Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The schematic is

Simple MOSFET test

The wave form is as follows, yellow line is channel 1 which is gate voltage and blue line is channel 2 which is drain voltage.

Gate and Drain voltage

The 148kHz PWM signal is totem pole output of UC3845, MOSFET is n-type (IRFZ44N) and 12.4V power.

The drain voltage drops fast when gate voltage is on, but it increase very slowly (about 2.5us) when gate voltage is off. If I replace the 1K resistor with lower value, the Vd increase faster. But this means more power consumption.

  1. Why Vd increase slowly?
  2. For the speed up of Vd increase, is there any other way than decreasing Vd resistor(1K)?
share|improve this question
To make Vd increase quickly without a high power consumption, you might want to look at the way CMOS logic circuits work. – immibis Mar 21 '14 at 8:04
up vote 12 down vote accepted

The output capacitance of the MOSFET has a time constant with the 1K resistor that is around 1usec. You might think it should be more like 360pF * 1K = 0.36usec from the value of \$C_{OSS}\$ on the datasheet, but that low capacitance is measured at a 25V \$V_{DS}\$.

enter image description here

Referring to Figure 5, we have:

enter image description here

As to what can be done about it-- it's a bit silly to be using a ~50A MOSFET to switch 12mA.

Incidentally, the reason it decreases fast is because the capacitance may be the same, but it's being discharged by the MOSFET drain, which can sink perhaps 100A with 12V on the gate. That means the switching speed is dominated by how fast you can drive the gate high through a 40 ohm resistor.

Use a smaller MOSFET and not only will the gate be easier to drive, but the output will switch snappily. Something like a 2N7000 is more than adequate for 12mA and it has perhaps 1/30 the output capacitance.

share|improve this answer
Can't this answer be simplified to "charging goes through the resistor, discharging does not"? – Simon Richter Mar 21 '14 at 10:09
@SimonRichter It certainly could, but it would not convey as the same information, nor answer the second question. – Spehro Pefhany Mar 21 '14 at 15:19
This a very nice answer. Not only does it cover charging Coss through the load, but also discharging through the drain. It also shows proper way of determining Coss value. And questions use of the Z44 in such a low current application. Very complete. – gsills Mar 21 '14 at 16:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.