mosfet makes my power supply ring

Discussion in 'General Electronics Chat' started by Gibson486, Dec 6, 2012.

  1. Gibson486

    Thread Starter Member

    Jul 20, 2012
    199
    12
    I am using a SI1488DH-T1-E3CT-ND mosfet and I am controlling it through a micro controller. The switch speed is pretty fast (about 6 us on, then it turns off for 50 us). The issue is that my PS gets lots of ringing (a decaying oscillation) now when the circuit goes. I was able to get the ringing down by adding a .1 uF ceramic cap to ground, but the ringing is still there (initially about 300mV, then it decays...it was originally a 1V initial peak!). What can I do to get rid of this ringing?

    Also, for the FET, the Vgth is very low, but I still cannot get a lot current from drain to source. Basically, I need the LED (LE UW S2W) to be as bright as possible, but it looks like I am getting the same voltage drop at the drain no matter what current sourcing resistor I use. It seems like it is really half as bright as it should be.
     
  2. tindel

    Active Member

    Sep 16, 2012
    568
    193
    Can you post a schematic?
     
  3. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    You are only turning on the LED 5/56 of the time. This provides a much lower average voltage to the LED than just turning it on steadily with a switch.

    One way to make up for this is to make the short pulses a much higher voltage.
     
  4. praondevou

    AAC Fanatic!

    Jul 9, 2011
    2,936
    488
    Why don't you increase the duty cycle?
    What does "current sourcing resistor" mean? Is that the resistor in series with the LED?
    The source of the MOSFET is connected to GND, then between + and the drain is the LED/resistor combination connected.
    What gate resistor value are you using?

    As for the ringing, you need to provide schematic. An unappropriate layout can also be a reason for ringing, so if possible post a picture of it.
     
  5. Gibson486

    Thread Starter Member

    Jul 20, 2012
    199
    12
    Here is a link to circuit....

    [​IMG]

    All of this is on a breadboard.....
     
  6. Gibson486

    Thread Starter Member

    Jul 20, 2012
    199
    12
    That sounds right...maybe I can put another fet in front so I can get 12V to gate...you know, like a darlington, but with a fet? On ly problem is that this fet is 8V max.
     
  7. thatoneguy

    AAC Fanatic!

    Feb 19, 2009
    6,357
    718
    Remove bell from power supply. :D

    See sigline. You DO have bypass caps, don't you? 0.1μF mono/MLCC ceramic + >22μF electrolytic on power bus, as close to controller Vdd and Vss as possible? (See Sigline)

    If bypass caps are in place, a shot in the dark would be adding a higher "gate load" to the mosfet with a 0.01μF from gate to source?
     
  8. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    If you are really using a 360 ohm resistor in series with the LED, the current to the LED is way too low. Ignoring the FET drop, (12-3.5)/360= 24 ma. This LED has close to a 3.5 volt drop at 700-800 ma if I read the datasheet (page 9), correctly.
    You could use a 12 ohm at (yes) 10Watt resistor.
    Can your PS supply 1 amp or more @12v ?
    Don't do that, might blow out the 8V gate. It isn't needed, doesn't work like a darlington.
    Edit: I was unclear - meant that you should drive the LED, not the MOSFET

    What is the voltage out from the MCU ?
     
    Last edited: Dec 7, 2012
  9. praondevou

    AAC Fanatic!

    Jul 9, 2011
    2,936
    488
    Can you post a picture of it showing where power supply comes in and where bypass caps are located?

    You also should use a gate resistor, e.g 10R to start with. Not using gate resistors can easily lead to ringing when switching the MOSFET. The gate to source resistor isn't absolutely needed, you can leave it if you want but within reasonable limits it won't make a difference in LED brightness (as you've seen yourself)
     
  10. Gibson486

    Thread Starter Member

    Jul 20, 2012
    199
    12
    Ringing is gone. Now, it just dips 100mV about the length of the pulse. I sprinkled bypass caps to get it down to 100mV. What else can I do?

    edit: I also put a resistor in series with the gate instead of using a pull down at the gate. That fixed the issue alot!
     
    Last edited: Dec 7, 2012
  11. tindel

    Active Member

    Sep 16, 2012
    568
    193
    I agree putting a resistor on the gate is mandatory. Without it you're basically shorting out your uC output when you try to turn it on, this is probably causing oscillations in your uC, not your FET, because your uC probably isn't rated for short circuit current. Although a FET is generally considered a transconductance amplifier (voltage in, get's current out), it takes current to charge and discharge the FETs internal capacitance before it starts to work like a FET... The phenomenon is explained well here: http://www.irf.com/technical-info/appnotes/mosfet.pdf in the 'gate charge' section.

