I am trying to generate a 65 V square wave very similar to the post found here: Mosfet PWM signal not square

My load is in a slightly different configuration (I've tried them both ways), but the problem I am having is very similar to the original poster in the link above.

Basically I'm switching a FET that is in connected to a 65 VDC supply through a 10 kΩ resistor and placing my load across the FET. Here is my circuit:

enter image description here

Here are my scope shots; the blue is voltage across the load and the yellow is the gate signal:

enter image description here I'm inquiring how I can make that rising edge across the load fast and sharp. The rounded edge makes me unable to achieve 65 V when I increase the switching frequency since the capacitance in the load and my resistor cause the voltage across the load to rise too slowly. The screenshots show about a 60 Hz frequency, but I'd like to get up to about 10 kHz.

My load is a PLDC diffusing film and there is about 1 μF worth of parasitic capacitance built into the material which I cannot remove. I am also aware that I can reduce the 10 kΩ resistor at the drain of the FET, but since my peak voltage is 65 V, reducing the resistor begins to put quite a bit of power on that resistor which I'd like to avoid.

The only other way I can think of to make this happen is by putting another FET in parallel with the drain resistor and syncing and inverting the gate signal of this other FET such that it will turn on when the other one turns off and vice versa.

But is there a better way than this?

I tried putting the load in place of the 10 kΩ resistor with the voltage developing across the load being less. In regards to the FET not being turned on completely, I am using a frequency generator and have increased the turn-on voltage to 10 V with no different results so I don't think that is the issue, see my scope shots below. Yellow is the gate drive, blue is the voltage across the FET, purple is the high (65 V), and red is the voltage across the load.

enter image description here

  • 2
    \$\begingroup\$ You're going to need a push-pull as you suspect. You're then going to need to prevent shoot-through (if both transistors were to be on simultaneously). You will also need to watch the peak current in and out of the capacitor so you may need a little resistance between the output and the load. Have a look at half-H-bridge designs. \$\endgroup\$
    – Transistor
    Dec 4, 2019 at 22:40
  • 4
    \$\begingroup\$ Can you put the FET in series with the load (between load and GND) and ditch the resistor entirely? \$\endgroup\$ Dec 4, 2019 at 22:40
  • \$\begingroup\$ Are you simply trying to turn the power on and off to the load, if so do as @evildemonic said. \$\endgroup\$
    – HandyHowie
    Dec 4, 2019 at 23:09
  • 1
    \$\begingroup\$ You don't mention the intrinsic resistance of your load, only that it has about \$1\:\mu\text{F}\$. If you are looking for \$10\:\text{kHz}\$ (which to me implies edges that are about 10% of that in order to be close to square), then this implies about \$7\:\text{A}\$ or so, continuous (which implies a rail that is much higher than \$65\:\text{V}\$ to me.) If passively charging from a \$65\:\text{V}\$ through a transistor, \$3\tau=\frac1{10}\cdot 100\:\mu\text{s}\implies R_\text{SRC}\le3\:\Omega\$. And that's only if you are comfortable with 20% of your cycle time spent in rising and falling. \$\endgroup\$
    – jonk
    Dec 4, 2019 at 23:21
  • \$\begingroup\$ Measuring the resistance across the load gives 2 Mohms. And strangely enough, when I remove the 10K and put the load in its place as was suggested, the voltage developed across the load is even less than in my original configuration. \$\endgroup\$ Dec 4, 2019 at 23:28

4 Answers 4


At 10kHz, you'll need that 1μF film capacitance to charge in well under half a cycle of 50μs, say 10μs. This implies that whatever drain resistor you use to charge the film will have to produce a time constant of less than

$$ \begin{aligned} \tau &= RC \\ \\ &= 50\mu s \\ \\ R &= \frac{50\mu s}{1\mu F} \\ \\ &= 50\Omega \end{aligned} $$

That will certainly dissipate a lot of power while the MOSFET is on:

$$ P = \frac{V^2}{R} = \frac{(65V)^2}{50\Omega} = 85W $$

That will be 43W on average, over many cycles. At lower frequencies the resistor can be larger, with a commensurate reduction in power dissipation, but the problem remains that during half of each cycle, the resistor has 65V across it, and is getting hot.

