question you asked in your comment, about what happens if you short circuit the source to ground, I will add something to my answer above. \$\endgroup\$
    – Daniele Tampieri
    Jun 24, 2019 at 9:05
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With this arrangement D4 clamps negative Vgs and the other diodes provides a load impedance above Vgs(th) if less than 1.8V , then you can drive it with a current source with pull up or a voltage source or even a capacitor with diode negative clamping with PWM if you thought this was an advantage for isolated DC driving of the gate at some different ground potential. A series R is needed to allow Vgs=>3Vt to ensure low RdsOn so a diode+ suitable Zener would be better than 3 diodes.

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  • \$\begingroup\$ thank you for the answer. Can you further explain your answer. Your first sentence seems to be very long and i cannot understand i clearly. can you give me some good resources (a link, a book etc) to learn more about MOSFET gate driving. \$\endgroup\$
    – Damith Pavithra
    Jun 22, 2019 at 9:47

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