High-Side MOSFET pFET driver

Discussion in 'General Electronics Chat' started by PicStar218, Aug 8, 2011.

  1. PicStar218

    Thread Starter New Member

    Jul 2, 2011
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    Hi guys. New to the forum, got a quick question... So the attached image is a modified version of a circuit that SgtWookie posted in this thread:

    http://forum.allaboutcircuits.com/showthread.php?t=33116

    Oh and sorry for the new thread, apparently the old one is too old to be bumped... So anyway, looking at that circuit, I have a few questions:

    1. Since Q2 is used to pull the gate Hi, what purpose does R5 serve...? I can't understand why S.W. included it; when I delete it and run the simulation again, nothing changes. :confused:

    2. What purpose does D1 serve...? I can't figure it out. I know it's necessary, I just can't figure out what it's doing. Just toying around and doing some trouble shooting, when I replace the diode with a piece of wire, and run the simulation again, suddenly the gate voltage is VERY slow at going Hi. This can be seen in the second image - notice how the gate voltage lags at pulling up. I'm sure this has something to do with the gate capacitance, I just can't figure out what's happening.


    Any help would be greatly appreciated!!
     
  2. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    D1 pulls the gate of M2 to GND when Q1 is conducting.

    R5... still thinking. He may have added it because in his first post on the other thread the resistor was necessary because Q2 didn't exist.

    Wait a bit , sooner or later he'll respond.
     
    Last edited: Aug 8, 2011
  3. PicStar218

    Thread Starter New Member

    Jul 2, 2011
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    Yeah but why is the actual diode necessary? As opposed to just a piece of wire/trace?
     
  4. praondevou

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    you would short base to emitter.
     
  5. assembler_C

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    Mar 19, 2008
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    D1 is a shottky diode. It is used to speed-up the HI-to-LOW transition (thus speeding-up the MOSFET conduction transition) when Q1 is turned-on. When Q1 is turned-off, D1 is reverse-biased and allows Q2 to speed-up the LOW-to-HIGH transition (speed-up the MOSFET's turn-off transition). R4's purpose is to limit the MOSFET's transition currents.

    Being a shottky diode, D1 allows a higher DC bias on the MOSFET when Q1 is conducting.
     
    Last edited: Aug 8, 2011
  6. SgtWookie

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    Praondevou gets a cookie for giving the right answer. ;)

    If Vbe=0, then the transistor will be turned off; then the current path to charge the gate would be through R3+R4 in parallel with R5; roughly 917 Ohms, leading to a very slow turn-off time for the MOSFET.

    R5, the 10k resistor, was included in case there was a failure in the remainder of the MOSFET driver; like if R4 happened to burn up. It'll turn the MOSFET off if nothing else is connected.
     
  7. PicStar218

    Thread Starter New Member

    Jul 2, 2011
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    Thanks guys! I can't believe I wasn't seeing that before. Of COURSE you need the diode, otherwise the B - E would never be forward biased, and Q2 would never conduct... So by replacing the diode with a piece of wire, I was essentially negating Q2 entirely, and the Gate turn-on became slow in the simulation because the circuit was relying on R3/R5 to pull-up the Gate voltage.

    I sure love understanding the basis for how and why something works :). I'm probably just OCD, but I can't stand relying on empiricism alone, without understanding the underlying concepts. Which brings me to R5... I still don't get it!! I tend to agree with the other poster, I think S.W. probably just left it in the circuit from a previous design, even though it's obviated by Q2. Hopefully he sees the thread and voices a response though.
     
  8. praondevou

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    Jul 9, 2011
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    Thanks for the cookie.

    I'm trying to simulate this in Multisim with 10kHz to 33kHz but I don't see a difference in using Q2 and D1 or not (on the output pulse)... Why is this? I'm trying to find out if Q2 is really contributing to the gate charging or not.

    He responded. :)
     
  9. PicStar218

    Thread Starter New Member

    Jul 2, 2011
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    Oh hey, you made it!! Haha thanks for the response. Yeah it totally just clicked why the diode is needed; I can't believe I didn't notice that. And thanks for clarifying R5 - as long as it's not vital to the circuit operation, that's all I needed to know. Once I hammer out the basics of my circuit I'll go back and sprinkle some safeguards where I think necessary.

    Actually, since you're here, maybe I can pick your brain on that as well... The circuit I'm designing will be running an 8A DC motor/pump. So obviously there's going to be some inductance/back EMF considerations. A few things I'm pondering:

    1. I assume I need a flyback diode, correct? If so, should I wire it directly across the load (i.e. between ground and the pFET Drain/output)? I'm kinda leery about that, because it seems like then it'd be momentarily applying a high voltage to the pFET Drain, which could reverse drive it...

    2. Should I add a Zener between the Gate and the FET Drain/output? Most example circuits I've seen use a Zener in this fashion, to "protect the Gate from overvoltage." I'm a bit confused on this, because I don't know how the Gate could possibly be subjected to such an overvoltage.

