Problem with driving P-channel MOSFET..

Discussion in 'General Electronics Chat' started by hazim, Jan 5, 2013.

1. hazim Thread Starter Active Member

Jan 3, 2008
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Dear all.
I hope everybody is ok.

I have a small problem in this circuit. The circuit is designed to switch off the load when the input voltage decreases under 12V, this is set by R3 pot. and to regulate the voltage at output when the input voltage exceeds 24V, so when the input voltage is in between 24V and 30V (which will be the maximum input voltage) the IRF5305 will be off and the current will drain across the leaner voltage regulator circuit (U3, TIP35C...), the 24V is set by R1 pot.

Now as I said, the circuit is working fine but there is a small problem facing me. U1A's output is used to sink current in order for the PNP transistor to work, LM358 needs a minimum sink current of 10mA and maximum of 20mA, I think the 2N2906 is a good choice. U1A sinks current as it's set, when the supply voltage is 24V or lower, calculating at maximum voltage (24V) for RB will give approximately (24V-10V)/0.015A=933Ω.. so I used 1KΩ resistor...

Now IC will be approximately 150mA, which means 0.15Ax24V = 3.6W, when calculating the values of RC and RE I found that they should be about 100Ω and 60Ω, 2 to 5 watt resistors.. this is my problem. So as I need the PNP transistor for only switching the p-channel ON and OFF, I need to dissipate that power and use high wattage resistors, and this seems to be ridiculous!

It seems that I need an "inverse darlington" here

Any idea?

Regards,
Hazim

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2. tubeguy Well-Known Member

Nov 3, 2012
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I'll mention a couple things.
You could eliminate Q4, change the 620 ohm res to a 47k, and drive the P-fet directly thru a 4.7k res like you are doing for the N-fet if you are just turning them on/off. Add a 47k pulldown to the gate of the N-fet also.
I believe you can eliminate the 1k R18.
What's the purpose of Q3 ?

Other issues. R4 @ 100 ohms will turn on U4? TIP35C at 7ma current, bypassing (I think) the 317.

Last edited: Jan 5, 2013
3. Ron H AAC Fanatic!

Apr 14, 2005
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Good advice and good questions, tubeguy.

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4. hazim Thread Starter Active Member

Jan 3, 2008
419
13
tubeguy, first I think you mean the 62 ohm resistor, there is no 620 ohm resistor. I think you didn't understood why I used the PNP transistor, it's because I need the MOSFET to close when the voltage increase above 24V, i.e:

Vss>24 , then U1 output is "high" , then Q4 is closed , then Ic=0A , then Vgs=0V , so Q1 will be closed and the linear voltage regulator works.. what you suggested is what I did in the first version of the circuit, that makes the MOSFET be closed under 24V and opened above 24V, in opposite for what I want. Remember that the source is connected to Vss, and the zener is a must because without it Vgs may become -30V which will damage the MOSFET.

The 1k R18 is to prevent burning the IC, because the source voltage may exceed 30V a little, where the IC's maximum supply voltage is 32V. (Maybe the value 1k isn't good, should look for the approximate LM358's current drain at 30V)

Q3 will prevent current to pass through the circuit if the supply voltage polarity is reversed, so it works as a protection.

For the voltage regulator part, actually I didn't understood it completely, especially for the resistor values. I see the 100 Ohm will make U3 just control the output's voltage, and leaving the current supply for the TIP35C, I don't know why to use a smaller resistor, say 0.7V/0.7A= 1 Ohm 1Watt resistor, so U3 will supply 0.7 or even 1A, where TIP35C is supplying 10A at the same time for example, and TIP35C is capable of supplying much higher current, then why to use U3 to supply current? I think I even could use the TO-92 package of LM317, or LM317L)

Regards.

Last edited: Jan 5, 2013
5. tubeguy Well-Known Member

Nov 3, 2012
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Yes I meant 62 ohm.
Q4 is connected as a voltage follower so when U1 is HI so will be the gate on the P-fet. The 62 ohm pullup is just much too low and requires unnecessary current as you pointed out yourself.

No mention was made of the 30 volt supply being maybe a little higher.
I did not say to remove the protection zener, only Q4. You might instead want to use a zener/resistor to hold the IC supply to 30VDC

The 100 ohm resistor drops .7 volts at 7ma which turns on the TIP35 and bypasses the 317. It appears to be intended for current limiting, so should limit above the max load current.