During each charging, the energy transferred into the capacitor would be:

$$ E = \frac{1}{2}CV^2 = \frac{1}{2}\times 1\mu F \times (65V)^2 = 2.1mJ $$

Even if you used a another transistor to charge it, in place of the drain resistor, and switched off the existing lower one during that half-cycle, due to the resistive nature of that charging route, the same amount of energy would be dissipated in the transistor, 10000 times per second, for a power of:

$$ P = 2.1mJ \times 10000s^{-1} = 21W $$

The same power would be dissipated in the other transistor, during discharge half-cycles.

While this isn't an answer to your question, at least it should be clearer why using transistors and resistors to charge 1μF to 65V, and then discharge it, 10000 times per second, is not trivial.

I'll keep thinking about this. I am wondering how simple it would be to implement charge pumps, switch mode, or an oscillating tank (maybe a Colpitts/Hartley setup) to make the system more energy efficient. This is certainly an interesting problem.


I can think of 2 filter approaches:

1) an impedance controlled RLC filter for maximum power transfer thus 50% efficiency, using C as load

2) A RLC filter based on user's rise time requirement at given f, Vpp
- thus dV/dt =Ic/C and dI/dt=V/L with low Q <2 and lower loss - but higher peak power demand for same energy stored in C load.

Using 1)

  • with a period of 100 us cycle and 10us rise time at 50% efficiency
  • compute the average power dumped in R.

    • E=0.5CV² = 0.5 * 1uF * 65V² = 2.1 mJ = 2.1 W-us means 21 Watts in 10us
    • thus here is a quick & dirty design

enter image description here

  • \$\begingroup\$ That's some cool software, what's that called? \$\endgroup\$ Dec 5, 2019 at 1:15
  • \$\begingroup\$ Falstad Javascript in any browser tinyurl.com/t8ytelg It's worth the learning curve \$\endgroup\$ Dec 5, 2019 at 1:18
  • \$\begingroup\$ Awesome, thanks for the tip! \$\endgroup\$ Dec 5, 2019 at 1:20

I am also aware that I can reduce the 10K ohm resistor at the drain of the FET, but since my peak voltage is 65V, reducing the resistor begins to put quite a bit of power on that resistor which I'd like to avoid.

You can avoid some of the power loss, but not all of it. Whenever a capacitor is charged or discharged through a resistor there will always be a power loss, as only 50% of the energy goes into changing the charge on the capacitor. The other 50% is dissipated in the resistance.

I simulated your circuit in LTspice with a 65V supply, IRF530 and 10kΩ pullup resistor, driven with a 58.65Hz square wave. Unfortunately with 1uF capacitance the curve didn't match your display, so I fitted it by eye - which required 330nF. With these values the 10k resistor dissipated 254mW and the MOSFET drew a peak current of 85A when discharging the capacitor (way over its rating!).

Then I increased the frequency to 10kHz and lowered the resistor value to 60Ω. This produced a similar waveform but the resistor dissipated 42W, 35W of which was due to it being pulled down to ground while the FET was turned on.

To remove that unnecessary loss I created a half bridge circuit with 2 MOSFETs, putting the 60Ω resistor in series with the capacitor and switching it alternately between Ground and +65V. Voltage on the capacitor then swung from +5V to +60V, and the resistor dissipated 12W. The rms power supply current was 320mA. Since that voltage swing wasn't quite enough I then reduced the resistor value to 30Ω, which got the voltage swing within 0.5V of the supply rails. This increased resistor power dissipation to 14W and rms current draw to 490mA.

Once you get to the point where the capacitor is fully charging and discharging, speeding it up further won't significantly increase power consumption (unless you increase the frequency). The only bad effect is increased peak current. With resistance reduced to 15Ω the waveform starts to look reasonably 'square' and power dissipated in the resistor is still only 14W, but rms power supply current increases to 700mA. The IRF530's now draw a peak current of ~4.5A each, which is within their safe operating area.

If 14W is still too much loss then the only other option is to use some inductance to form a tuned circuit with the capacitance. Any energy put into a tuned circuit will circulate around it, increasing the voltage without wasting power. However this will make the result critically dependent on the capacitance, inductance and driving frequency, as well as any other parasitic elements (film resistance, wiring inductance, dielectric effects etc.).


Look up “complementary gate driver integrated circuits”, use two high voltage FETS in a totem pole configuration. The data sheets will have example circuits.


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