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I have designed this Buck Converter which convert 60VDC to 12VDC at 10A. Switching Freq 100KHz. Facing MOSFET too much heating issue. MOSFET ON time and OFF time is decided by a UC3845B based circuit. MOSFET gate is biased with 2.2R resistor and pull down of 5.1K is there any way to reduce MOSFET heating? I've increased MOSFET rating to 110A 80V. Previously was 75V 75A but no succcess.

Edit 1: Updated Schematic for better understanding.

Edit 2: Previously tried this INFINEON MOSFET. Heating was less.

Then used this ST MOSFET. Heating was more in ST MOSFET

enter image description here enter image description here

Hello, here is an update, can I use below circuit as bootstrap? here in place of 5V Input either I can use 60V directly or 12V from output.

enter image description here

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    \$\begingroup\$ You need to show how the MOSFET gate drive works as a circuit and note that an N channel MOSFET driven as a source follower may not be very efficient unless your circuit jumps through a few hoops to produce sufficient gate drive voltage. \$\endgroup\$ – Andy aka Apr 6 at 15:19
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    \$\begingroup\$ The MOSFET you are using does not return anything in a search; include a link to the datasheet. \$\endgroup\$ – Peter Smith Apr 6 at 15:22
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    \$\begingroup\$ If the UC384x is referenced to your GND, no wonder why the MOSFET won't be properly turned on. You need to install a simple bootstrap architecture to do that or use a transformer. If you confirm you don't have any of these, I'll suggest a simple driver I have successfully tested in a separate answer. \$\endgroup\$ – Verbal Kint Apr 6 at 15:48
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    \$\begingroup\$ Could you use different ground symbols for port 4 and 6, and give the ports more descriptive labels please? \$\endgroup\$ – Oskar Skog Apr 7 at 8:11
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    \$\begingroup\$ That’s a source follower. You need a high side gate driver with either bootstrap or floating supply for it. \$\endgroup\$ – winny Apr 7 at 9:05
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If it is confirmed the MOSFET needs a better drive then the best is to advise a suitable circuit. A simple bootstrap circuit described by Monsieur Balogh in a TI application note can do the job nicely in a cost-sensitive application. As noted in some of the comments, the UC384x was not really meant for hi-side drive - unless you make it entirely float and tie its GND pin to the MOSFET source and supply the IC via the rectified buck output - but this little circuit does the job fine:

enter image description here

Below is the circuit I tested with component values:

enter image description here

This is excerpted from a seminar I taught in an APEC seminar in 2019.

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    \$\begingroup\$ And, what about the IC's supply? Need a pre-regulator for that. So even more parts... that cheap old IC is starting to get expensive. Not a good approach really. \$\endgroup\$ – hacktastical Apr 6 at 22:45
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    \$\begingroup\$ It's indeed not a panacea for a buck converter supplied from a 60-V source. It is however a possible solution if you wanna tinker with a buck and the only controller you have on hand is good old UC384x. \$\endgroup\$ – Verbal Kint Apr 7 at 6:23
  • \$\begingroup\$ @VerbalKint Sir, yes, in my circuit, same is done that rectified output is given as power source to IC \$\endgroup\$ – Hrishikesh Dixit Apr 7 at 7:53
  • \$\begingroup\$ Sir, VDRV pin Voltage same as VCC or VIN i.e. 60VDC will work? \$\endgroup\$ – Hrishikesh Dixit Apr 8 at 12:58
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    \$\begingroup\$ Probably if you add also a resistance between the gate and the source of \$Q_7\$ to force the Zener conduction with a few mA. \$\endgroup\$ – Verbal Kint May 11 at 6:21
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MOSFET ON time and OFF time is decided by a UC3845B based circuit

The UC3845B has a push-pull type output but, unless this chip is powered from a supply that is several volts higher than the 60 volt rail (as shown in your circuit) you won't efficiently drive the MOSFET in your buck regulator. Given that the UC3845 is only rated at a maximum of 36 volts, you are likely driving the MOSFET very, very ineffectively and it will get very warm on load.

The gate voltage needs to exceed the main supply voltage by around 10 volts in order to get source and drain ohmically connected at a low value. This is a problem with source follower MOSFET configurations and they way around it is to use a proper "high-side MOSFET driver" chip.

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  • \$\begingroup\$ Sir, the max voltage reached to 72 V as it is an EV. 3845 has an internal zener of 36V. What should be the Gate triggering voltage? \$\endgroup\$ – Hrishikesh Dixit Apr 7 at 7:46
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    \$\begingroup\$ You now have two circuits drawn with different packages and both have a ground connection that appears to be at contradictory places. If you want help with this you are going to have to explain why you don't have one consistent circuit (can lead to errors also) and, if the two circuits you now show are what you have always had, what waveforms you are getting. @HrishikeshDixit \$\endgroup\$ – Andy aka Apr 7 at 9:33
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I’m guessing the N-FET isn’t being turned on all the way. It's being biased in linear mode and so has a substantial IR drop across it, which is being shed as heat. You don't want that.

