H Bridge High Side Mosfet Driver

Discussion in 'General Electronics Chat' started by bigbigblue, Mar 15, 2006.

  1. bigbigblue

    Thread Starter Active Member

    Mar 15, 2006
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    Hi,
    I am in the process of building a biploar stepper motor driver, which uses IRF540 MOSFETS for the high and low side switching. The current limiting mechanism is based on figure 4.11 from the "!Jones on stepping motors" we page at this address : http://www.cs.uiowa.edu/~jones/step/current.html.

    I say based on it, because I have a slightly different arrangement for the voltage sensing - I use an IRC540 current sensing MOSFET instead of a resistor. I also have a Microchip PIC driving 2 * DACs to provide the reference voltages for the comparators and the logic is all 4000 series with a VCC of 12v.

    I originally had 4000 series chips with a Vcc of 12v driving the gates of the MOSFETs. This gave me problems because when I did the design I hadn't realised the problem of driving high side MOSFETS in a bridge - (I am a beginner at electronics and have learnt everything the hard way over the last couple of months working on this design - so please be gentle with me :)

    Anyway - to the nub of my problem. I have all the logic working to provide the logic level signals for driving the MOSFETS, but am having very little success in driving the high side.

    Looking at the outputs of the logic gates which provide the switching signal to the MOSFETS on a 'scope, they seem to be switching at about 420KHz - which seems to be a lot faster than the norm (which I believe is about 20KhZ).

    I have the ouputs from the logic gates going to an IR2304 half bridge driver, which I would have thought would solve my problem by providing a high enough current to drive the MOSFET gates with a floating high side drive - but it does not work - even though reading an application note from International Rectifier suggests that the IR2304 will work up to the 100's KHz frequency range.

    So down to my questions :

    1. Has anybody any experience they could share with me on using the IR2304 in this way?
    2. The IR2304 is a "bootstrap mode" driver - I have been unable to find an explanation of what this means - can anyone help?
    3. I have heard that this type of "bootstrap" driver will not work unless there is switching going on (i.e where the current is permanently switched on) - is this true?

    Any help would be much appreciated. I am so close to getting everything to work and I hope this is the last hurdle.

    Best Regards
    Pete
     
  2. Papabravo

    Expert

    Feb 24, 2006
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    Congratulations on an excellent and coherent statement of the problem. The IRF540 is an N-channel MOSFET. This means that the silicon channel through which the current flows is made of silicon doped with N-type impurities. The N-type impurities have extra electrons in their outer shells which allow electron currents to flow. In contrast P-type impurities have a lack of free electrons but an excess of "holes". These "holes" behave like slow current carriers with a positive charge. Note carefully that "holes" in the semiconductor lattice are not the same thing as protons!

    Look carefully at the picture of the three electrodes on the data sheet. Your problem is with the nodes labeld "G" for gate and "S" for source. The parameter of interest in turning on an N-channel MOSFET is called "Vgs(th)" In the case of the IRF540 this parameter from the data sheet is a minimum of 2.0 Volts to a maximum of 4.0 Volts. This means that in order for the IRF540 to conduct any appreciable current, the Gate must be 2.0 to 4.0 Volts above the Source.

    If the Drain is connected to +12 Volts and some small amount of current begins to flow the Source will be some tens of millivolts below +12 Volts. In order for this to happen the voltage at the Gate needs to be 14.0 to 16.0 Volts. If you want the IRF540 to really turn on hard then you need a Vgs of 10V. This means you need to get the gate to 22.0 Volts or so. Vgs = Vg - Vs = 22 - 12 = 10 Volts.

    In order to solve this problem many designers build what amounts to a small relatively low current switching regulator in the boost configuration or a diode capacitor diode voltage doubler. What is being "bootstrapped" is the Vcc voltage of +12 Volts into the gate control voltage of +22 to +24 Volts. The amount of DC current required to turn a gate on and off is very small, but what does a gate look like electrically? It looks like a pretty big capacitor. So our gate driver needs to have a big chunk of current for just long enough to charge and discharge the gate capacitence. After the gate gets to a steady state the current requirement drops to a very small value. If you look at what happens on the gate during a transition you will see the voltage rise to a plateau, stay there for a while, and then proceed to its final value.

