MOSFET gate driver current requirement for 2000 volt MOSFET

Question

I am experimenting with a flyback converter that uses an Infineon IMYH200R100M1H MOSFET.

How much drive current is needed to drive this MOSFET gate? I am predicting 8 amperes of drive current is sufficient for this. Correct me if I am wrong.

Answer 1

Following the comments by Hearth and by loser8106250, if you need to switch ON and OFF your MOSFET at a given frequency \$f\$, you should charge (and obviously discharge) the gate G of the device \$f\$ times per second. Now you know from the datasheet the total gate charge \$Q_\text{G}\$ needed by the device to pass from interdiction to the fully ON conduction state, therefore you can set a minimum value for the gate current \$I_{\text{G}}\$ by using it and the \$f\$ datum, i.e. $$ I_{\text{G}_\min}\ge Q_\text{G}\cdot{f}\label{1}\tag{1} $$ On the other hand, the maximum value of \$I_{\text{G}}\$ is instead estimated by considering that, in order to minimize switching losses, you may require a well defined transition time \$t_T=\min\{t_\text{ON}, t_\text{OFF}\}\$ in order for the MOSFET to switch sufficiently fast between its two conduction states: this implies the following upper limit $$ I_{\text{G}_\max}\le \frac{Q_\text{G}}{t_T}\label{2}\tag{2} $$ Thus, as a design rule of thumb, joining \eqref{1} with \eqref{2} we can say $$ Q_\text{G}\cdot{f} \le I_{\text{G}}\le \frac{Q_\text{G}}{t_T}\label{3}\tag{3} $$

Edit: how to choose the transition time? In answering Vikram's comment below, I remark that the shorter the transition time, the lower is the drain power lost during the transition. Thus, as a rule of thumb, I initially chose the (lowest) values of \$t_\text{ON}\$ and \$t_\text{OFF}\$ stated in the data sheet, remembering that $$ \begin{split} t_\text{ON} &= t_\text{d(on)}+t_r\\ t_\text{OFF} &= t_\text{d(off)}+t_f \end{split} $$ Then I chose the least time between the one specified and this is the transition time \$t_T\$ I use in the calculations.

Final notes

• The described procedure is of course a "first order" one. It may be that the transition speeds are too high from the point of view, say, of the EMC radiated power or any whatsoever reason, including the unavailability of his speed high current drivers: then, in those case I choose a higher transition time suitable for these "second order needs" and use the above formula accordingly.
• Note that while formula \eqref{1} is defines the lowest limit to the RMS value of the gate current, formula \eqref{2} defines its upper (repetitive) pulsed peak value. This means that, considering the MOSFET in the OP, $$ I_{\text{G}}\le \frac{Q_\text{G}}{t_T} \simeq\frac{55\text{nC}}{5\text{ns}}=11\text{A} $$ but this current is the peak current the MOSFET drive should supply for at least \$5\text{ns}\$: the \$8\text{A}\$ value is thus more than adequate, leading to a transition time that is only the 38% higher than the one stated in the datasheet (as \$t_\text{d(on)}+t_r\$).

Edit2: a practical example. Stemming of the first and the second comments of Vikram, I'll sketch a design procedure evaluating for the described application. Let's set \$f =100\mathrm{kHz}\$ and correspondingly consider a cycle time of \$T=\frac{1}{f}=10\mathrm{\mu s}\$ as the characteristic time of the flyback converter: the device considered is the Infineon IMYH200R100M1H MOSFET. as in the OP. At first we see that $$ Q_\text{G}\cdot{f}=5.5\mathrm{mA}, $$ thus the minimal gate current is not a problem with this design data: any sufficiently performing driver satisfies this requirement. Then we turn to considering the MCP1415/16 already used by the asker in a preceding application: this device is characterize by a typical drive current \$I_\text{PK}=I_{\text{G}_\max}=1.5\mathrm{A}\$ thus it allows a maximum transition time of $$ t_T \le \frac{Q_\text{G}}{ I_{\text{G}_\max}}=\frac{Q_\text{G}}{ I_\text{PK}}=\frac{55\text{nC}}{1.5\mathrm{A}} \simeq 37\mathrm{ns} $$ Since the cycle time is \$T=10\mathrm{\mu s}\gg 10 t_T\$ you can use this driver for the analyzed application: on the other, if you have a driver capable of delivering a \$I_\text{PK}=I_{\text{G}_\max}=8\mathrm{A}\$, the same calculations above give rise to a transition time \$t_T\le 6.875\mathrm{ns}\$ which is near the maximum allowable performance of the device, thus even the \$8\mathrm{A}\$ driver is more than acceptable. This in turn implies that choosing one or the other drive is now only a matter of other design specification, notably cost, power efficiency, EMI and the like, which however are outside of the question topic.

Answer 2

As per the data sheet of the MOSFET, the test condition is given at gate resistor 2 ohm and Vgs change as (18-(-2))=20V. So, as per this, the peak gate current requirement is 10 Amp.

Disclosure: I work as a support engineer with Infineon at the time of writing.