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I am outputting a PWM signal, but the problem is that I can't generate a complementary PWM signal with dead time because this timer channel doesn't support it. So I want to use/buy an integrated circuit (IC) that generates a complementary PWM signal if I feed the PWM signal into it (with dead time). Is there such an IC available on the market?

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  • \$\begingroup\$ Simple ccts use Diode fast OFF and R slow ON into Ciss to affect dead time. \$\endgroup\$ Commented Mar 31, 2017 at 16:55
  • \$\begingroup\$ I did but couldn't find one, that is why I asked to see if someone knows. Thanks. \$\endgroup\$
    – user203
    Commented Mar 31, 2017 at 17:06
  • \$\begingroup\$ I can't help you with a dedicated PWM circuit. To be honest, I thought it often came down to what's in microcontrollers or using a CPLD/FPGA. Is the latter something you'd entertain? You can get what you like in a small cheap IC that way if you're in a position to develop and use it. \$\endgroup\$
    – TonyM
    Commented Mar 31, 2017 at 17:22
  • \$\begingroup\$ FET bridge driver ICs such as IR2103 have dead time built in. \$\endgroup\$ Commented Mar 31, 2017 at 17:55
  • \$\begingroup\$ Possible duplicate of Isolating PWM From H-Bridge \$\endgroup\$
    – user16222
    Commented Mar 31, 2017 at 18:13

2 Answers 2

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A typical way of producing what is required (deadtime & complimentary level) is via an R-C network (to produce a delayed waveform) and then feeding the two waveforms into suitable logic gates

enter image description here

for completeness, a similar reply by Andy aka

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  • \$\begingroup\$ +1 for using my picture and note that to properly make it complementary one output signal needs to be inverted. \$\endgroup\$
    – Andy aka
    Commented Mar 31, 2017 at 18:33
  • \$\begingroup\$ @Andyaka well I remembered you recently posted on this topic and to distract away from the rapid poster I went and linked that. I had to check by Masters dissertation w.r.t. what I did for an SR drive to redraw it \$\endgroup\$
    – user16222
    Commented Mar 31, 2017 at 18:51
  • \$\begingroup\$ what happens when a very high or very low dc signal is fed into the circuitry? :) \$\endgroup\$
    – dannyf
    Commented Mar 31, 2017 at 19:13
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    \$\begingroup\$ you get a response similar to your typical reply here @dannyf \$\endgroup\$
    – user16222
    Commented Mar 31, 2017 at 19:14
  • \$\begingroup\$ sorry that my comment seems to have offended you. you have a great circuit here. I hope it worked. \$\endgroup\$
    – dannyf
    Commented Mar 31, 2017 at 20:23
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take the pwm signal and run it through an invertor / NOT gate.

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    \$\begingroup\$ That doesn't add the dead time the OP wants. \$\endgroup\$ Commented Mar 31, 2017 at 17:52
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    \$\begingroup\$ You've answered a question that you haven't understood, as @OlinLathrop pointed out. And an inverter would actually add a little dead time, even when spelled properly. \$\endgroup\$
    – TonyM
    Commented Mar 31, 2017 at 17:58

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