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I'm trying to get rid of this heat problem , I know that I have to use heat sinks according to it's specification , but the problem here is there's no space for that heat sink. enter image description here

So I do have to need a method to calculate and get a idea about it's switching loss to compare these two MOSFETS to select the better one. The design that I speak is a push-pull topology and it does drain 20A peak current , switching frequency is 50KHz and how could I calculate the power dissipation due to the switching losses?

I have compared these two MOSFETS, enter image description here Is that calculating the operating point from that graph is a good idea here? Regarding the 20A in mind so there will be 1V between Drain and the source. So which means approximately 20W as heat on it's peak. Regarding the Ipeak reaches linearly so approximate heat dissparation is around 10W per two MOSFETS, so 5W per a one. So do I still need a heat sink there?

What other options that I should concern,

  1. Voltage ringing.

  2. Hi voltages causing by leakage flux. (use a clamp drive)

  3. Reverse and forward switching delays. (use a gate drive)

5W is still too hot for a design regarding without heat sink. Could somebdoy suggest any other MOSFET which does the job with less heat. Ron is very small device.

EDIT: My supply voltage is 24V and I am not using a gate drive here. I'm using 5V TTL output to drive these two MOSFETS.

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    \$\begingroup\$ Can you add a dead zone in the push-pull? \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 3 '13 at 4:50
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    \$\begingroup\$ My gut feeling is that at 50 kHz the switching losses will be considerable, even with a very good gate driver (what do you use?). For choosing a MOSFET we need some parameters you did not mention, starting with the supply voltage. If you have no place for a heatsink at all my gut feeling is: tough luck, at 20A and 50 kHz you'll need one with any MOSFET you can find. But I'm not a specialist in this field. \$\endgroup\$ – Wouter van Ooijen Oct 3 '13 at 5:38
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    \$\begingroup\$ A "dead zone" is a period when both transistors are off. Having one will reduce losses since it will be impossible to have both transistors even partially on at the same time. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 3 '13 at 6:13
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    \$\begingroup\$ Which IC? Another note: you mentioned "Ipeak reaches linearly". Dissipation is proportionally to I^2, so you can't simply halve the dissipation! I get the impression that you are way beyond your known territory, and you should leave playing with 20A to someone who is more familiar with this stuff. (I think I am more familiar with this than you are and I would not dare to design something like you seem to try!) \$\endgroup\$ – Wouter van Ooijen Oct 3 '13 at 7:18
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    \$\begingroup\$ I see you now mention 5V TTL to drive the FETs. DON'T! 5V is way to low for these FETs, and a TTL output will be way to slow. At least get a decent MOSFET gate driver, and feed it 12V or 15V. \$\endgroup\$ – Wouter van Ooijen Oct 3 '13 at 7:21
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Driving with TTL is a bad idea. At 4D drive the IRFZ44 is only guaranteeing to pass 250 micro amps. You need to follow the guidelines in the spec: -

enter image description here

The spec says it has a test to check rise times and fall times - they use a 10V pulse with an output impedance of 12 ohms.

You cannot expect to get anywhere near this performance from TTL at 4 or 5V. The gate input capacitance is 1.5nF and this needs something like a 1 or 2A drive (off the top of my head) to get the device to switch on and off at the rate you are likely wanting.

EDIT to include drive current into the gate.

It's easiest to start off with Q = CV then differentiating we get \$\dfrac{dQ}{dT} = C\dfrac{dV}{dT}\$ where

\$\dfrac{dQ}{dt}\$ equals charging current into the gate capacitor of 1.5nF.

The voltage on the gate needs to change about 10V in 20 ns hence \$\dfrac{dV}{dt}\$ = 500,000,000.

Therefore charging current (to be supplied by gate drive) is 1.5 \$\times 10^{-9} \times 500,000,000 = 0.75A\$. This means your driver ought to be able to deliver 1 or 2A as previously mentioned.

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  • \$\begingroup\$ sir using the I=CVf (f=frequency) I got the 6mA gate current for 400KHz.Where is the wrong now? \$\endgroup\$ – Standard Sandun Oct 3 '13 at 9:47
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    \$\begingroup\$ @sandundhammika The CR time for 12 ohms and 1.5 nF (gate capacitance) is 18ns. At this rate of voltage build-up on the gate, use Q=CV which becomes \$\frac{dQ}{dt} = C\frac{dV}{dt}\$ which equals current. dV/dt is 10V in maybe 20ns = 500E6 and therefore current = 500E6 x 1.5E-9 = 0.75A - this is the current you need to drive into the gate to get decent switching speed. \$\endgroup\$ – Andy aka Oct 3 '13 at 12:39
  • \$\begingroup\$ oky thank you sir. So if I use 50KHz then it would drain 20A, peak current as I calculated. Is this is a problem with the gate drive or still I need to reduce the on time? How could I calculate the power dissipation on time ? using Ron on the datasheet? Is it stable to use Ron there? \$\endgroup\$ – Standard Sandun Oct 3 '13 at 12:57
  • \$\begingroup\$ as the datasheet shows , it could not withstand 20A over 150C , where it would terminally runaway.So I still need to think about it as Woulter told that kind of current needs an heat sink? \$\endgroup\$ – Standard Sandun Oct 3 '13 at 12:59
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    \$\begingroup\$ Assuming you have at least a 10V gate drive and this gate drive is able to supply peak currents of at least 1A (pretty normal for MOSFETs) then you can use the graph of expected rise time and fall times (shown in my answer) to estimate the drain current and voltage during these changes and calculate the power loss. Personally I'd use a simulator like microcap and let it do the math for you. \$\endgroup\$ – Andy aka Oct 3 '13 at 13:01

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