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### Topic: Comments requested on MOSFET high-side driver (Read 9617 times)previous topic - next topic

#### polymorph

#45
##### Feb 03, 2015, 11:10 pmLast Edit: Feb 03, 2015, 11:11 pm by polymorph
They both specify current ratios of 1:10 for base and collector current when specifying the saturation VCE and VBE.

Also note the actual gain with the CE not quite saturated at 1.0V, hfe is 40 at 500mA for the 2N2222, and 30 at 100mA for the 2N3904.
Steve Greenfield AE7HD
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#### nickgammon

#46
##### Feb 04, 2015, 02:35 am
BTW, that rule of thumb of base current being 1/10th collector current is in the datasheet for the 2N2222 and 2N3904. It isn't that the gain is 10.
OK, I reworded that paragraph to get rid of the "rule of thumb" and back it up with figures. I hope the rest comes out as making a reasonable amount of sense.
Please post technical questions on the forum, not by personal message. Thanks!

#### michinyon

#47
##### Feb 04, 2015, 04:05 am
How would you factor that in?
When your circuit is turned on,  you want the gate voltage of the FET to be as negative as possible, with respect to the 12V supply.   For a given amount of current that you can sink through Q1,   this means the higher the value of the pull-up resistor,   the better.

But when you turn Q1 off,  you want the gate voltage of the FET to rise as quickly as possible.   This would be best achieved with a smaller value for the pull-up resistor.

One solution is to have another transistor to actively pull up the FET gate as fast as possible,  but that would be overcomplicating things.

The data sheet that was linked there,  implies a gate charge of 180 nanocoulombs for a Vgs of -10 volts,  which is about what you get if you can pull the collector voltage of Q1 down to about 2 volts.

It is possible to calculate how much current has to flow through your pull-up resistor for what period of time,  in order to discharge that capacitance.   For that to be useful,   you'd also need to figure out how quickly Q1 really turns off.

#### michinyon

#48
##### Feb 04, 2015, 04:16 am
On the subject of using the varistor instead of the diode,  I am doubtfull.

When the Mosfet is turned off,   the current flowing through the load, I,  drops rapidly.   If the load is an inductive load,   this will tend to cause the potential at the connection between the mosfet and the load to be large and negative,  with respect to the ground.

If you are using a diode,   then as soon as the potential at the connection between the mosfet and the load falls to -0.7V with respect to the ground,     then the diode will start to conduct forward,   feeding current back into the entry side of the load.    This mitigates the rate dI/dt at which the load current is dropping,   and therefore mitigates the negative potential at the connection between the mosfet and the load.

The reason you don't want a large negative potential there,  is that there is still +12V on the supply side of the mosfet.  If you got say -50V at the connection between the mosfet and the load,  you would have 62 V across the mosfet and exceed its forward breakdown voltage.

If you had a perfect diode,  using the diode limits the maximum forward voltage across the mosfet to 12.7 volts.   It will be worse than this for real diodes with actual turn-on time.

If you use the varistor,   the varistor is not going to conduct until the potential at the connection between the mosfet and the load falls to about -18 or -19 volts,  relative to the ground.    When that happens,  you are going to have a forward voltage of more than 30 volts across the mosfet.    It seems to me,  that is not such a good outcome.

#### nickgammon

#49
##### Feb 04, 2015, 04:36 am
Thanks for the comments. Can you put figures on a way to choose a diode? We've seen comments so far that some diodes either don't have the current-handling capacity, or switch too slowly.

Given a datasheet, eg. for the 1N4001, what parameters would you say are relevant to this application, and for what sort of motor? I presume at some stage (eg. higher voltage) it would be time to choose a different diode.
Please post technical questions on the forum, not by personal message. Thanks!

#### michinyon

#50
##### Feb 04, 2015, 06:04 am
According to the datasheet that was linked to ( which might not be the same mosfet you used ), the gate charge at Vgs=-10V  would be 180 nanocoulombs.   It is not actually linear,  but if it was,   that would be be 18 nC/V === 18 nF  ( by the definition of a farad ).

To consider how long this takes to discharge through the resistor R2,   consider the timeconstant of an RC series circuit is T=RC.   Here,  if your R2 was 1k,   then your time constant would be 1k x 18 nF = 18 microseconds.   If you think it takes about 3 time constants to discharge your capacitance to a negligible value,  looks like about 54 microseconds.

If your R2 was 10k,  it would take ten times that long.

#### michinyon

#51
##### Feb 04, 2015, 06:25 am
Looking at the vishay datasheet for 1N4001 diode,  this is what I see:

The maximum reverse voltage that the diode is going to see,  is 12V.    That would be when the motor is running,   the mosfet Vds is as small as you can make it,   and the ground is good.   The maximum reverse voltage for the diode is 50V,  or 35 V RMS  ( which as far as I can tell is more or less the same thing ),  so you are good there.

When the mosfet interrupts the flow of current though the motor,   the maximum circulating current through the diode starts at a maximum value of whatever the load current was,    and falls from that value back towards zero.     The diode current is not going to be more than the load current.

Looking at the datasheet,  the peak forward current for the diode is around 30 to 50 amps depending on the duration - in the order of a few milliseconds.         If your load was 2 ohms,   you'd be looking at a maximum motor current around 6 A ?   That would mean a peak diode current of less than 6 amps,  for less than a millisecond.     It would appear that is not going to overload the diode.

The diode also has an average forward rectified current of 1 amp.   Depending on the duty cycle of your PWM  and the inductance of the motor,   you might come close to this.      You'd have to have 6 amps ( peak ) flowing through the diode  for the equivalent of 1/6 of your total duty cycle,    to create a forward current (  and diode heating load ),  as large as 1 amp.

If you know the inductance L of the motor,   and you assume that when the mosfet turns off,  the +ve terminal of the motor falls to say -0.8 volts relative to ground,   enough to turn the diode fully "on",      then you can estimate dI/dt,   and therefore estimate how long the diode current event lasts.

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