    The IR2304 needs some kind of a switching signal to make the boost regulator or voltage doubler generate the gate control voltage. If you can post your schematic, we may be able to offer further suggestions.

    Good Luck
     
  3. Papabravo

    Expert

    Feb 24, 2006
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    I've had a chance to review the IR2304 datasheet and I have some further comments. In the block diagram on page 1 of the IR2304 data sheet you will notice a small and insignificant diode with no part number or value which I will call D1, and a capacitor from VB to VS also with no value which I'll call C2.

    At the dawn of time Vcc charges C2 through D1 up to (Vcc - 0.7) Volts or (Vcc - 0.25) Volts if D1 is a Schottky diode. Refer to the functional block diagram on page 4 of the data sheet. There is no obvious way of going from VS to ground but it must be there.

    At some later time a transition comes into the HIN input. This is detected by the "Pulse Generator" which turns a transition into a pulse. This pulse goes through the pulse filter and either sets or resets the the SR flip-flop. At the same time Vs is disconnected from ground and it rises to VS while VB rises to VS + (Vcc - 0.7). This gate control voltage can now be routed pin HO to control the MOSFET gate.

    The logic from the SR flip-flop to the P-channel/N-channel totem pole output seems backwards to me. I think the output of the SR flip-flop should be labeled Q bar. So undervoltage detect turns on the lower MOSFET connecting HO to VS. In any case pin HO is connected to either VB which is at VS + (Vcc-0.7) which turns the upper driver on(Vgs(th) = (Vcc -0.7)), or it is connected to VS which turns the upper driver off(Vgs(th) = Vs - Vs = 0).

    I'm not clear from the data sheet if VS goes back to ground which would allow the charge on C2 to be replenished from Vcc in preparation for the next turn on cycle.

    Some additional points.
    1. If you implement the circuit in the IR2304 data sheet with a 600V supply make sure the diode will have a reverse breakdown rating exceeding 600Volts.
    2. Make sure the capacitor across VB/VS can withstand the inrush current.
     
  4. bigbigblue

    Thread Starter Active Member

    Mar 15, 2006
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    Papabravo,
    Many thanks for such a comprehensive and helpful reply.

    Am I right in thinking that the capacitor between VB and VS, plus the diode form "DC Restorer" circuit?

    Does the MOSFET effectively latch on until the voltage on the internal gate capacitor discharges to below the threshold voltage? The reason I ask is that I can forsee a problem when there is a long "space" and a short "mark" in the signal from the gate driving logic. In this case the voltage on the "bootstrap" capacitor will start to drop and the MOSFET may get into the linear region (rather than being switched on hard), with a drop in the current through the motor winding and the MOSFET starting to get hot.



    Are you referring to a signal other then the HIN / LIN signals? Unfortunately I do not have an up to date schematic, only PCB layouts - I have been building up the design section by section as I go along. I need to update the schematic when I am nearer a final design.


    Yes - I was aware of point 1. How do I decide what value capacitor to use - and what type should it be - i.e is a tant or electrolytic OK.

    I wonder if you would mind commenting on another idea I had for the high side drive? In my circuit I utilise an op-amp as a non inverting summing amplifier to generate a reference voltage for the lower comparator (so I have a Vref of say 2.00v on one comparator and 2.03v on the other).

    I was thinking that I could use a similar arrangement to create a high side driver. If I used a summing amp to add my high side gate switching signal to the Mosfet Source Voltage I would generate voltages of either Vsource or Vsource+12v (I am using 4000 series gates with Vcc of 12v).

    I understand I would have to provide an additional supply for the op-amp and that the max motor supply voltage would have to be approx 13.5v below the opamp supply voltage (as most op amps will only drive to 1.5v below their supply voltage), but this would suffice for my application (for the moment anyway).