    Again, any help would be greatly appreciated... Any insight at all concerning inductive-load driving would be helpful. I understand why flybacks are needed for inductive loads, but I'm still not 100% on the whole reverse-voltage/back-EMF issue. For instance, since the pump uses permanent magnets... Once the voltage is disconnected, and the pump output shaft continues to spin, doesn't that make it operate like a generator? So it will be reverse-biasing the pFET? Again, I don't fully understand heavy DC motor characteristics.
     
  10. PicStar218

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    Jul 2, 2011
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    I know!! He must've responded while I was typing. I'm a slow poster!!!
     
  11. PicStar218

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    Jul 2, 2011
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    You know that's funny, I noticed the same thing. The actual output doesn't really change whether you use Q2 or simply a pull-up alone. But the GATE voltage DEFINITELY shows a change, and that's what I'm concerned about, seeing as a slow Gate ON/OFF speed will greatly increase the FET's ON resistance and cause it to overheat. Considering I'm pulling 8A, I want the Gate switching speed to be as fast as possible.
     
  12. PicStar218

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    Jul 2, 2011
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    Here's the sim with D1 reinstated. Notice how quickly the gate switches now, compared to the other sim I posted, where I had replaced D1 with a piece of wire.
     
  13. praondevou

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    Well, that's really funny then. Multisim doesn't do it, there is no difference in gate rise time between D1 or D1 short, i.e. Multisim ignores Q2...:confused:
     
  14. PicStar218

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    Jul 2, 2011
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    Hmm... yes, that's very strange indeed... The difference between my two sims is literally just replacing D1 with a short. I just ran both sims again to double check.

    Eek. This really makes me question the accuracy of simulation software :(. I'm using LTspice IV btw.
     
  15. SgtWookie

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    Jul 17, 2007
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    The original simulation from the old thread was run at 100kHz, 50% duty cycle. Better to simulate at a higher frequency to see where the problems may lie.

    I've added a 2nd simulation in with the original; the 2nd sans the upper transistor and diode. The upper pink plot is the power dissipation in M1 (original circuit) and the lower red plot is the power dissipation in M2 (without the transistor/diode).

    I changed the load resistors so that load current was about 8 Amperes. I changed the transistor from the original 2n2222 to a 2n3904; that resulted in a slight improvement. A 2n4401 was slightly better than the 2n3904.

    The average power dissipation without the upper transistor/diode is over 2.1 times as high as with it; ~6.5W vs 13.6W

    With lower frequencies, the difference in power dissipation would not be as noticeable in a simulation, as a significantly lower percentage of the time would be spent in transition.

    As far as the differences in simulations between MultiSim and LTSpice - I haven't seen the MultiSim simulation, and I don't know what the models being used in the MultiSim simulation look like.
     
    Last edited: Aug 9, 2011
  16. praondevou

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    Where does this french word come from? :)

    I tried it again with exactly the same circuit, apart from the 5817 being a 5820. There is no difference in gate voltage signal wave form with or without diode/transistor... There has to be something wrong with the models they use... Maybe it's better to move to LTSPICE too... :confused:
     
  17. SgtWookie

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    Well, like I mentioned before - perhaps the models used for the transistors, diodes, or MOSFET are different, or they're using a very different algorithm for the SPICE simulation. LTSpice isn't perfect, but it's pretty good.

    I've never used MultiSim, so I really can't comment on it.
     
  18. SgtWookie

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    Jul 17, 2007
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    At what voltage?

    Just wire it across the motor terminals, cathode towards more positive. Use a diode rated for at least the motor current, and mount it on a heat sink. It should be a fast diode; a slow diode will result in excessive power dissipation and low motor power.
    Sure, you could add that.
    The gate can wind up going much more positive or negative than the source terminal due to inductance of the interconnecting wiring.

    Any current that the motor generates will be provided a path via the reverse-EMF diode.
     
  19. PicStar218

    Thread Starter New Member

    Jul 2, 2011
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    Holy smokes... You never cease to amaze me S.W.!! Sorry I've been MIA for a while, I'm currently "in between homes" so to speak... So give me a bit to digest your last few posts and I'll get back to you. That power calculation is brilliant btw, I would have never thought of that or figured out the syntax without that pic.
     
  20. SgtWookie

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    I used the super-secret key press. :cool:

    Holding down the ALT key while hovering the cursor over a component will change the cursor to a thermometer. If you then left-click on the component, you will get a similar formula and a plot in the plot pane.

    This is a heck of a lot easier than having to type in the formula, which is what I was doing before I found out about using the ALT key.

    After you get the formula showing in the plot pane, you can hold down the CTRL key, and left-click on the formula to get the average and integral power for the time range that you have displayed. The simulation has to be completed or stopped for this function to work.

    By the way, this is also a handy way to get an idea of the efficiency of a circuit. You can get the average power input from any power sources, and then the average power outputs. Efficiency% = Power_Out*100/Power_In
     
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