[SOLVED] How to protect my MosFETs???

[KerimF]

Fortunately you do a minor mistake once a while... otherwise I would likely have a serious complex :grin: since I keep doing a lot

 

[faraj]

Yes you are right, I'm obviously not as good as my PC in multitasking (doing two thing at the same time).
Ignore my post, it applies only to the four Nmosfet version.

Alex

so, the circuit in post #7 with 12 VDC supply and without 1k resistors in parallel with 10k resistors will work fine without blowing my transistors?

 

[alexan_e]

Added: I suppose that behind R10 and R9 there is a rail-to-rail output.

What do you mean?
It doesn't matter if the transistors are driven will a 3 or 5 or 12v voltage.

Alex

 

[FvM]

so, the circuit in post #7 with 12 VDC supply and without 1k resistors in parallel with 10k resistors will work fine without blowing my transistors?

During gate voltage transistion, you'll get a "shootthrough" current peak of about 10 - 15 A at 12 V supply. If the transition is fast enough, this won't damage the transistors but cause excessive losses.

 

[alexan_e]

so, the circuit in post #7 with 12 VDC supply and without 1k resistors in parallel with 10k resistors will work fine without blowing my transistors?

Yes your lower mosfets will work fine with 12v supply but there is still a problem in your circuit because you can not apply deadtime to the mosfet.

Alex

 

[KerimF]

so, the circuit in post #7 with 12 VDC supply and without 1k resistors in parallel with 10k resistors will work fine without blowing my transistors?

Oh no!... please let us be with the one on post #17
Just to be sure, did you read my post #19? Any comments?



---------- Post added at 13:16 ---------- Previous post was at 13:13 ----------

What do you mean?
It doesn't matter if the transistors are driven will a 3 or 5 or 12v voltage.
Alex

I meant not an open collector as for my ATMEL C51 (or open drain) :wink:

 

[alexan_e]

OK, you meant push-pull output, I misunderstood the rail to rail output as referring to the voltage level.

Alex

 

[faraj]

hi everyone, i made the circuit and tested it with resistive load and it is working properly, but i've not tested it with inductive load. i'll let you know the results after
testing the system with inductive load.

 

[faraj]

Here I have another problem now. I use the circuit in post #17 and I have a resistive load about 20 mA and I switch the polarity of the load. when switching in low frequency (lower than 40HZ) there is no but as I increase the frequency the PowerMosfets heat up and in a high frequency like 5 KHz the heat up very fast and extremely. I checked the datasheet of the IRF640 which I'm using in this circuit and it says 1MHz. Now, how could it happen? A 16 Ampere/1 Mhz PowerMosFETs cannot stand a resistive 20 mA/5 KHz load? and how can i cool them down because i want to use them for driving an inductive 8 A/ 1 KHz load and in these conditions heatsinks just won't be enough for cooling.

thanks in advance for taking time.

 

[KerimF]

Please let us remember always that the hardware AND software should run in harmony for the controller to give what we expect.

For example, which is the dead time you have decided on?
I mean the time when both Q5 and Q6 are open (post #17) hence Q1 and Q3 (P-MOS) are off.

As you know, a MOSFET has a relatively high input capacitance, mainly the power ones.
When Q5 or Q6 is turned on (becomes saturated, its collector close to ground), the driven Cgs will be discharged rather quickly.
But when Q5 or Q6 is off, Cgs will have to charge via a 10K resistor (R5 or R6) and the PMOS will not turn off very fast.
So if there is no dead time at each transition, each leg will act as a short circuit between 12V and ground for a very short time.
The effect of this malfunction (caused by the software which is not compatible with the circuit) will be noticed more when the frequency is increased.

