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Here I am attaching step down buck converter circuit. While testing, mosfets on high side get hot extremely high even for 100W load. Low side has no problem. All mosfets are fixed on very big heat sink. Anyone please help me to overcome this problem. Step down Buck Converter

Components used are as per given in circuit. PWM frequency is 50KHz & max duty is 80%. Gate voltage observed is always >10v.

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    \$\begingroup\$ It's probably my insomnia, but I can't seem to understand the sequence of operation in the given schematic. The low side doesn't seem to do anything. Regardless, based on what you are describing, I would suspect that the Mosfets themselves aren't switching fast enough. You might need more gate drive or lower frequency. We'll need part numbers to really find out. \$\endgroup\$ – Sean Boddy Sep 28 '15 at 9:22
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    \$\begingroup\$ I have given part number of each component in the pic. Mosfet used is p80nf55. we have tried with p80nf10 & IRFZ44n. But result was same. Low side & High side is working as per pwm from IR2110. During 'on' time of pwm high side will turn on & low side will turn on during 'off' time of pwm. current through load is always same direction. \$\endgroup\$ – Thahseen K Muhammed Sep 28 '15 at 10:05
  • \$\begingroup\$ Does the IR2110 have anti-shoot-thru circuits or are you implementing that in the signals that feed the chip? \$\endgroup\$ – Andy aka Sep 28 '15 at 10:35
  • \$\begingroup\$ Make it SIX PHASE, solve everything with elegance. \$\endgroup\$ – Avijit Das Apr 30 '16 at 8:49
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the IR2110 high side driver simply isn't capable of providing enough drive current to drive all of these MOSFETS in parallel. Each mosfet has an input (gate) capacitance of typically 7000pF at turn on with the 10R gate resistance (ignoring the diode) the required input current is initially > 6A - the IR2110 is only rated for 2A - you will be getting a very slow turn on of the MOSFET's, hence lots of heat. Try one IR2110 per MOSFET

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  • \$\begingroup\$ Thank you. But I think, adding IR2110 for each MOSFET is quite impractical for us. Is there any other remedy available? We are already driving 12 mosfet from one IR2110 in other circuit and it is working okay. \$\endgroup\$ – Thahseen K Muhammed Sep 28 '15 at 11:08
  • \$\begingroup\$ You have to find a way to turn the MOSFET's on and off faster, essentially this comes down to, more gate drive current - more or different driver, or reduced gate capacitance on the MOSFETS - but 7nF is not unusual for high power MOSFETS. What MOSFETS did you use in your previous design? \$\endgroup\$ – Icy Sep 28 '15 at 12:55
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    \$\begingroup\$ @ThahseenKMuhammed, If everything is as you assume it to be, conditions will be in a known state. Otherwise, your assumption is invalid. This is science. And for the life of me I can't imagine why you would assemble 6 pairs of 80 amp mosfets, enough to handle 11 Kilowatts in your given application, and not want more driver ICs. \$\endgroup\$ – Sean Boddy Sep 28 '15 at 15:57
  • \$\begingroup\$ @ThahseenKMuhammed, the problem is almost certainly being caused by tiny differences in the mosfets and their connections, leading them to not be switching at the same time, which will be especially true when having to share a single drive source. Depending on how the drive signal is distributed, additional differences accrue as the conductor to the gate changes. Additional driver ICs should make the rise and fall time fast enough to fix this, provided you don't have any shoot through problems. \$\endgroup\$ – Sean Boddy Sep 28 '15 at 16:26
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    \$\begingroup\$ Is the inverter using the same switching frequency as the buck converter? Also in the inverter you probably have no possibility of a direct current path from supply, through high side, then low side back to supply. It is most likely feeding two halves of a transformer either through one set of FETS or another? In which case slow turn on / off may matter less. \$\endgroup\$ – Icy Sep 29 '15 at 13:25
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The short answer to why are the high side FETs heating up (as already given by I Chodera) is switching loss due to insufficient gate drive. But, let's use some Baby Math to give a rough analysis of what that means.

