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I am constructing a 20 W buck converter. The voltage is bucking but the current is not, the current that the power supply is producing is going straight through to the load. Please help.

I am suppose to apply a 12 V input (1.66 A maximum) and get a 5 v output with a maximum of 4 A of current.

schematic

simulate this circuit – Schematic created using CircuitLab

Also is it necessary to use the ground terminal (green) of the bench power supply when testing?

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    \$\begingroup\$ Ah the generic green terminal on the generic power supply feeding a circuit that is driven by the one size fits all generic duty cycle. \$\endgroup\$
    – Andy aka
    Commented May 1, 2018 at 12:54
  • \$\begingroup\$ How are you driving your MOSFET? What do you mean by "the current is going straight through to the load"? It sounds like you might be operating this as a linear regulator, dissipating significant power in your FET. \$\endgroup\$
    – Hearth
    Commented May 1, 2018 at 12:55
  • \$\begingroup\$ Should i connect the green ground terminal of the bench power supply to the black negative terminal when testing? \$\endgroup\$
    – Raees Khan
    Commented May 1, 2018 at 12:55
  • \$\begingroup\$ I am using a TC4428A MOSFET driver which is connected to an ATTINTY85 output. The ATTINY85 is used to generate a PWM at 250 kHz \$\endgroup\$
    – Raees Khan
    Commented May 1, 2018 at 12:57
  • \$\begingroup\$ There's no particular reason not to ground your circuit, but also no particular reason to ground it, from what we know from your question. So, it doesn't really matter whether you connect ground to negative or not, from what I can see. \$\endgroup\$
    – Hearth
    Commented May 1, 2018 at 12:58

2 Answers 2

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TC4428A is a LOW SIDE driver, that mosfet is not in a low side configuration, what you need is a high side driver and boost capacitor (You have to get the gate up above the supply rail by 10V or so to switch the mosfet on fully).

As it is your gate can only swing up to the supply rail, which will leave the source Vgs below the supply, thus in the linear region and dissipating considerable power.

You also want some input capacitance and some high frequency output capacitance, a few uF or so of MLCC would probably do.

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  • \$\begingroup\$ +1. Was considering this myself, but didn't know enough to be confident making it an answer. \$\endgroup\$
    – Hearth
    Commented May 1, 2018 at 14:35
  • \$\begingroup\$ Thanks, I will try using the TC4432 driver as well as a MIC5018. \$\endgroup\$
    – Raees Khan
    Commented May 1, 2018 at 14:40
  • \$\begingroup\$ MIC5018 is 10V abs max, not a good choice on a 12V rail! The 4432 lacks the oh so useful boost pin, you can probably emulate it, but it will be an annoying fiddle. Something like a UCC27212A would probably be ok, just use the high side half of the part. \$\endgroup\$
    – Dan Mills
    Commented May 1, 2018 at 15:16
  • \$\begingroup\$ I am unable to get the UCC27212A at my local electronic store, will a IR2112 MOSFET Driver work? @DanMills \$\endgroup\$
    – Raees Khan
    Commented May 1, 2018 at 23:02
  • \$\begingroup\$ The 2112 is a part really designed for high voltage applications, it has a rather weak peak output current on the high side driver, but if you can live with limited ability to slew a high capacitance gate I see no reason to think it is not workable. You will need an external bootstrap diode and cap of course. \$\endgroup\$
    – Dan Mills
    Commented May 1, 2018 at 23:10
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The output current needs to be supplied by the input power supply, in bursts. It's only the average input current that's supposed to be 1.66A, the actual current will cycle between none, and about 4A.

If the input power supply has overcurrent protection, it may shut down on these peaks, messing with your converter. You need an input capacitor across the power supply, which can source these short term current peaks.

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