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### Topic: Calculating heat dissipation when self clamping MOSFETs (Read 8985 times)previous topic - next topic

#### jtw11

##### Mar 09, 2013, 09:40 pmLast Edit: Mar 10, 2013, 11:47 am by jtw11 Reason: 1
Hi guys,

I'm drive a couple of solenoids with MOSFETs, and so of course - it's an inductive load.

Besides the plethora of clamping techniques out there; Zeners to ground, Zeners to supply, RC snubbers, RCD snubbers, lossless inductive snubbers etc - it would appear more and more devices are beginning to be sold as 'self clamped', in that they have a Zener from the source to the gate, to switch the device on when source voltage exceeds the Zener voltage. (and of course a Zener from the gate to ground, to prevent overvoltage)

However, I'd like to replicate this myself as these devices tend to have higher on resistance values than really available with 'normal' MOSFETs - I guess a side effect of squeezing everything onto the same die. So, these devices have the following internal structure - so i've replicated it as a schematic.

The question is, knowing the energy stored in the inductive load - does anybody have any idea how to calculate the extra power dissipated in the MOSFET when it switches on to dissipate the voltage spike?

EDIT - I just found an app note regarding this...

Quote
The active clamp feature is shown in Figure 5. During the off-state, the power MOSFET is turned back on when Vds > Vzener +Vf,diode + V threshold,MOSFET. Also, there are two large resistances, the resistor and the MOSFET, within this current path to help dissipate energy. (Note that during the active clamp the FET is in a linear, or high-resistance, mode.) The load demagnetizes quickly during active clamp because the stored energy in the load is dissipated across a large po- tential [Vcc - Vclamp]. The larger the difference, the faster the demagnetization. The energy dissipated by the IPS during the Active Clamp sequence is Vclamp x ?Ids(t)dt.
The energy stored by the inductance is [Vclamp - Vcc] x ?Ids(t) dt. Thus, it is important to notice that during the active clamp sequence, the device dissipates more power than the load. (The current through the load and the IPS is the same but the voltage across the IPS is higher than the one across the load).

Here's the app note for anybody interested http://www.irf.com/technical-info/designtp/dt99-4.pdf

#### dc42

#1
##### Mar 10, 2013, 11:37 am
Power mosfets can tolerate absorbing a certain amount of energy in avalanche mode (without any clamping or flyback diodes). Look at the figure on the datasheet for repetitive avalanche energy, in the absolute maximum ratings section. In practice, this typically means you can use an unclamped power mosfet to drive a small transformer, but not a large motor or solenoid.

if you use a self-clamping zener between gate and drain, then the energy from the inductive load is still dissipated in the mosfet. The difference is that the mosfet is operating in the linear region, not the avalanche region. There will still be a spike in the die temperature. Whether the energy it can absorb in this mode is greater than the maximum avalanche energy is difficult to say - I've not seen any figure given for maximum linear mode energy burst given on a datasheet, nor on the thermal capacity of the die. So this arrangement is not a generic substitute for a flyback diode.

If you connect a flyback diode across the inductive load, then the flyback energy is mostly dissipated in the inductive load, with a little dissipated in the diode. So no worries about the energy-absorbing capacity of the mosfet. The disadvantage is that the current in the inductive load decays more slowly, which can be a problem in some applications.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

#### jtw11

#2
##### Mar 10, 2013, 11:45 amLast Edit: Mar 10, 2013, 11:53 am by jtw11 Reason: 1

Quote
The disadvantage is that the current in the inductive load decays more slowly, which can be a problem in some applications.

This is my exact problem, a flyback diode directly across the inductor isn't possible here - the protection must lie at the switch itself.

It seems to me, there are just so many ways to clamp the drain voltage, that I'm lost for choices.

Perhaps a solution would be to have the active clamp Zener switch on a second small transistor that connects the drain of the power MOSFET to a power resistor in a low side configuration?

EDIT - The Zener wouldn't actually have to switch a second transistor on at all, the drain could permanently be connected to the power resistor, in series with a Zener to ground - when the Zener voltage is exceeded the back EMF is burned off in the power resistor.

#### dc42

#3
##### Mar 10, 2013, 11:52 am
Here are some other ways of speeding up the current decay:

1. Resistor in series with flyback diode. This makes the current decay faster, but also increases the voltage that appears at the drain of the mosfet. For example, if you choose a resistor with a value equal to the resistance of the inductive load, then the current will decay twice as fast, but the peak drain voltage will be twice the supply voltage plus the forward voltage of the diode.

2. As (1) plus capacitor in parallel with the resistor. This allows you to get a faster current decay than with a resistor alone, for the same peak drain voltage. The optimum value of the capacitor depends on the load inductance.

3. Power zener diode in series with flyback diode.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

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