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