    I have actually tried this, but it didn't work due to the op amp I was using (LM324) having a bandwidth of only 1Mhz - my switching is taking place at approx 420KHz- as it is a square wave, although the fundamental frequency is within the unity gain bandwidth of the op-amp, the first harmonic is only just below 1MHz and all the other harmonics are way above 1MHz - which results is an output waveform which looks nothing like a square wave ! Is there a rule of thumb as to how many times higher than the fundamental frequency of the input the unity gain bandwidth of the opamp would need to be to give me an output which closely matches the input? In other words do I need a 3, 10, 20, 50, or 100MHz opamp for this to work correctly?

    I hope all that makes sense !

    Thanks once again for your help - it is very much appreciated.
     
  5. Papabravo

    Expert

    Feb 24, 2006
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    My recollection of the term "DC Restorer" comes from television circuits and that it refers to taking an AC signal and "restoring" a DC offset. Although it is hard to see, this is working more like a "charge pump" or a "voltage doubler".
    ...
    The MOSFET is not latched on, what is happening is that there is an intrinsic delay in transfering charge from the gate capacitence to the surroundings. N-channel MOSFETs are faster than P-channel MOSFETS in this regard. That is why designers go to such lengths to use N-channel MOSFETS for high side switches.
    ...
    In my first response I was speaking in general terms, without having reviewed the IR2304 datasheet. After looking at the data sheet I realized that the switching regulator analogy was not entirely appropriate. I still have some questions about what is going on with the part, which the datasheet provides only ambiguous clues at best. The basic idea is still to charge C2 through D1 and then to add that voltage difference to VS to produce the gate control voltage. The exact machinery for doing that remains obscure.
    ...
    Generally when picking capacitors I like to pick a working voltage that is 2 times the largest voltage drop I expect across the capacitor. In non timing applications where an accurate charge discharge rate is not required the capacitence value is generally not critical. In this case you want to find a value that will charge at a sufficiently rapid rate so that the gate control voltage is developed in time to be useful. I'd start with a time constant that was one-tenth of the fastest switching period of the motor drive. The type of capacitor depends largely on the value. Up to 1uF ceramic capacitors work quite well.
    ...
    This opamp thing might work but I'm having trouble visualizing what is going on. If you have the extra supply that is above the motor voltage then you don't need the IR2304. With respect to switching going on at 420kHz, which is what led me to believe that a switching regulator was involved. I don't think there is a stepper motor on the planet that can move at that speed. Even if there is one that can move at that speed, do you know that you must ramp the speed up to the final velocity so as to avoid a potentially large acceleration? Do you also know that there are mechanical resonances which will cause the motor to stop and vibrate as the torque drops to near zero if you try to run at those speeds? Typical resonaces I have encountered are at 200 Hz and 1.5 kHz.
    ...
    I'm not sure why you are considering harmonics, I think the inability of an LM324 to handle a square wave at 420 kHz is a slew rate limitation. If this is a gain bandwidth issue then I would suggest a unity gain bandwidth of at least ten times the 420kHz. In your case that would be about 5 MHz. I know it may be hard to believe but such parts may not exist, if they do exist they will be expensive, if you still insist on using them, they will be tricky to get working. You will need to have careful layout, frequency compensation, and impeadance matching in your circuit design. As you increase the frequency of operation it becomes increasingly difficult to build amplifiers with gain much less the gain required of an opamp.

    I hope this helps
    Good Luck
     
  6. bigbigblue

    Thread Starter Active Member

    Mar 15, 2006
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    No - it isn't the best datasheet I've ever seen - in fact I think it has at least one error in it!

    Thanks for that veru useful information.

    I'm intrigued by your second sentence here - I do have an extra supply available, but I'm not sure how I would interface my signal logic to utilise it (sorry for being so dumb, but as I mentioned before I am pretty new to electronic design). If I get a chance I'll draw up a schematic of the summing amplifier setup I an considering and post it here.