    Your resistance that you put on your gate should be a function of your uC maximum output current to obtain your maximum turn-on speed. If your uC can only source 25mA (WAG - so look at your datasheet) then I'd make the total drive resistance something close to 5V/25mA = 200ohms, any lower and you risk damaging your uC because you're exceeding your uC output source current. You can make this resistance larger, you just sacrifice turn-on speed.

    However, please note that you're 1k gate-source resistor is taking 5mA by itself, so if your maximum uC current is 25mA then your maximum gate current would be 25mA-5mA = 20mA so you really need 250ohms instead.

    If you decide you need more current to turn on the FET faster, then there are ways to get more current by using a BJT to drive more current to your FET or by using a higher current source buffer chip.

    Note that with your current schematic a 5V output will be sourcing 5mA all the time into the 1k Gate-Source resistor. I'd probably make that resistor 10k just to bring down the current draw from your uC (and in turn, your power), but to have adequate pull-down on the FET. Then you can lower your gate resistance to get it to switch a bit faster.

    Your LED current seems low, as others have pointed out, so I won't touch on it.
     
  12. nigelwright7557

    Senior Member

    May 10, 2008
    487
    71
    You can get ringing if you don't use a gate resistor.
    The output of the micro is trying to drive a capacitor and rings.
     
  13. JMac3108

    Active Member

    Aug 16, 2010
    349
    66
    You should still use a pull down on the gate to make sure that the mosfet does not turn on when the gate is not driven, for example when your micro-controller is first booting and the pin has not yet been setup. Remember, N-channel mosfet gates will float high and turn on the mosfet when the gate is not driven. Its not critical in this circuit because you may not care if the LED flickers at power-up. But if you ever use a mosfet in a power application and this happens, you stand a good chance of starting a fire :D

    The series resistor should be very small, something like 10 ohms, therefore the pull down, which should be large >10k) will not affect the gate voltage.

    Also, you may find that the ringing goes away when you build the real circuit. Breadboards have a lot of inductance which will cause the mosfet gate to ring.
     
  14. tindel

    Active Member

    Sep 16, 2012
    568
    193
    There has been 10 ohm gate drive suggested from two people... I'm trying to understand the logic of using such a low impedance gate drive resistor. Surely the uC can't handle driving that kind of current even for short periods, especially if you're duty cycling like the OP indicated - regardless of oscillation.

    To clarify my previous post - I suggest putting the gate drive resistor between the gate and pull-down, not between the pull-down and uC - that way you have no loss in gate-drive voltage after the FET has been charged with no effect on the gate pull-down resistor.

    I know I'm new here - but I'm not new to electronics, and I'm learning more every day. Maybe I can learn something from somebody else's experience. Maybe I have overlooked something in my previous design.
     
  15. JMac3108

    Active Member

    Aug 16, 2010
    349
    66
    Tindel,

    The gate drive resistor is so small compared to the pull-down that is has no affect on the gate voltage. Typical values are 10 ohms for the gate drive resistor and 100K for the pull-down. Think about a voltage divider with a 10ohm on top and a 100K on bottom...

    There are several purposes for the gate drive resistor. Sometimes its used to damp ringing in the gate circuit. But mostly I use a gate resistor in switching applications to reduce radiated emissions to pass FCC testing. The resistor slows down the charging of the mosfet gate capacitance and slows down the edges of the switching waveform. This reduces the amount of high frequencies radiated by the circuit. Sometimes in switching power supply design you have to do this. Its a tradeoff though ... reducing eniisions to pass FCC at the expense of decreased efficiency in your circuit.
     
  16. tindel

    Active Member

    Sep 16, 2012
    568
    193
    A larger gate resistance lowers di/dt, improving RE... of course you may not be able to lower your di/dt depending on your switching frequency, power loss, heatsink mass, or other factors.

    I think we're saying the same thing. Just two different ways! Ha!

    I do worry about reliability if the gate is consistently seeing 500mA though. 10 ohms just sounds too low to me. I wish my home lab was better so I could test things like this. I don't have enough time at work. :rolleyes:

    For the OP: In this application I'd still recommend a resistor in the couple hundred ohm range between the pull-down resistor and the gate to protect the uC output drive circuitry, just be aware that your power consumption of the FET will increase and switching time will slow, but should still be adequate for your application, as long as you heatsink appropriately.
     
Loading...