A common way to prevent damage from PS reversal is a series schottky/diode on the + input

Last edited: Jan 5, 2013
6. tubeguy Well-Known Member

Nov 3, 2012
1,157
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The 10W load requires 333ma at 30v so you need the LM317.
The 100 ohm resistor should be re-calculated to provide about a 0.7 volt drop at maybe 1/2 amp.

Edit: Forget the limiting reference. Its a bypass circuit. Not needed.

Last edited: Jan 5, 2013
7. hazim Thread Starter Active Member

Jan 3, 2008
419
13
what you said that Q4 is connected as a voltage follower, and the 62 ohm works as a pullup resistor is right, but as I said before, I choosed it as 62 ohm specificly by calculating the required Vzener/150mA or 9V/0.15=60 ohm, 150mA came from Ib times 10 or 15mA x 10 = 150mA. The base current 15mA is specified by LM358 datasheet, where the sink current should be between 10mA and 20mA.

I'll assume the supply voltage may exceed 30V a little, but anyway this isn't my problem, and neither I have a problem in the voltage regulator circuit, I said that the circuit is designed to be able to supply 10A as an example, and the 10W 30V bulb isn't the real load.

The load will drain around 10A at times, and this requires a big high current and wattage schotky diode, but the way I used the MOSFET Q3 makes Vds almost equal zero which means there will be no real loss, it works great and is a nice idea.

Again, my problem is difinite there around the pnp transistor.

8. Ron H AAC Fanatic!

Apr 14, 2005
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You're reading the datasheet incorrectly. The LM358 does not have a minimum load requirement.

9. tubeguy Well-Known Member

Nov 3, 2012
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As RonH pointed out there is no minimum LM358 sink/source current requirement. Current spec is the maximum available. Why would the same LM358 be able to directly drive the N-Fet through 4.7k OK, but not the P-Fet ??

Look at the LM317 datasheet. Can it supply 10 amps ??

You know what... just for fun ... why don't you just try what is suggested.

Last edited: Jan 6, 2013
10. hazim Thread Starter Active Member

Jan 3, 2008
419
13
Dear. I've used 2N3906 instead of PN2906, as I'll use it for lower current... I used the resistor values as in the following picture. I'm not sure if that's ok, and put it here for you to see and comment. tubeguy, of course the 10A will not be drained through LM317 but TIP35C.. Anyway lets finish from the first point.
Regards,
Hazim

• p-channel.png
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Last edited: Jan 13, 2013
11. tubeguy Well-Known Member

Nov 3, 2012
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Hello Hazim,

Don't see the picture, would you please re-post.

Thanks

12. Ron H AAC Fanatic!

Apr 14, 2005
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Here are instructions for posting attachments:

13. hazim Thread Starter Active Member

Jan 3, 2008
419
13
Sorry. I edited the post and now you should see the picture.

14. tubeguy Well-Known Member

Nov 3, 2012
1,157
197
Well, still no picture.
Before posting you can preview the post to ensure that your attachments work. And make sure the fornat is one that is allowed,

15. Ron H AAC Fanatic!

Apr 14, 2005
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It's in post #10. (He said he edited the post.)

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16. hazim Thread Starter Active Member

Jan 3, 2008
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You should see it in post #10... but again here it is:

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17. Ron H AAC Fanatic!

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Why do you think you need the emitter follower? You can remove Q4 and R19, change R2 to 4.7k, and it will still work. In fact, it will probably switch faster.

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18. hazim Thread Starter Active Member

Jan 3, 2008
419
13
I don't remember what I was thinking when I put the transistor . Anyway, yes I see no problem without it, as you said. Thank you for helping.

19. hazim Thread Starter Active Member

Jan 3, 2008
419
13
Dear.

Back to this thread again, I still not sure about the resistor values of this part of the circuit, the part in the picture below. How to calculate the values of R4, R14 and R13. Note that the input voltage and output voltage is mentioned in the picture, and the output current is 8A, and I'll put another TIP35 in parallel with the existing one with a good heatsink.

Regards,
Hazim

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20. hazim Thread Starter Active Member

Jan 3, 2008
419
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Calculating according to the equation Ic=10(Ib), if Ic of TIP35 is 8A, then calculating for Ib=0.8A which is approximately equal to Ic of PN2907, then Ib of PN2907 is 0.08A.
Ic(max) of PN2907 is 0.8A, I may use higher current transistor instead of paralleling two PN2907's, even though the actual current maybe 6A not 8A, assuming 8A is the maximum current.