    What is switching at 420KHz is the "chopper" control signals which my circuit generates and which are used to switch the MOSFETs on and off to maintain the set current through the motor coil - I'm not trying to step the motor 420,000 times per second :-0

    Yes I was aware of the issues of mechanical (and electrical) resonance. I'm not sure I understand everything I read, but I do know it is an issue. However I need to get the motors spinning first :)

    The reason for mentioning harmonics is that I understand an "ideal" square wave is made up of a sine wave of the fundamental fequency plus every harmonic of the fundamental sine wave. Therefore if you have a 420KHz fundamental, the first harmonic is at 840KHz (within the unity gain bandwidth of the particular opamp I am using), the second harmonic is at 1260KHz, which is above the bandwidth of the opamp, therefore if the slew rate were infinite (!), the opamp still could not faithfully reproduce the input square wave.

    I think you may be right - I will try to find an op-amp with a better slew rate.
    <!--QuoteBegin-Papabravo
    @Mar 16 2006, 02:10 PM
    I hope this helps
    [/quote]
    It does - thanks :)
     
  7. Papabravo

    Expert

    Feb 24, 2006
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    A simple method is to use an NPN switch in place of the IR2304. Connect the most positive supply through a resistor to the MOSFET gate. Connect the MOSFET gate to the collector of an NPN switching transistor like a 2N3904 or a 2N2222. Connect the emitter to ground. The transistor acts as an inverter so you may have to change the type of 4000B series gate you are using. The 4000B series logic gate with the enable and the chopping signal inputs can drive the base of the transistor through a resistor. The value of the resistor should be (12V - 0.7V)/some reasonable base current like 500 microamps. When the NPN transistor is off the pullup resistor will turn the MOSFET ON. when the transistor is on the gate is at ground and the MOSFET is most definitely off. One problem with this circuit is that turn on and turn off times will not be symmetrical because you have a passive pullup and an active pulldown.

    It was a bit difficult to understand what you were doing without the last explanation on the role of the 420 kHz chopping circuit.. Are you sure that the IR2304 is designed to work with the HIN input being chopped? Again the poor datasheet contributes to our confusion. There might possibly be an application note for this part that would shed some light on the subject.
     
  8. bigbigblue

    Thread Starter Active Member

    Mar 15, 2006
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    I THINK (but I'm not sure) that there may be another problem with this approach. The IRF540 Mosfets I am using require a Source to Gate differential of 10V to be switched hard on, and will only tolerate a maximum Source to Gate voltage of 20 V.

    If I utilise a 12V supply for the stepper motor, I need a minimum of a 22v supply to ensure the MOSFET is fully on when the source is at 12v (ie 12V motor supply + 10V Source to Gate differential), However, when the source falls close to the 0V rail, which I understand it is at the instant it is switched on (ie the motor coil is a virtual short to 0V - if you ignore its resistancel), the Source to Gate differential will be 22V (i.e higher than the maximum allowed). If I used a 20v motor supply, then when the source is at 12v (i.e when the coil is fully energised), the Source to Gate differntial would be a maximum of 8V - insufficient to turn the MOSFET hard on. Or have I misunderstood something?

    <!--QuoteBegin-Papabravo
    @Mar 16 2006, 08:32 PM
    Are you sure that the IR2304 is designed to work with the HIN input being chopped?  Again the poor datasheet contributes to our confusion.  There might possibly be an application note for this part that would shed some light on the subject.
    [/quote]
    No - I'm not sure the IR2304 is designed for use with a chopped HIN - however I think it needs the HIN / LIN lines to pulse in order to operate the charge pump. For sure the datasheet doesn't make it clear. I did find an application note at this address : http://www.irf.com/technical-info/appnotes/an-978.pdf but my head almost exploded trying to understand it.

    I will put together and post a schematic of my op-amp based idea. BTW I have just managed to get a couple of opamps with a much better slew rate (13V/uS) than the LM324's I was using, I'll try out my idea tomorrow.

