Originally posted by bigbigblue+Mar 19 2006, 02:24 PM--><div class='quotetop'>QUOTE(bigbigblue @ Mar 19 2006, 02:24 PM)</div><div class='quotemain'>I've had a look at the datasheet for the IRF9Z34 P Channel MOSFET from IRF (datasheet can be found here : http://ec.irf.com/v6/en/US/adirect/ir?cmd=...ductID=IRF9Z34).
It will pass 18amps - more than enough for my application, with an RDS(on) of 0.14 ohms (twice that of the n-channel IRF540 I am currently using).
At 8 amps, the IRF540 will dissipate 4.928 watts, whereas the IRF9Z34 will dissipate 8.96 watts (ie twice as much heat). The reason I am having a problem with the N channel MOSFETS is that they got hot due to a significant proportion of their operation being in the linear area - but I doubt whether they are dissipating as much as the P channel will when it is hard on. Therefore I am not sure I gain much from using a P Channel MOSFET (other than ease of driving them).
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That is the only gain, ease of driving, no need to generate higher gate voltage. As you already have 32V supply handy and you don't need to generate higher gate voltage, then the advantage is not there anymore.
There are two main sources of power dissipation in switching, the conducting (saturated) power loss and the switching power loss. To minimise the conducting power loss you need to ensure the device is in saturation (enough gate voltage) and select the lowest Rds(on) you can afford. To minimise the switching power loss basically you need to ensure that the device switches as fast as possible (not counting ZVS or ZCS schemes).
If you have a scope, you could check the switching waveform and see which one is your main problem, conducting (i.e. device not in full saturation) or switching (i.e. slow switching transition).
If you could capture the switching waveform and post it here that would be very helpful in diagnosing why the MOSFET gets hot.
One easy mistake to made at high frequency switching is to mount the MOSFET too far from the gate driver circuit. Long track or mounting wire on the gate would slow it right down.
Ignoring the capacitance variations due to different Vds and Vgs and feedback through the Miller capacitance, that amount of charge is correct. However, I much prefer to say that in order to raise the Vgs to 10V, the gate needs approx. 69nC of charge. To be fully on, the MOSFET is also affected by dynamic (transient) characteristics of the channel and the circuit (this is related to the t(on)/tf and t(off)/tr you mentioned below).Originally posted by bigbigblue@Mar 19 2006, 02:24 PM
I wonder if I could run by you my understanding of the gate charge and how much current is required to achieve a hard switch on in a given time ?
In the IRF540 datasheet, it states :
Total gate charge = 69nC (nano-Coulombs ?)
Gate to source charge = 13nC
"Miller" charge = 37nC
(all at Id = 29A, Vds = 80V, Vgs = 10V.
So, as I understand it, I need to shove 69nC of charge into the Gate at 10V above the Source Voltage in order fot the gate to be fully open. Is this correct?
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The gate charge value is very useful in calculating the gate drive loss and charging/discharging currents, not very useful in calculating the ton/toff times.
Yes, charging the gate to 10V.Originally posted by bigbigblue@Mar 19 2006, 02:24 PM
Ignoring the CR time constant of the gate capacitance and any losses for a moment, to charge the Gate in :
1 nS requires 69 Amps of charge current
10 nS requires 6.9 Amps of charge current
100 nS requires 690mA of charge current
1 uS requires 69mA of charge current
etc etc
Again, is this correct (I think it is OK as 1 Amp = 1 Coulomb / Second) ?
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Is this 1.3nF the ciss? You need to differentiate the ciss with total gate charge. Have a look at the typical ciss measurement conditions/circuit and compare it to your circuit. ciss is more useful in AC analysis.Originally posted by bigbigblue@Mar 19 2006, 02:24 PM
The input capacitance of the IRF540 is 1300pF ( = 1.3nF?). Assuming I have a 32V power supply to provide the Gate voltage and the max Source voltage is 12v, and I have a 15 v zener between Gate and Source (so I can guarantee the gate is no more than 15V above the Source).
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You've said it yourself, ignoring the tf and the tr. In fact, when you drive the gate sufficiently hard, it is no longer the deciding factor in tr and tf. The datasheet usualy gives these values, such as tf = 20nS and tr = 100nS with near perfect (very large current capability) gate drive under specific Vgs, Vds, Id and load type. These would be the best timing that you would get under similar conditions. Your load type (inductive in your case) and circuit would also affect the switching time.Originally posted by bigbigblue@Mar 19 2006, 02:24 PM
As I have a 32V supply, to limit the current to the Gate to 690mA (assuming I want to charge the Gate in 100nS) I would need a current limiting resistor of 46.38 ohms (=32 / 0.69).
Now, taking the (CR) time constant of the gate into account:
46.38 x 1.3 nS = 60nS (to achieve 66% of the charge)
or 120nS to achieve 88.4% of the gate charge (66% + (66% of 34%))
etc etc
So, I should be able to get something close to my wished 100ns turn on time. I am of course ignoring rise times and turn on delays of the MOSFET here. Or have I got this badly wrong ?
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<!--QuoteBegin-bigbigblue@Mar 19 2006, 02:24 PM
Now to a problem I forsee. As I have a 32v supply and a current limiting resistor of 46.38 ohms (giving a 690mA current), the power dissipation of that resistor is 0.69 x 0.69 * 46.38 = 22 Watts approx - one big resistor - and I need one for each high side MOSFET Gate.
All comments on the above are most welcome.
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[/quote]You do not take into account the fact that the 690mA current only flows for a very very short duration (a few nS?) with relatively low duty cycle. Once the gate is charged the current ceases to flow (in fact it starts decreasing the moment the gate starts charging). The average power dissipation is, therefore, much much lower than the 22W you are suggesting (ignoring any current shunted by the zener). There is a way to calculate this, but you have to get hold of the thermal transfer characteristics of the resistor that you are going to use (quite difficult to get and a waste of precious time). The easiest way is just trial and error, calculate the shunted current and find the appropriate resistor power to sustain that power and just add 0.5W to start with. Increase the rating if it was too hot to touch during operation. Use metal film if you can.