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Why are gate resistors heating?

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LAN_403

Tonic

Sat Jul 05 2008, 10:25PM

Registered Member #528 Joined: Fri Feb 16 2007, 10:32PM
Location: Warsaw, Poland
Posts: 166

Gate driver consists TC4422 and TC4421, each driver has tantalum and ceramic capacitors. They supply the GDT (Ferroxcube 25mm OD, 3C90 material, 1:1:1 ratio, trifilar wiring, 15 turns with UTP wire), then half-bridge of IRFP250's.

The problem is that gate resistors heats slowly (it takes 30-60 seconds to get pretty warm). The 15V Zener, Schottky and ultra fast diodes are fine - each condunts current in one way and digital multimeter said each of them has the same voltage drop, around 700mV. There are no any shorts in gate-source. Neither on PCB. Gate waveforms seems to be fine, it's hard to measure exactly, but fall and rise times are about 80-100ns. Gate resistors are fine (except the fact they're heating), they have 10ohm resistance. The peak-to-peak voltage is something aroung 26V (-13V to +13V). There are photos:

Test configuration (don't mind unsoldered gate wires):

Waveforms with unsoldered gate wires and soldered source wires (with unsoldered source wires, I have a high-frequency ringings, but gate resistor seems to be enough):
1215296202 528 FT0 Unsoldered Gate Wire

With soldered gate wires:
1215296202 528 FT0 Soldered Gate Wire

Now, the question for 100 point - what the heck is going? :)

EDIT: Forgot to mention frequency - it's 250kHz.

GeordieBoy

Sat Jul 05 2008, 10:47PM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881

Your gate drive waveforms are perfect and everything is operating as it should. It is normal for the gate damping resistors to heat up as they dissipate the energy removed from the MOSFET gate each time it is discharged. If the drive is continuous at 250kHz this can be a few watts so they will get warm.

Tonic

Sat Jul 05 2008, 11:22PM

Registered Member #528 Joined: Fri Feb 16 2007, 10:32PM
Location: Warsaw, Poland
Posts: 166

Ah, that's so. I've used 1W resistors, so it explains why I barely can touch them. I'm going to buy 3W resistors and forget about it. Thanks for clarifying :)

HV Enthusiast

Sun Jul 06 2008, 12:02AM

Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
Location:
Posts: 3068

Basically your power dissipation in those resistors is going to be 0.5*CV^2*freq.

Take that power dissipation, and then divide up among the impedance of your gate driver and the gate resistor in that ratio.

101111

Sun Jul 06 2008, 12:05AM

Registered Member #575 Joined: Sun Mar 11 2007, 04:00AM
Location: Norway
Posts: 263

I like your setup. What are you making?

Tonic

Sun Jul 06 2008, 11:42AM

Registered Member #528 Joined: Fri Feb 16 2007, 10:32PM
Location: Warsaw, Poland
Posts: 166

Ace7one wrote ...

I like your setup. What are you making?

Mainly SSTC, but it depends how it will succeed :) I might use half-bridge for induction heating as well, or Jacob's ladder.

Dr. Shock, thanks for equation, it's sure useful.

GeordieBoy

Sun Jul 06 2008, 12:05PM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881

EVR wrote:

Basically your power dissipation in those resistors is going to be 0.5*CV^2*freq

The power dissipated in the gate resistance is actually TWICE that figure for unipolar gate drive (i.e. between 0 and +V) and EIGHT times that figure for bipolar drive (i.e. swinging between -V to +V)

The reason for this is as follows:

In order to put a certain amount of energy into the gate capacitance to turn on the MOSFET the same amount of energy must be dissipated in the charging resistance during the charging process. Then when it comes to turning the MOSFET off that stored energy must be removed from the gate capacitance. This stored energy is also dissipated in the gate damping resistance. So you get dissipation equal to the change in stored energy on BOTH the rising and falling edges of the gate drive waveform.

This makes the dissipation equation become:

P = C V^2 * freq

where V represents the TOTAL voltage swing. This point becomes important where bipolar drive is being used. If the MOSFET gate is being driven between -20V and +20V that causes 4 times as much dissipation in the gate drive resistor than if it was just being driven between 0V and 20V. Something worth keeping in mind. (Of course there are reasons why one might want to negatively bias the gate during the off-time such as combatting Miller effect or increasing noise immunity.)

Dissipation in squarewave gate-drive circuits rises rapidly with voltage-swing and with frequency. Energy is wasted in order to charge the gate, and then this stored energy is wasted again when discharging it! This is one of the main motivations behind going to resonant gate drive for frequencies above a MHz or so. In this case the gate charge is recycled, dramatically reducing drive power requirements and dissipation.

-Richie,

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