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MOSFET Gate driver IC

Discussion in 'General Electronics Discussion' started by KX36, Oct 17, 2011.

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  1. KX36

    KX36

    14
    0
    Aug 26, 2011
    Hi all. I'm trying to use a microcontroller's PWM (i.e. logic level out) to drive a high side N-channel MOSFET (source follower) and thereby provide a higher voltage (30V), higher current (up to 5A) PWM supply to a resistive load (heater filament).

    Block diagram:
    http://dl.dropbox.com/u/32751654/PWM driver block diagram.png

    Please tell me if I'm going down a blind alley.

    Having a look at the available MOSFET gate driver ICs, they seem to be mostly intended as half H-bridges for motor control or for controlling switched mode power supplies. One I have seen is National Semiconductor LM5109 http://www.national.com/ds/LM/LM5109B.pdf

    Would I be able to use just the high side half of this IC for my purpose? Will it still work without the low side connected to a MOSFET? The datasheet says "Unused LI input should be tied to ground and not left open", so presumably it can be, but something else I read seemed to imply that the low side MOSFET is necessary to keep the bias voltage on the bootstrap capacitor.

    Can someone please confirm that this is a sensible way of translating a logic level PWM signal to something that can drive these voltages and currents, and whether or not I could use this IC in this way? If not, any suggestions for a different driver circuit or IC suitable for a high side PWM would be appreciated.

    Cheers,
    Matt
     
  2. D_Hambley

    D_Hambley

    11
    0
    Sep 1, 2011
    Matt, I just looked at DigiKey's site and typed in "high side driver". There are hundreds of them. Fairchild FAN7371 for example is one of them. Most of them need a voltage source equal to the amount you need to drive the gate of your MOSFET; This isusually at least 8V minimum, worst-case. You can drive most of them right from a low-level logic signal.
     
  3. OLIVE2222

    OLIVE2222

    690
    25
    Oct 2, 2011
    You can also bypass the driver by using your microcontroler (guessing 5 volts supplied) to directly drive a so called logic level FET. A query on digikey or mouser site should help.
     
  4. KX36

    KX36

    14
    0
    Aug 26, 2011
    D_Hambley,
    Indeed, Shopping on RS (not digikey or mouser as I'm in UK), They have a whole section of MOSFET drivers. And filtering by ones that are compatible with TTL input and include driver type "high", sorting cheapest first. Leaves me with plenty of choice. I just went through the list and stopped at the first one that the datasheet seemed reasonable in terms of meeting these criteria plus the voltage and providing a fair bit of current to charge the MOSFET gate capacitance quickly. (The main point of it is to minimise the switching time in the MOSFET). When you say a voltage source, do you mean the input signal voltage has to be of that magnitude or it just needs a supply voltage of that magnitude to operate? This datasheet recommends Vdd to be between 8-14V and I have a regulated 12Vdc supply I can use for that if this is the case. Vdd contributes to the bootstrap voltage which drives the gate.

    OLIVE2222,
    When you say bypass the driver, do you mean just find a FET that can operate directly off the microcontroller's logic level output? The first problem with this is that the mosfet needs to switch very quickly, and this requires charging and discharging its parasitic capacitances quickly, requiring a high current charge and discharge path. A microcontroller pin's not suitable for this. The second problem is that since this is a high-side driver, the MOSFET's source is not grounded, and in fact is very close to the drain voltage which is connected to the power rail so the gate voltage needs to be sufficiently higher than the source, not just higher than ground; and these high side drivers appear to use a bootstrap capacitor and diode to produce a bias voltage for this that's higher than the supply voltage.

    I don't really understand this bootstrap circuit, which is probably the main reason I'm having problems (I'm only an amateur with no formal education and not having worked with such circuits before). The absolute max rating for Vdd is 18V, the max HV supply voltage is 100V and the max bias voltage on the bootstrap capacitor is 108V. I'm assuming it's a charge pump, although I don't know if the lower MOSFET being present and saturated while the upper MOSFET is in cut-off is necessary for the charge pump to work.
     
  5. OLIVE2222

    OLIVE2222

    690
    25
    Oct 2, 2011
    Yes, by bypass the driver I mean connecting the fet Gate directly to the microcontroller. (that why the are called logic level FET) Just add a resistor from the gate to the ground to ensure it's stay off at power up.
    Normally you don't need to worry about switching time to drive a heater filament as load

    note that you can oder at mouser and digikey from UK:
    http://uk.mouser.com/
    http://www.digikey.co.uk/
     
  6. KX36

    KX36

    14
    0
    Aug 26, 2011
    Although you can order from mouser or digikey in UK, it isn't economically viable when there are local alternatives within the UK and EU. Delivery from US takes about 3 weeks, there's an extra shipping charge of around £16 ($20). There is generally also import tax and, depending on your luck, customs fees and delays and courier handling charges depending on the courier. It's only useful when you need a specific product not available in Europe.

    As I have said, this is a PWM circuit [that switches on and off in the kHz range to efficiently provide a variable voltage at a high current (up to 5A)], and on top of this, the dissipation in the MOSFET is proportional to the switching times, so yes, switching time is important in this circuit.
     
  7. KX36

    KX36

    14
    0
    Aug 26, 2011
    To go part way to answering my own original question, I now understand the bootstrap circuit better.

    When the power MOSFET is off, its Rds is high and its source is near ground potential, so the bootstrap capactor charges through the bootstrap diode to the drive voltage (Vdd; +12V if I use that regulated +12V supply as I mentioned.) minus the diode forward voltage drop (0.7V) = +11.3V.

    When the MOSFET switches on and its Rds falls to a very low resistance, the source voltage is pulled up to near the HV rail voltage (+30V in this case). Because the bottom of the bootstrap capacitor is connected to the source pin, the top of the capacitor will be at Vdd-Vdiode+Vsource (+12-0.7+30=41.3V). This is the voltage that the driver outputs to the gate, so Vgs is +11.3V even though the source is floating at the HV rail voltage, and the MOSFET remains turned on. The boostrap capacitor discharges to provide the current to charge the MOSFET input capacitance and also slowly discharges through leakage in the driver and MOSFET, so it will not hold the MOSFET on indefinately, and therefore can't drive a PWM with a 100% duty cycle.

    So the short answer to my original question is that the lower MOSFET is not needed for the high side circuit to work, as long as the source falls to at least the MOSFET saturation voltage (8V in this case) below Vdd during the cycle so that the bootstrap capacitor can be charged.

    I have another question though:
    How do I calculate the necessary bootstrap capacitance? I've been reading into the theory, but I haven't found any equations. I hope this question is more able to attract an expert to point me in the right direction than my original one.

    Edit:
    Found a formula for C(bootstrap), although I need to know lots of charges and leakage currents. I'll have to trawl through a few datasheets to see which of these I have.
     
    Last edited: Oct 19, 2011
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,401
    2,777
    Jan 21, 2010
    Just a quick note (from India)...

    If you were to use the microcontroller pin to drive a mosfet directly (and in this case it would have to be a P channel low side switch) instead of using a driver, you may have problems due to the the fact that the microcontroller pin can only supply a very small current. This will charge up the gate capacitance relatively slowly and lead to relatively slow switching.

    Gate drivers (mosfet dirvers) are often capable of several amps of drive current.
     
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