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I am trying to check if I've appropriately sized a gate driver for a SIRA50ADP-T1-RE3 MOSFET.

The fastest it would expect it to switch is at ~20 kHz and the driving voltage (Vs) is 10V. From the datasheet I found that:

  • gate charge (Qg) can range from ~98-150 nC
  • gate resistance (Rg) from ~0.2-1.2 ohms

Using these values I estimated the gate capacitance (Cg) as 9.8-15 nF via Cg = Qg/Vgs. I used the capacitor charging equation to estimate its behavior when charging:

$$ V_{g} = V_{s}\left (1- e^{\frac{-t}{R_{g}C_{g}}} \right ) $$

Solved for t at 99% of Vs: $$ t = -R_{g}C_{g}* ln(1-\frac{0.99V_{s}}{V_{s}}) = -R_{g}C_{g}* ln(0.01) $$

From this I estimated that it would take ~9-83 ns to charge with 'unlimited' supply current (requiring upwards of 50A peak). I estimated the average current by integrating Vg/Rg:

$$I_{g,avg} = \frac{V_{s}}{RP}\int_{0}^{P}e^{(\frac{-t}{RC})}dt = \frac{VsC}{P}e^{(\frac{-P}{RC})}$$

which also matched a numerical approximation. However, the values calculated by this method would not be particularly useful in sizing the driver except for "ballpark" figures, since the bulk of charging/discharging happens early and at higher than average current. (Ex: the average current for the 83 ns charge time is only 1.8A, but supplying only 1.8A would increase charge time to ~386 ns)

Instead, a new charging time estimate can be found by assuming a known current and calculating the equivalent gate resistance: Rg = Vgs/Ig

Conclusion: For a small drive current of ~1A it could take ~451-691 ns to charge the gate. Assuming the switching frequency gives us half of one period (1/20000 Hz)/2 = 25000 ns, then the 1A driver should have plenty of time (2.8% of available)...?

Please let me know if any of my assumptions are wrong, or if I've missed something important.

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    \$\begingroup\$ TLDR; Unless you need 50A load current, it'll be fine. Don't get too hung up on the gate charge at 10V though. On the "Gate Charge" plot (page 3, bottom left) see the flat region? That's the "Miller Plateau" (FET gates have rather non-linear capacitances), once you're past that (about 4V/40nC) you can call the FET "on" and any more Vgs just makes it even more "on" (and reduces conduction/i2r losses). Your actual turn-on time will likely be quite a bit faster than 700ns. A good rule of thumb: 1C = 1A*sec, so think of 150nC as a constant 1A for 150ns. \$\endgroup\$
    – Sam
    Sep 10, 2023 at 1:11
  • \$\begingroup\$ What switching loss is permissible, and what loop dynamics are possible given project constraints (switching loop length/area/width, amount/size/proximity of bypass caps, component package type, etc.)? What load impedance/range (inductive, hard switched, etc.)? What conduction loss for that matter; is the transistor well chosen? \$\endgroup\$ Sep 10, 2023 at 2:04

1 Answer 1

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To determine the gate charge required to turn on this MOSFET, I would suggest applying the chart showing Vgs vs Gate charge, rather than trying to estimate the gate-to-source capacitance as a "lumped parameter" value. The chart shows that Qg is determined primarily by Vgs, and does not vary much with Vds.

enter image description here

Having determined Qg, the next consideration is: What is the peak current required from the gate driver? This will depend on the maximum permitted switching time for your application, and one of the factors to take into account here is the maximum permitted power loss due to switching. This depends on the application (eg: the switching frequency) and the type of load seen by the drain of the MOSFET. The energy loss within the MOSFET per switching cycle can vary dramatically depending on the type of load presented to the MOSFET drain terminal.

For example, if the MOSFET drain sees a purely resistive load, then both Vds and Id rise and fall together during each of the switching transitions, both from off-to-on, and from on-to-off. Compare this to an inductive load: during the turn-on transition the drain voltage is clamped high while the drain current rises; and during turn-off, the drain voltage is again clamped high while the drain current falls. Thus, for same load current and supply voltage, the inductive load causes a much higher energy loss per switching cycle than a resistive load.

If the load is inductive, specifically a diode-clamped inductive load, then there is another very important consideration for determining switching time, in particular, the off-to-on time. That consideration is the switching performance of the diode that is clamping the inductive load.

In the case where this MOSFET is one member of a totem-pole arm where the upper and lower devices are the same type, then the clamping diode is just the intrinsic body diode of the MOSFET. Back in the early days of MOSFETs, the body diode was poorly characterised on the datasheets until the MOSFET manufacturers learned from bitter painful experience that it was critical to the success or failure of their customers' products. These days, most MOSFET datasheets do a good job of characterizing the body diode for the intended application of the MOSFET. The corollary of that is: avoid using the MOSFET for any application where the application does not specifically match the test conditions used to characterise the body-diode in the datasheet.

With that in mind, here is the relevant part of the datasheet for the device nominated:

enter image description here

In your application, if the MOSFET body diode is used to clamp an inductive load, then my recommendation is to not exceed the value of di/dt mentioned here under the test conditions, specifically: di/dt = 100 A/μs. The value of gate resistance required to achieve this limit should be provided by the MOSFET manufacturer. A good first guess for this would be the values provided in the datasheet section dealing with switching times, where Rg=1 ohm.

However, there are a couple of concerns with the datasheet that raise some red flags as to how suitable this product is for clamped inductive loads.

The first is that the body diode reverse-recovery characteristics are specified at a junction temperature of just 25C. It is common knowledge that reverse-recovery performance degrades significantly at higher temperatures.

The second is that none of the switching time characteristics are specified with a clamped inductive load; the test conditions are with resitive loads, eg:
VDD = 20 V, RL = 1Ω
ID = 20 A, VGEN = 10 V, Rg = 1Ω.

My recommendation is to contact the manufacturer of this MOSFET to confirm that it is suitable for the application you have in mind, and to obtain more data that may not be in the datasheet but which may be crucial to your application.

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  • \$\begingroup\$ Thanks for the detailed response. It does happen to be an inductive load (low-side driven DC motor), though the plan is to have a diode across the motor to suppress voltage spikes. It sounds like the MOSFET body diode would not be such a concern then? \$\endgroup\$
    – Mandias
    Sep 10, 2023 at 3:02
  • \$\begingroup\$ @Mandias Depends on the complete wiring schematic. If the motor drive is a totem pole made with two instances of this MOSFET, then sorry to say that the MOSFET body diode will remain a concern unless you take very deliberate steps to prevent it from carrying current. In this case, simply placing a separate diode at the motor winding may not be sufficient, that will simply result in one diode being in parallel with the other, and they will then share current in accordance with their forward I-V characteristics. || Best to share the complete wiring schematic if you need further assitance. \$\endgroup\$ Sep 10, 2023 at 3:38

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