    Once again thanks for you patience and assistance :)
     
  9. Papabravo

    Expert

    Feb 24, 2006
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    I found the same application note and it definitely has more information. You need to work through it patiently. If you look at the 600V case in the IRF540 datasheet you need to remember that node voltages on the MOSFET are relative to each other and not necessarily relative to ground.

    If Vd is approximately Vs at 600V then
    Vg would needt to be between 610V and 620V to turn the part ON.
    Right?
     
  10. bigbigblue

    Thread Starter Active Member

    Mar 15, 2006
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    Here is the schematic for my proposed opamp based High Side MosFet driver.

    The HIN and LIN signals are the chopped High Side and Low Side driver signals from my driving logic. The AC motor symbol is simply meant to represent one of the stepper motor coils as the load.

    The OpAmp has a 32V supply and the GND is common between the driving logic, the OpAmp supply and the motor supply.

    The opamp is configured as a non inverting voltage summing amplifier, so that the O/P from the Opamp is the sum of HIN and the Source voltage on the High side MOSFET. As HIN swings from 0V to 12v, the ouput of the opamp should swing from the Source Voltage to (Source Voltage + HIN). As in the circuit I have attached The motor supply is 12v, the max value of (High Side Source Voltage + HIN) = 24V, within the 32v supply of the opamp.

    As the opamp can source a reasonable current, I am hoping that it will be able to drive the gate to open quickly.

    The disadvantage is that the max motor supply voltage is limited to the opamp supply voltage -10v (less whatever the opamp ceiling voltage is, assuming the opamp isn't a rail to rail type).

    I tried this without success as the opamps I was using either didn't have the bandwidth, or thgeir slew rate was too slow to follow the HIN signal.


    What do you think?

    EDIT: I forgot to add a current limiting resistor between the opamp output and the High side MOSFET gate in the schematic.
     
  11. n9352527

    AAC Fanatic!

    Oct 14, 2005
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    The summing opamp circuit is your problem. Forget the complicated scheme and just use the opamp as a comparator with threshold of half HIN. Limit the maximum Vgs with a zener across g-s after the current limiting resistor that you mentioned above.

    If I were you, I would forego the opamp and design a driver with transistor switch and resistor like Papabravo suggested. Papabravo also mentioned unsymmetrical tf and tr, which is not really a problem in application like this. You could calculate for worst case slope resistor value and that would take care of that.

    May I ask why do you switch both high-side and low-side? What's the temporal relationship of HIN and LIN?
     
  12. bigbigblue

    Thread Starter Active Member

    Mar 15, 2006
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    Any chance you could explain why? I know it doesn't work, as I downloaded the Multisim free edition and simulated the circuit and I can see it requires an opamp with a massive bandwidth (300 MHz +) and HUGE slew rate (200V/us + !!!!) - which probably doesn't exist, and if it did would cost a fortune, but I don't understand why it needs such a high bandwidth and slew rate when the chopping input is only (?) running at 420 KHz.

    What a great idea - which I don't think I would have ever thought of. Thanks, I'll give it a try.

    I'll also give this a try, but I haven't yet managed to convince myself this will work as I want.


    <!--QuoteBegin-n9352527
    @Mar 17 2006, 04:34 PM
    May I ask why do you switch both high-side and low-side? What's the temporal relationship of HIN and LIN?
    [/quote]

    They don't chop at the same time. The high side chops when the motor coil is drawing power and the low side chops when it is generating power - it is designed to limit the current in both cases. So when the coil is drawing power, LIN is permanently high and HIN "chops" and vice versa when the coil is generating power.

    I don't understand what is meant by the phrase "temporal relationship" I'm afraid - my apologies for being dumb :)
     
  13. n9352527

    AAC Fanatic!

    Oct 14, 2005
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    You are effectively adding a negative feedback through the source of the MOSFET to the positive input of the opamp. It suppresses the opamp output increase and that is why you need a fast opamp to even makes a small dent.

    I always use transistor switches to drive MOSFETs in my designs. They are cheap, easy to design and very versatile. You could almost adapt them to any occasion and requirement.

