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This is my totem pole circuit:

enter image description here

It works well, but R32 is too hot.

This is scope:

enter image description here

I changed R32, R34to 470 ohm, but I got the same result:

enter image description here

I think the resistor value is the problem. How can I fix this?

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    \$\begingroup\$ Why all that circuitry to drive that NFET gate? \$\endgroup\$ Commented Jun 7, 2023 at 4:59
  • \$\begingroup\$ This is to reduce the heatof the FET as much as possible \$\endgroup\$
    – KimJiHoon
    Commented Jun 7, 2023 at 5:09
  • \$\begingroup\$ Isn't that a low gate voltage FET? Seems like they are selling it on that idea, anyway. Also, you are capping the gate voltage to 12 V with the zener. Still seems like way more than is needed. What's the PWM1 source look like? What voltages and what's its impedance? \$\endgroup\$ Commented Jun 7, 2023 at 5:22
  • \$\begingroup\$ PWM from MCU, i found this circuit from 555 timer ic (motor pwm speed controller module) but if i using this MCU's pwm pin directly fet get's to hot \$\endgroup\$
    – KimJiHoon
    Commented Jun 7, 2023 at 5:23
  • \$\begingroup\$ Oh. Okay. I/O pins from MCUs will vary somewhat. But if I lack details, I will tend to assume 100 Ohms as a guess until and unless corrected later (with a compliance current of at least a few mA.) What's the MCU voltage rail? Better yet, what's the MCU? \$\endgroup\$ Commented Jun 7, 2023 at 5:25

2 Answers 2

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When Q11 is switched on, the voltage across R32 is the full 24V of the power supply. In this state it will dissipate power:

$$ P = \frac{V^2}{R} = \frac{(24V)^2}{1000\Omega} = 0.58W $$

That's enough to cause it to heat up quite a lot. R32 will have to be rated for power dissipation greater than this. If it's a 1W model, or above, then it will be fine.

If you are unable to use a larger power rated resistor (for size concerns, perhaps), then using that same formula you can work out what minimum resistance is acceptable, given some maximum permissible power dissipation. For example, if you want to use a standard 250mW (\$\frac{1}{4}W\$) resistor:

$$ R_{32} = \frac{V^2}{P} = \frac{(24V)^2}{0.25W} = 2.3k\Omega $$

Round up to the nearest available value, from the E12 series, for example:

$$ R_{32} = 2.7k\Omega $$

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  • \$\begingroup\$ i changed R32, R34 to 2K7 FET get's no heat at all (27°C) R32 is about 37°C i used R32 1608 smd type, if i use 2012 smd size is that okay? \$\endgroup\$
    – KimJiHoon
    Commented Jun 7, 2023 at 5:22
  • \$\begingroup\$ @KimJiHoon I calculated 2.7k based on 250mW maximum. Both the resistors you have specified are lower than that. I don't mind helping, but this is easy research and arithmetic you are perfectly capable of doing yourself. Find out maximum power dissipation for 2012 and 1608 resistors, and recalculate. You don't need R34, I don't know why you included it in the first place. Power dissipation in R32 will vary with PWM duty cycle. Measure temperature at 100% duty cycle, which is when R32 dissipates most power. Temperature measurements of R32 at anything but worst case are meaningless. \$\endgroup\$ Commented Jun 7, 2023 at 6:00
  • \$\begingroup\$ but R32->2K7, R34->1K, when set to this value, the most stable result was obtained and i don't know why \$\endgroup\$
    – KimJiHoon
    Commented Jun 7, 2023 at 6:07
  • \$\begingroup\$ I just tried running it using a separate 12V power supply Everything is the same except that there is no heat so i think using a large wattage resistor seems to be a better way than adding a separate regulator thank you \$\endgroup\$
    – KimJiHoon
    Commented Jun 7, 2023 at 6:26
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You have three options here.

  1. Increase the resistance of R32. This means it will pull less current, lowering the power dissipated in the resistor. The higher output impedance of the first stage will lower the peak output current you can achieve and the bandwidth of the circuit, but it should be fine for PWM at a frequency this low.

  2. Use a higher-power rated resistor for R32. If you use a resistor designed to be used at higher powers, designed to dissipate more heat, it won't get as hot because it can transfer the heat to the environment faster. Depending on just how much heat we're talking, you may need to add a heatsink.

2a) If R32 is through-hole, raise it up on its leads instead of mounting it flush to the board. This will help it exchange heat with the environment.

  1. Just accept that R32 is going to get hot. As long as it's rated for the power it's dissipating and stays within its rated temperature range, it'll be fine, even if it feels hot to the touch. Things start to feel painfully hot when they're around 50 °C, but your average resistor is perfectly happy at much hotter temperatures than that--it's not uncommon for resistors to be rated to 150 °C, 175 °C, or even 200 °C or higher. (do consider the temperature rating of your PCB though. Most FR-4 has a Tg of 125 °C or 150 °C, and you don't want to keep it over its Tg for long periods of time if you can avoid it.) Resistors are just about the least temperature-sensitive components out there.

I would go with option 1, personally. Option 3 is also fine, but option 1 also decreases power consumption.

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  • \$\begingroup\$ i increased R32 to 2K7, 4K7, 10K. Resistor heat about 37~40°C. but fet is getting to hot 50°C. i removed R34. is R34 necessary? \$\endgroup\$
    – KimJiHoon
    Commented Jun 7, 2023 at 5:11
  • \$\begingroup\$ @KimJiHoon R34 is not necessary. \$\endgroup\$
    – Hearth
    Commented Jun 7, 2023 at 13:34

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