Gate Drive:

Not only are the high side FETs hot, but the IR2110 should be pretty toasty too, driving all those FETs. Power lost in the gate circuit, and that will be mostly in the IR2110, is:

\$P_{\text{Gate}}\$ = \$V_{\text{drv}} f_{\text{pwm}} Q_g\$ = (12V)(50kHz)(6)(150nC) ~ 0.5W

Since the thermal resistance of the IR2110 is ~ 100C/W, it will have a temperature rise above ambient of ~50C. Let's say ambient is 50C, so that junction temp is 100C. Who cares? Well, \$R_o\$ (output resistance) of the IR2110 is ~8Ohms at 25C, but it is a MOS device, so resistances at 100C are about 1.5 times those at 25C. With a heat elevated \$R_o\$ of ~12Ohms (plus about 3Ohms of additional gate circuit resistance), the IR2110 will not be driving as hard as you think.

Switching Loss:

Most of the switching action in a topology like this takes place while the gate drive processes the Miller plateau charge (\$Q_{\text{mp}}\$). Normally the rising and falling times (\$\tau \$) are different, but here they will be equal (and later combined, 2 \$\tau \$) because, Baby Math. So, the FETs switching time is going to be:

\$\tau \$ ~ \$\frac{2 R_g Q_{\text{mp}}}{V_{\text{drv}}}\$ = \$\frac{2 \text{(15 Ohms)(6)(75 nC)}}{\text{10V}}\$ ~ 1.4uSec

For the loss calculation, peak inductor current will be used as a simplification.

\$P_{\text{sw}}\$ = \$I_{\text{pk} } \tau V_{\text{ds}} f_{\text{pwm}}\$ = \$\text{(24V)} \text{(9A)} \text{(1.4uSec)} \text{(50kHz)}\$ ~ 15W

From the equations you can see that reducing the loss could be accomplished by:

  • Improving the gate drive.
  • Use fewer FETs, another way to reduce \$\tau\$.
  • Lowering the switching frequency.

If 100W is all the power to be processed, it makes no sense to have 6 parallel power modulators. One would do.

A good reference for loss in Synchronous Bucks is Fairchild AN6005.

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  • \$\begingroup\$ Okay. But actually what is remedy against this? \$\endgroup\$ – Thahseen K Muhammed Sep 30 '15 at 5:23
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    \$\begingroup\$ @ThahseenKMuhammed, if that one bulb really is the only load, go down to one set of Mosfets, and consider lowering the switching frequency. \$\endgroup\$ – Sean Boddy Sep 30 '15 at 5:29
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    \$\begingroup\$ gsills and I have already told you! more gate drivers, fewer FETS or lower switching frequency. \$\endgroup\$ – Icy Sep 30 '15 at 16:49
  • \$\begingroup\$ @SeanBoddy,100W bulb is just used for testing.Our actual load is 400W compressor. \$\endgroup\$ – Thahseen K Muhammed Oct 1 '15 at 4:25
  • \$\begingroup\$ @ThahseenKMuhammed, at 12 volts that is still only a current of 33 amperes. Even if you needed the buck converter to be able to handle the start surge, you could probably get away with four sets of MOSFETs. My entire discipline tells me that you need to model an actual resistive inductive load, analyze for switching losses as above and select a more appropriate switching frequency. \$\endgroup\$ – Sean Boddy Oct 1 '15 at 4:33
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Upgrade the drive circuit using a single bjt or cmos totem pole to avoid multiple IR2110 shoot thru error. The gate drive MUST be dimensioned to provide +/-20V at 5A, at 200KHz. Redimension gate resistors within +/-20% to align switching times using a scope. Anything serious can't go wrong, and this is a >5KW power processor. Alternatively, a you can make this a six phase buck converter by using six IR2110, that should be the easiest way since you're using a microcontroller, period.

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