    Could you elaborate more on this driving scheme? What do you mean by drawing power and generating power? Where from? How? If it is a coil and one end of the coil is switched off, it didn't really matter what the other end of the coil is doing.

    <!--QuoteBegin-bigbigblue
    @Mar 17 2006, 06:37 PM
    I don't understand what is meant by the phrase "temporal relationship" I'm afraid - my apologies for being dumb :)
    [post=15105]Quoted post[/post]​
    [/quote]

    Oh no... I'm the one that should apologise. I have a bad habit of using inappropriate phrases because I'm used to them at work. Temporal relationship is the timing relationship, basically what HIN and LIN do in respect to each other and time.
     
  14. bigbigblue

    Thread Starter Active Member

    Mar 15, 2006
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    OK its time to own up - I don't really know I'm afraid - I just based it on figure 4.11 in the classic "Jones on stepping motors" web page (which can be found here : http://www.cs.uiowa.edu/~jones/step/current.html) I actually have my design working up to the point of driving the H-Bridge properly.

    I tried the comparator suggestion this evening - didn't work :) the comparator wasn't fast enough. I'll give the transistor drive a try tomorrow - I have a few transistors lying around that could do with getting fried ;-)

    <!--QuoteBegin-n9352527
    @Mar 17 2006, 10:07 PM
    Oh no... I'm the one that should apologise. I have a bad habit of using inappropriate phrases because I'm used to them at work. Temporal relationship is the timing relationship, basically what HIN and LIN do in respect to each other and time.
    [/quote]
    No prob - I learnt something new - the meaning of "temporal relationship" :)

    Hope you don't mind me referring you to Jones on stepping motors again - I'm not sure I can explain the relationship clearly - sorry (I'm one of those damned newbies !)

    Many thanks for your help
     
  15. n9352527

    AAC Fanatic!

    Oct 14, 2005
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    OK. Is your design unipolar or bipolar? Jones has two switches for unipolar with the power coming from the CT or a full h-bridge with four switches to reverse drive bipolar winding. None of them is similar to yours, which is a single winding with unidirectional current flow and two switches. Can you do a full diagram of power drivers/switches and all step motor windings and post it here?

    Are you referring to running and braking power?
     
  16. bigbigblue

    Thread Starter Active Member

    Mar 15, 2006
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    Bipolar - the schematic I posted was just to illustrate my idea to use the summing amplifier - it does not look anything like my driver - which is ALMOST identical to figure 4.11 in the Jones on Stepping motors (a full H-Bridge with 4 switches). This is why I have the high side drive problem.

    <!--QuoteBegin-n9352527
    @Mar 17 2006, 10:47 PM
    Are you referring to running and braking power?
    [/quote]
    I think so :)
     
  17. n9352527

    AAC Fanatic!

    Oct 14, 2005
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    Have you considered using P MOSFET instead? For your application, where the current is not that large, the disadvantage of using P MOSFET is not that great. The advantage is certainly very attractive, no need for Vsupply + 12V extra voltage drive for high-side switches. Much simpler gate drive circuit and certainly easier and cheaper to design.
     
  18. bigbigblue

    Thread Starter Active Member

    Mar 15, 2006
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    The hybrid motors I have, when connected in parallel mode, will take just over 8 Amps per phase. Are P channel MOSFETS suitable for use at this current?
     
  19. n9352527

    AAC Fanatic!

    Oct 14, 2005
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    Take a look at On Semi or IRF range of > 8A P MOSFETs and note the price differences with similar N MOSFETs. They are not that pricey, and whatever extra you would spent on the P MOSFET you surely could gain back from saving on simpler gate driver circuit.

    Personally, I would start thinking of using N instead of P MOSFETs for current above ~20A or high voltage application where the Vgs(br) of P MOSFET becomes an issue.
     
  20. bigbigblue

    Thread Starter Active Member

    Mar 15, 2006
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    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).

    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?

    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) ?

    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).

    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 ?

    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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