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picture 1

picture2

In the first picture, the capacitors are used in driving an inverter. In the second one the capacitor is used at the input of an amplifier.

What are the functions of each capacitor?

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  • \$\begingroup\$ What horrible tool is this, what does the text even say? 33u and 100n respectively? What more? \$\endgroup\$ – Lundin Jul 3 '19 at 10:56
  • \$\begingroup\$ i really just follow the schematic, don't understand why those value are selected \$\endgroup\$ – rahul Jul 3 '19 at 11:33
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Your first circuit uses an IRF2110 MOSFET driver.

The capacitors you refer are there to smooth the voltage that the IRF2110 generates to drive the FET gates. The IRF2110 is used to make a higher voltage in order to drive the gates of the FETs turn the FET fully on. FETs require a certain voltage differnce between gate and source in order to switch fully. If your circuit can't provide the needed voltage, you use something like the IRF2110 to generate it using the available voltage.

There are two capacitors because capacitors aren't perfect. If they were, a single large capacitor would be all you need. Because of imperfections (large capacitors act a bit like an inductor at high frequencies) you use a large capacitor to catch the low frequency noise and a smaller capacitor to catch the higher frequencies.

In your second circuit, the capacitor you marked functions as a filter. It works together with the 1.2M resistors (two in series, so 2.4M total) to form a low pass filter with a cutoff frequency of around 30Hz. Any frequency above 30Hz will be reduced in amplitude.

I don' know why you need that filter with that cutoff there because the rest of the circuit is missing (it would help if you had explained where you got the circuits.)

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  • \$\begingroup\$ great, excellent explanation, i use the second circuit to set up the offset value, so arduino can read, but got much thing i don't understand \$\endgroup\$ – rahul Jul 3 '19 at 11:30
  • \$\begingroup\$ @rahul: Check back. I had to correct my answer. \$\endgroup\$ – JRE Jul 3 '19 at 11:33
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The caps are needed in this highside drive circuit to hold charge for the Gate of the highside N channel fet .The cap value needed is a function of on time .In these bootstrap circuits the cap can only gain charge when the highside Fet is off .When the highside fet is on charge bleeds away due to the gate source pulldown resister and due to chip losses .

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  • \$\begingroup\$ why is it hooked parallel with another capacitor? \$\endgroup\$ – rahul Jul 3 '19 at 10:41
  • \$\begingroup\$ @rahul.Big cap little cap for wide frequency range before big MLC caps \$\endgroup\$ – Autistic Jul 3 '19 at 11:04
  • \$\begingroup\$ well thanks for the explanation :) \$\endgroup\$ – rahul Jul 3 '19 at 11:38

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