Gate drivers suggestions

Discussion in 'General Electronics Chat' started by ak52, Sep 15, 2015.

  1. ak52

    Thread Starter Member

    Oct 15, 2014
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    Hello Everyone,
    A few months back i had completed a solar charge controller project,with all of your help ')
    But there were a few problems with my circuit,i would like to start rectifying those issues as i am now making another charger with a higher rating.

    One major mistake was using the IR2110,as a gate driver(high side and low side).
    It was suggested that this was not a suitable gate driver as it is mostly used in induction motor drive systems.
    In my new design i would like to use only one N channel mosfet(or a parallel combinations of the mosfet to increase current ratings),on the lower side so that the gate driver has the same ground as a reference as the rest of the circuit.(so that there is no separate charge pump circuit required.)
    Could you please suggest me a suitable gate driver,my input to the gate driver would be from a PIC 3.3 to 5v and gate driver output should be 12volts.
    I have made a crude hand drawing of what i need if that helps.
    Untitled.png

    Thanks in advance,
    AK
     
  2. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    5,790
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    What you show is a high-side switch, which would need a P-channel FET if you want to avoid using the charge-pump driver in the IR2110. To use an N-channel FET you would insert it below and in series with Rsh. (The IR2110 would still be ok as a low-side driver).
     
  3. PeterCoxSmith

    Member

    Feb 23, 2015
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    The "earth" connection on the gate drive needs to be a floating ground to allow the drive to float on the potential at the drain of the mosfet.
     
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  4. ak52

    Thread Starter Member

    Oct 15, 2014
    145
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    Yes sir.the drawing does show a p channel mosfet,
    Perhaps i should have mentioned this earlier(my bad),but i would like to do two things :
    1.Use an N channel Fet
    2.Use a low side shunt.
    As you said ,i can shift the battery upwards ,but that would mean using a high side shut as well?

    Also when using a p channel Fet i cannot have the same ground point as the gate voltage should be slightly above the source voltage

    Ps: I haven't yet taken a final call on using N or P channel Fets,i'm leaning slightly towards the N channel Fets,but if using a P channel is simpler i will definitely go with them.
     
  5. crutschow

    Expert

    Mar 14, 2008
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    Just understand that the gate voltage of a MOSFET is always referenced to the source and that will help you understand the drive voltage needed.
    An N-MOSFET needs the gate voltage to be more positive than the source (+10V Vgs) for standard N-MOSFETs.
    A P-MOSFET needs the gate voltage to be more negative than the source (-10V Vgs) for standard P-MOSFETs.

    So, in your circuit, it's the relative voltage between the gate driver and +Vcc that determines the gate drive for a P-MOSFET.

    If you want to use an N-MOSFET with the source grounded, then you would need a different circuit.
    What are the +Vcc and battery voltages?
     
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  6. ak52

    Thread Starter Member

    Oct 15, 2014
    145
    4
    +Vcc is my solar panel voltage which would be 192v
    Battery voltage is 120v(10 lead acid batteries in series)
    Could you please elaborate on the underlined text above,a circuit would be helpful.

    AK
     
  7. crutschow

    Expert

    Mar 14, 2008
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    Here's a standard Buck switching regulator with a grounded-source N-MOSFET switch.
    The limitation is that if you ground the MOSFET source then the solar panel and battery common cannot be connected to the Mosfet driver and MOSFET common (they must float with respect to each other).
    That means you must use an isolation circuit if you need to measure the battery or solar panel voltage with respect to the circuit common.
    If the battery voltage is needed for the buck regulator feedback loop then there are opto isolator circuits that can do that (here is an example).
    Buck.PNG
     
    Last edited: Sep 17, 2015
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  8. ak52

    Thread Starter Member

    Oct 15, 2014
    145
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    Thanks for the circuit,i looked at a few books and but in this case using N channel mosfets would mean a separate isolated circuit for gatedriver which would increase the complexity of the circuit.Using a P channel mosfet is simpler in this case,which brings be to my next question,it may seem quite basic but when using a P channel Mosfet,is a specialized gate driver really required?
    Can't i just do something like this:
    upload_2015-9-18_13-8-26.png
    Can can make my own gate driver using a couple of resistors and an npn transistor.

    Thanks in advance,
    AK
     
  9. PeterCoxSmith

    Member

    Feb 23, 2015
    148
    38
    mosfet gates are capacitive and need a big pulse of current to turn them on (and off). A resistive drive will be slow to deliver current with consequent slow turn on of the mosfet. And slow turn on leads to high dissipation in the mosfet as it transitions between the off and on states. Mosfet drives are fast active push pull devices that deliver the necessary high current pulse.
     
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  10. ak52

    Thread Starter Member

    Oct 15, 2014
    145
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    I see,so any high side gate driver will do the trick for me?Would i still need a charge pump circuit to boost the gate voltage close to the source voltage?