You are fortunate that your bridge didn't give a smoke yet

Kerim

 

[Raza]

Hello Faraj,
I do not think you need any protection for the MOSFETS as they have internal built in diodes and the Drain-Source voltage is 500 volts with a continuous drain current of 8 Amps. Only the supplied voltage should not exceed or the output circuit must never be short-circuited. I am using these MOSFETS since long and had never any problem except for my own mistakes.
Goodluck :smile:

 

[faraj]

Please let us remember always that the hardware AND software should run in harmony for the controller to give what we expect.

For example, which is the dead time you have decided on?
I mean the time when both Q5 and Q6 are open (post #17) hence Q1 and Q3 (P-MOS) are off.

As you know, a MOSFET has a relatively high input capacitance, mainly the power ones.
When Q5 or Q6 is turned on (becomes saturated, its collector close to ground), the driven Cgs will be discharged rather quickly.
But when Q5 or Q6 is off, Cgs will have to charge via a 10K resistor (R5 or R6) and the PMOS will not turn off very fast.
So if there is no dead time at each transition, each leg will act as a short circuit between 12V and ground for a very short time.
The effect of this malfunction (caused by the software which is not compatible with the circuit) will be noticed more when the frequency is increased.

You are fortunate that your bridge didn't give a smoke yet

Kerim

indeed it did give smoke but i was fast . now i put a 50 usecs dead time between turning on the outputs, but it still heats up not like before although but if i increase the dead time more than this i wont be able to get the desired frequency output. and when i think of heating in a low frequency i can't imagine how would it be with an 8 ampere load.

 

[KerimF]

indeed it did give smoke but i was fast . now i put a 50 usecs dead time between turning on the outputs, but it still heats up not like before although but if i increase the dead time more than this i wont be able to get the desired frequency output. and when i think of heating in a low frequency i can't imagine how would it be with an 8 ampere load.

First, let us agree that the unusual heating is not because of the load (mainly if it is resistive).
Second, let us play with some figures:
I checked the datasheets. The typical Cgs is similar to both P and N MOSFET and is about 1.2nF.

So the RC time constant = 10K * 1.2nF = 12 usec.
The typical threshold voltage is about 3V (for both).

V(t) - V(0) = [ (V(∞) - V(0) ]*[ 1-e^(-t/RC) ]

If we let the 12V node as the reference for the Cgs of IRF640 (Q1)

-3 - (-12) = [ 0 - (-12) )*[1-e^(-t/12e-6)]
9 = 12*[1-e^(-t/12e-6)]
t= 12e-6 * ln[ (12/(12-9)] =12e-6 * 1.386
t= 16.6 us

But on the other leg also the Cgs of IRF840 (Q2) has to be discharged slowly via R4 (10K). In fact, the discharge will be slower than the previous one because the drain voltage of Q1 will be off slowly unlike of the npn (Q5) collector. So the turn off delay of Q2 might be twice of 16us. So the total will be around 32 usec. Please note the calculations may not be accurate but I present them in a simple way to give an overall idea about what is happening.

Now you know how to decrease the turn off delay.
Your headache comes from the 10K resistors. So if we don't like to change the topology of the bridge, we can just decrease the 10K to 1K. Therefore:
R5 = R6 = 1K
R1 = R2 = 100R
R3 = R4 = 1K
This will reduce the turn off delays 10 times; from 32us to 3.2us.

After this minor change you will surely notice the difference in the heating.

Kerim

 

[faraj]

thank you Kerim, i'll do it and share the result.

 

[faraj]

ok it was a mistake, everything is right.

 

[KerimF]

I think (since you missed to say) in your simulated circuit you placed an inductive load. In this case when one leg is off, the load current continues to flow using the diodes of the other leg for a while (to discharge the stored magnetic energy). This gives the impression that the two MOSFETs of the other leg are automatically turned on for a while.
If you will find out that this is the case with you, we will discuss how to shorten the time of this discharge. You can verify this point by changing the load inductance. For instance, how do you determine the impedance of your real load?

Kerim

Added:
Okay

 

[faraj]

Finally, I tested the circuit with the inductive load and it's working perfectly and there is no problem. thank you everyone, specially KerimF