Art of Electronics, 3rd Ed, pg. 164-165 give an example of naively using a 4000-series CMOS logic gate to directly switch a power MOSFET, disregarding the significant gate current required to charge the MOSFET capacitance. They give the following values: VDD = +10V, C_mosfet = 200pF, maximum current source/sink capability of logic gate = +/- 1mA. Their conclusion is that it will take approximately 50us to switch the MOSFET on/off. I don't see how they come up with this number. If the logic gate is slew-rate limited, then it will take t = CV/I = (200pF)(10V)/1mA = 2us to turn on the MOSFET.

I checked the second edition and although they use different numbers, I'm still not able to come up with the provided answer. What am I doing wrong?

  • \$\begingroup\$ I don't have my copy of AoE nearby, but in practical terms, gate capacitance is not the key parameter for this calculation. You need to look at a parameter called "total gate charge" (or something similar), which can easily be an order of magnitude larger. \$\endgroup\$ – Dave Tweed May 2 '18 at 11:25
  • \$\begingroup\$ Probably they model the output impedance of the gate as 10kohm (10 volts/1 mA), then look at the resulting 10k/200 pF response, which will be much slower than your model suggests. \$\endgroup\$ – WhatRoughBeast May 2 '18 at 11:41
  • \$\begingroup\$ @DaveTweed Actually, they do proceed to mention that using the datasheet specified gate charge parameter is a more accurate method of approximating the rise/fall time. However, I still wanted to know how they got 50us using 200pF gate capacitance and 1 mA max. output current. \$\endgroup\$ – pr871 May 2 '18 at 12:41
  • \$\begingroup\$ @WhatRoughBeast That makes more sense, but that method gives an RC time constant of 2us, so 5 time constants is still well below the given value of 50us. The weird thing is that, when they mention that using the gate charge parameter (Qg) is a better way to do this sort of thing, they use t = Qg / I, which is more or less the same formula I used to get my wrong answer. \$\endgroup\$ – pr871 May 2 '18 at 12:48
  • \$\begingroup\$ @WhatRoughBeast: The time constant of that combination is still just 2 us, so what's your point? \$\endgroup\$ – Dave Tweed May 2 '18 at 12:48

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