    Many thanks,
    AK
     
  11. bertus

    Administrator

    Apr 5, 2008
    15,647
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    Hello,

    Most high side mosfet drivers will have the boost circuit.
    They are ment to be used with N-gate mosfets.
    The advantage of N-gate mosfets over P-gate mosfets is the faster switching time for N-gate mosfets, wich leads to lower losses.

    Bertus
     
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  12. PeterCoxSmith

    Member

    Feb 23, 2015
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    Components like IRS2184 have the boot strap circuit built in so easy to implement. The only down side of a bootstrap circuit is that it must be refreshed, to charge up the capacitor, which occurs during the off period of the PWM. So you always have to have PWM running you cannot simply turn on the mosfet and leave it on. Bertus gives you some good advice about N vs P type devices.
     
  13. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    If the 10 kHz drive signal is always there (I assume it is PWM), then a capacitor-coupled bipolar drive will work. I posted one a few months ago to another thread, I'll dig around for it. But you seem intent not to use a charge pump-based gate driver chip. Any particular reason? If you had a bad experience with one, there are lots of others that might work better in your application.

    ak
     
  14. ak52

    Thread Starter Member

    Oct 15, 2014
    145
    4
    Adding to that,after searching for a long time, it is very difficult to find an equivalent high current P channel MOSFET, the best one i could find was one with source voltage of 150v and drain current of 25amps.
    Rds was also quite high compared to an N MOSFET.

    But i need one with at least 250v and 40 amps.
    So it would be best to use a N channel MOSFET.

    Yes it is a constant 10khz pwm signal with a maximum duty cycle of 98%,so that the MOSFET never fully ON.
    Nothing against a charge pump based gate driver as long as i get it to work.

    I have only used the IR2110 for my driving needs so far,but there is a problem with it, For driving a N-MOSFET on the high side the VS(High side floating supply offset voltage) pin should be brought down to zero for the driver to work correctly(Bootstrap arrangement),where in i would need a lower FET.Since my load here is a battery,when ever i switch on the lower FET I essentially short the battery through the lower FET.To avoid this i again need to put in a series diode to protect the lower FET.
    And adding a diode where30 to 40 amp current is flowing would mean a lot of heat losses as well.
    So ..its a bit complicated !!

    Also,
    I have had a look at a few high side gate drivers such as IRS2184 as suggested above,they too have the VS pin present,assuming that my load is a battery would i need to manually bring VS to zero volt here as well?

    Many thanks
    AK
     
  15. PeterCoxSmith

    Member

    Feb 23, 2015
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    If you have a problem with refreshing the boot strap circuit you could use an isolated DC DC converter to generate the VB voltage referenced to VS. You can use lowish cost Traco converters or I have used a transformer driver MAX253 with some success.
     
  16. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    I don't think so. The front page of the datasheet shows Vs connected to the high side FET Source, which is on the high side of the load. It could go to zero if the load were shorted when the low side FET comes on, but that is a different problem.

    ak
     
  17. crutschow

    Expert

    Mar 14, 2008
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    I don't understand. :confused:
    When the bottom FET is turned ON the top FET will be OFF, so there is no current conduction from the battery (the FET, or diode to ground are connected at the output of the top FET, see my circuit in Post #7).
     
  18. ak52

    Thread Starter Member

    Oct 15, 2014
    145
    4
    My thoughts precisely when i had a look at the data sheet,but it does't work for some reason :(
    This was discussed earlier i think in my previous project.For a 12 v input to the driver i get a 6 to 7v on VS with respect to ground.
    http://forum.allaboutcircuits.com/threads/lower-side-mosfet-keeps-blowing.110700/page-2#post-858053
    Won't the battery discharge through the lower FET if it is turned on?
    This was from one of my earlier drafts before i realized the mistake in it.
    Untitled.png
     
  19. ak52

    Thread Starter Member

    Oct 15, 2014
    145
    4
    PS: I had done a few experiments at the time,may be of some help to sort out this puzling issue,
    If we use the IR2110 in high side configuration only,if the load is simple resistor the MOSFET switches perfectly,
    if i remove the resistor and connect a battery ,the MOSFET dosn't switch at all!
     
  20. crutschow

    Expert

    Mar 14, 2008
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    If the regulator is working properly, the inductive reactance of L2 will keep current flowing into the battery when Q2 is ON (free wheeling portion of cycle).
    No back current will flow.
    That's the purpose of L2, to store inductive energy when Q1 is ON and transfer that energy to the output when Q1 is OFF.

    Note that the normal flow of this free-wheeling current is from Q2's source to drain (opposite of the typical N-MOSFET current direction). That works fine since a FET will conduct equally well in either direction when ON.

    If you don't understand all that, then you need to further read abut how a buck regulator works.
     
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