How bad is it? Let's assume the FET Vgs threshold is about 4.5V. Then when it's on:

  • Vd = 60V
  • Vg = 36V - 2V = 34V (limited UC3845B output swing)
  • Vs = (Vg - Vth) = (34 - 4.5V) = 29.5V
  • Vds = (Vd - Vs) = (60 - 29.5V) = 30.5V

If you're drawing, say, 1A from the supply, that approximately 30-35W you're dissipating in the FET, peak.

What keeps it from frying immediately is the stepping ratio that determines the on-time of the FET:

  • Vout/Vin * W = 12/60 * 35W = 7W

And that's at only 1A. Clearly, this isn't workable.

To fix it, the N-FET gate needs to be brought all the way above 60V, to at least 65V to 70V, to ensure the FET is fully turned on to its lowest Rds(on).

What happens when you do that? Here's the peak wattage shed in the FET:

  • Vd = 60V, Vg = 65V
  • Vs = about 60V (transistor is fully on)
  • Vds = (Vd - Vs) = 0, or close to it

So in theory almost no power gets shed in the FET. In reality, this will be:

  • Iout^2 * Rds(on)
  • 10^2 * 0.020 ohm = 2W peak at 10A output

With stepping ratio being 12/60, the FET is on about 20% of the time:

  • 2W * 12/60% = 0.4W

That's very manageable for a TO-220 FET.

How to do this? You need a bootstrap circuit to generate the higher gate drive voltage (about 5-10V above Vin), and a gate driver that accepts that voltage to make the above-Vin gate signal. The bootstrap voltage can (and usually is) generated from the inductor flyback via a diode and capacitor.

Problem is, the UC3845B is not a bootstrapped high-side driver. It's really designed to be a low-side driver for a flyback topology. Further, it's limited to +36V. For both reasons it’s a poor choice for this application.

You could mess about with making a bootstrap + level shifter, but why? Select a different device instead. Example: this 75V input dcdc controller from TI (bonus: it’s synchronous so your supply will be more efficient): http://www.ti.com/lit/ds/snvsai4/snvsai4.pdf

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  • \$\begingroup\$ what I understood is right now the gate triggering voltage is 4.5V which is output from 3845B. To reduce heat, I need to increase gate Trigger voltage to 60V or more. Is that right? Request you to please correct me if I'm wrong as I'm new to this. Thanks for your kind help. \$\endgroup\$ – Hrishikesh Dixit Apr 7 at 14:55
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    \$\begingroup\$ The N-FET gate needs to be at least 4.5V above the source, and preferably as much as 10V to ensure the lowest Rds(on) for the FET. This means you need to make a drive voltage of 64.5 to 70V (4.5 to 10V above Vin.) \$\endgroup\$ – hacktastical Apr 7 at 15:22
  • \$\begingroup\$ The ST MOSFET which I'm using has gate voltage of +/- 20V whereas of INFINEON is +/- 10V. Is it okay to give MOSFET gate voltage more than specified in datasheet? Need your guidance. Thanks \$\endgroup\$ – Hrishikesh Dixit Apr 7 at 17:26
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    \$\begingroup\$ In a word, NO. You cannot exceed the maximum Vgs, otherwise the FET will break down. The bootstrap technique references Vin, so you can limit the gate voltage with a zener diode or other method. The devices with internal bootstrap generation do this for you. \$\endgroup\$ – hacktastical Apr 7 at 17:29
  • \$\begingroup\$ Vd = 60V Vg = 36V - 2V = 34V (limited UC3845B output swing) Vs = (Vg - Vth) = (34 - 4.5V) = 29.5V Vds = (Vd - Vs) = (60 - 29.5V) = 30.5V Sir, apologies in advance to trouble you but, 3845 datasheet says that output pin will be at 13.5v, which means Gate of MOSFET will get 13.5 V. Can you please tell the above calculation Vg = 36V-2V = 34V? \$\endgroup\$ – Hrishikesh Dixit Apr 8 at 12:12
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Pls reduce the resistance in Vin to maintain 15V across the IC UC38xx . The MOSFET minimum required gate drive voltage is 10v other wise it will be in ohmic region and generate heat as well as reduce power due to low contuctance

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You can check if a heat sink is needed by calculating (at least theoretically), by checking the numbers in the datasheet.

Normally these are called Tj (Junction temperature) and all related heat dissipation figures are shown together.

Also read https://en.wikipedia.org/wiki/Junction_temperature.

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Designed output power = 12V x 10A = 120W

High efficiency say 95%, heat dissipate = 120W x 5% = 6W

You have to use a large heat sink with fan.

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MOSFET heating was mainly due to Inductor. Inductor Flyback or Back EMF was causing problem. As said in Answer by @Verbal Kint sir, the ground of 3845B was completely floating as it was tied to source of MOSFET. Previously used PQ 32X20 core inductor with 0.4mm wire (don't know turns). Temperature was upto 100 degree C. Maybe logically more turns could cause more Back EMF or Flyback. Now used a FeSiAl core with 1 mm wire gauge and 14 no of turns. Noe, temperature is upto 60 degree C. Maybe using a bootstrap circuit or a built in IC will do more effect. But PCB was designed and needed to solve the issue badly. Thanks everyone here for help. Specially Verbal kint sir and hacktastical sir.

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