What characteristics should I look for the PNP transistor in an active PNP turn-off circuit for driving large TO247 mosfets? It is intended to be after the IR211x driver output.
-I guess VCEmax around 25V is more than enough, as Vce=Vgs.
-A high max. collector current (2 to 10 amps) is desired for really fast turn off, right?
-What about Hfe min, transition frequency, etc?
I want to run it at 350KHz approx.
I have found some candidates such as :
MJD2955
ZXTP2012ZTA
MJD45H11TM
Do you think they are suitable? Would you recommend any other part number?
Thanks!
-I guess VCEmax around 25V is more than enough, as Vce=Vgs.
-A high max. collector current (2 to 10 amps) is desired for really fast turn off, right?
-What about Hfe min, transition frequency, etc?
I want to run it at 350KHz approx.
I have found some candidates such as :
MJD2955
ZXTP2012ZTA
MJD45H11TM
Do you think they are suitable? Would you recommend any other part number?
Thanks!
Don't the IR211x gate driver IC already have active gate turn off?
Also they can switch at 120ns..???
Also they can switch at 120ns..???
A 2n3906 is enought! The peak to discharge a gate is about fews ns....Except if you use a tick trace on your PCB, IR2110 is enought to discharge it. The only draw back, is the dissipation. Here you need a PNP discharge help...
Fred
Fred
Well, I have seen some peak before turning on in the Vgs voltage that I don't like, it grows (up to 6-7V or more) as duty cycle is increased. I believe it can be due to reverse transfer capacitance turn on, so I think that a PNP turn off will help keeping the Vgs voltage down. Right?
I also want my IR2110 working more relaxed...
I also want my IR2110 working more relaxed...
Well, about IR2010, nice, but "only" 200V max. Vcc. A bit tight if you want +/-90V rails, isn't it?
Adding an optoisolator (or similar) and mosfet driver is a good idea, but I want to keep it as similar to the original design as possible, by the moment (using only IR2110).
Adding an optoisolator (or similar) and mosfet driver is a good idea, but I want to keep it as similar to the original design as possible, by the moment (using only IR2110).
Follower using it's good idea indeed, ir2110 almost have not chance to live, after the mosfet blew up, bc807/817 will have much more chance even. On the other hand, why you need 2A($2) driver, if followers in use anyway? So, you can get more reliability for less money.
The idea is simply to update the existing design for faster turn-off, by the use of a typical active-turn off PNP transistor.
The turn-on is still driven by the IR2110, so you need 2A. This way the chip will work more relaxed, too, as it doesn't have to sink the high turn off current.
The turn-on is still driven by the IR2110, so you need 2A. This way the chip will work more relaxed, too, as it doesn't have to sink the high turn off current.
Consider Zetex ZTX851 and ZTX951, they exhibit excellent gain and Ft up to a few amps of Ic in a TO-92-like package. Problem with other devices is beta anf Ft drop at high currents.
In my experience, IR2113 can handle 4Mhz operation with these transistors as buffers.
In my experience, IR2113 can handle 4Mhz operation with these transistors as buffers.
Yes, I tried them as emitter followers, and they do a great job keeping the IR2113 cool although they get pretty hot themselves.
Aren't those a little sluggish to be used at such a high switching frequency (turn off delay 1100 ns) ? I know that this delay is when highly saturated but it still leaves questions open.
Regards
Charles
How about using IRFD110/9110 mosfet pairs in Source Follower configuration....I think the Vgs threshold would pose a problem in this case....Some big guys are using them in Common Source configuration.
When you say that you are referring to the typical configuration where the PNP collector is connected to the mosfet source, the PNP emitter is connected to its gate, with a fast diode between base and emitter, and the base is driven by the IR2113 with the gate resistor in series, right?
The saturation is no more a limiting factor, because the Gate's current is falling exponentially. When the transistor is switching on at the opponent side, the B-C junction's recovery is ended. The saturation happens only at a very little current value what is falling exponentially.
phase_accurate said:
I was expecting from you a better understanding of bipolar transistor operation... 😉 An emitter follower driving a MOSFET gate never becomes saturated.
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