Buck converter P-channel MOSFET switching on problem

Status
Not open for further replies.
goldsmith said:
Dear Friend , my mean is that the P mosfets are alike with N mosfets but with this difference that if the voltage across the GS of them become -15 volt the will be at saturation region .
Click to expand...
Yes, though with FETs it's the linear region, not saturation.
so with this considerations , the efficiency equation and loss power equation will be equal with NMOSFET . isn't it ?
Click to expand...
That equation for conduction losses is valid while the FET is conducting. You can see from the waveforms where Vgs is pulled negative and the FET is on (the red trace is gate voltage). But the ringing has nothing to do with those conduction losses. So I'm still waiting for you to explain why you believe the ringing affects dissipation/efficiency.
 

Dear Friend
Hi again
How are you ?
at first step , i should ask that what is your mean by " Yes, though with FETs it's the linear region, not saturation. " ?
if the mosfet become at linear region , it will be like a dissipative element. and it is not good.



and as you can see at his attachments, the mosfet is not completely off . and at this time the dissipation can become linear . and it is not good .

And about Vgs . the red voltage isn't the gate voltage . see below of that ( VR2-1) it is before 1.5k resistor . and if the voltage across the GS junction exceed from 15 volt it will be destroy or you can see it's explosion!

Best regards
Goldsmith
 

This is sort of a nit pick on my part. The operating regions for FETs are named differently than BJTs. In switching circuits, BJTs operate in saturation region, FETs operate in linear region. Though most people incorrectly use the BJT region names for MOSFETs.
MOSFET - Wikipedia, the free encyclopedia

and as you can see at his attachments, the mosfet is not completely off . and at this time the dissipation can become linear . and it is not good .
Click to expand...
I don't see this. His Vgs is clearly near 0V during the ringing.
[quote[
And about Vgs . the red voltage isn't the gate voltage . see below of that ( VR2-1) it is before 1.5k resistor .[/quote]Yes you're correct, but the actual gate voltage will be almost exactly the same, just a little slower. It will still be ~0V during the ringing (though you can see there is a bit of ringing due to capacitive coupling. Still not enough to turn on the FET).
and if the voltage across the GS junction exceed from 15 volt it will be destroy or you can see it's explosion!
Click to expand...
I don't think most simulators would care about the max Vgs. It certainly won't explode.

Now if you were just talking about the slow switching time of Vgs affecting switching losses, then I would be agreeing with you. But you're saying that there's some relationship between the PFET operation and the DCM ringing, but you haven't been able to explain it at all... please understand that I just don't want people to get the idea that DCM ringing is a problem, because I'm quite sure it isn't (unless maybe the ringing tends to be at a certain frequency which is bad for the rest of the system).
 

My mean by saturation was this : the mosfet turning on and the VDS , will going to decrease about zero ( approx) .

By the way : the behavior of mosfet is alike with BJT but with some little differences .


And about "Now if you were just talking about the slow switching time of Vgs affecting switching losses, then I would be agreeing with you"

Yes my was this .
And about ringing : when the voltage of ringing is lower than max , it means that mosfet is not complete turned off . and thus the current will have the component alike with ringing , thus the loss power will be that ring voltage * that current ring ( with avg quantity ) and thus , the dissipation prediction will not simple at that time . this was my another mean.
 

goldsmith said:
My mean by saturation was this : the mosfet turning on and the VDS , will going to decrease about zero ( approx) .
Click to expand...
Right, that's the term most people use, but it's technically wrong. I know what you meant. That's why I said it was a nitpick

And about "Now if you were just talking about the slow switching time of Vgs affecting switching losses, then I would be agreeing with you"
For clarity, I've recreated the simulation myself so we can look at this in detail:


Here is a look at all the relevant waveforms:

Notice that Vgs is ~0V during the ringing. So the FET is definitely in cutoff. But still there is definitely some small current flowing through it.

Here's a closeup of the PFET current, and the PFET Vds:

You can clearly see that the current and voltage are 90 degrees out of phase, meaning that the current is not flowing through a resistive channel, but rather through the parasitic capacitance of the FET. Thus it dissipates practically no power. I went ahead and plotted the FET power as well, and it averages out to around 200uW over the entire ringing (around 25nJ). This is an utterly negligible amount of power, compared to normal switching losses in the FET and diode (it accounts for about 0.1% of power losses).
 
Dear Mtweig
Again Hi
You're completely right . i think it was a misunderstanding from me . And Thank you for your patience . And i wish success for you at all of the levels of living .
Best Regards
Goldsmith
 

Dear goldsmith,

I am interest to your statement that possible to make high voltage buck converter with n channel mosfet. Now, I am designing a buck converter which Vin=150V, Vout=14V. The problem on my circuit is slow P mosfet turn-off delay time. Any suggestion are appreciated.

Best Regards
purnomo16
 

Dear goldsmith,

Thanks for your reply.

If I use N channel in buck converter, regard to common circuit, I have to connect output part (inductor-diode) at source pin. In other hand, drain pin is connected to the high voltage source (100-150VDC). To turn on this mosfet, gate voltage is 10-20V. At the same time, the output on source will not more than 20V all the time eventhough voltage source is reduced. so, I think that buck converter will not work properly. how?

regards

Purnomo16
 

Hi again
You shouldn't drive it with those classical ways ! you can use a PNP transistor and an NPN transistor to drive your nmosfet , or use a simple boot strap driver like IR21xx ( 2130 /2113 /2110 .... etc ).
Best Wishes
Goldsmith
 
Dear Goldsmith

I have replaced to IR2117 of Mosfet Driver. It's work. Thank you

However, my buck converter will become halt if its output connected or pre-connected to 12 lead acid battery.

My buck specification, Vin=60 V, V for IR2117 = 12 V, Inductor=200 uH, Freq=100 kHz, Mosfet=IRFP260.

Any suggestion are welcome..

Thank you very much

Purnomo
 

Hi again
And what is the value of your out put capacitor ?
BTW : do you have any feed back path ?
Can you show me the schematic , please ? thus i can help you , as well as possible .
Best Wishes
Goldsmith
 

Thank you for reply

Output capacitor is 220uF and it has no feedback path. I will upload my circuit ASAP. However, It is a simple buck converter, similar to basic circuit of buck. I think that the mosfet was not switch on since source pin got 12V. How?


goldsmith said:
Hi again
And what is the value of your out put capacitor ?
BTW : do you have any feed back path ?
Can you show me the schematic , please ? thus i can help you , as well as possible .
Best Wishes
Goldsmith
Click to expand...
 

purnomo16 said:
Dear Goldsmith

I have replaced to IR2117 of Mosfet Driver. It's work. Thank you

However, my buck converter will become halt if its output connected or pre-connected to 12 lead acid battery.
Click to expand...
This is not surprising, and it has to do with the operation of the IR2117. In order for it to operate, the bootstrap capacitor must be charged first (meaning the source of the FET must be brought close to circuit ground). If the load voltage starts at zero, then the boost capacitor can charge through the buck inductor, and it should work fine. But if the output voltage is already high, then this can't happen, and the bootstrap cap will never charge.

This isn't really a problem with any specific part of the circuit. It's an expected behavior.

There are a couple things you can do:
1) Use synchronous rectification. This will allow you to forcibly charge the bootstrap capacitor with a low side FET
2) Use a gate driver that doesn't require bootstrapping, i.e. a driver with a truly isolated, regulated power supply.
3) Just don't start the circuit operation with the battery attached.
 
Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…