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Why not to use a P-Channel MOSFET in this circuit? It makes more sense, to me since the MOSFET act as a resistor when turned on which is better then the constant volage drop of a PNP.

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See Figure 2 of the transistor's datasheet - under some conditions the "constant voltage drop" is only about 0.05V. –  Brian Drummond Jan 9 '13 at 11:56
    
not mine ;) I will be saturating it and consuming about 50mA –  mFeinstein Jan 9 '13 at 13:31

3 Answers 3

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I suppose it could be made to work with a P-channel MOSFET, but I'm not sure what real advantage that would give. Is the voltage drop of about 0.2 volts over the transistor a problem? Probably not, and a P-channel MOSFET is likely more expensive than a PNP BJT, and this circuit is simple and works great.

To highlight one design hurtle to using a MOSFET, what happens \$V_{self}\$ is less than \$V_{bus}\$, but not less enough to reach \$V_{th}\$ of your MOSFET and turn it on? Also, the gate would be driven though a resistor here. If \$V_{self}\$ fails, will you be able to drive the gate low enough and fast enough to maintain power?

Not unsolvable problems by any means, but why bother thinking of them when the BJT works fine?

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The BJT will make the PIC's Vcc be 4.8V and my inputs are 5V so I have to add resistors to make sure I dont blow the PIC's inputs since it's clamping diodes could be triggered. Another problem is the current consuption of the gate, since it will need some current to make the PNP saturated. The problem you spoke about the Vself being not less enough could happen as well in the BJT AFAIK. I also dont like the Vdrop of the diode...but I guess I will just have to accept it, since using 2 MOSFETs will be a lot more complicated since I will need to activate a N-Channel. –  mFeinstein Jan 9 '13 at 4:19
    
VDD should be close to the 5 volts from VBUS. As for watching out for the clamping diodes, where are the devices that are driving the inputs getting their power? Resistors on inputs are often a good idea for signals that are coming from off the board in any case. –  supercat Jan 9 '13 at 5:19
    
They are off the board, probably with another power source...It's a good idea to keep the resistors, I just wanted to be 100% safe and not rely on the clamping diodes at all –  mFeinstein Jan 9 '13 at 6:24
    
@mFeinstein: if these are real problems for you, again, there isn't a reason you can't use a MOSFET. The point is, that will introduce problems of its own. If you need a more ideal solution which doesn't drop the voltage, is robust in a wide variety of faults, and has very low power consumption, you are probably better off looking for a power management IC. –  Phil Frost Jan 9 '13 at 12:28

By looking at the circuit from a functionality point-of-view, your right, you could use a suitable P-Channel MOSFET, and you would get a lower voltage drop.

However, there maybe other reasons why a PNP transistor was chosen (was this circuit designed for production?), such as price, availability, robustness (MOSFET gates are susceptible to punch-through), package size, or the fact it could of been designed 20 years ago.

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I'd use a BC327-40* if the resultant base current required for the required transistor saturation was acceptable. If not then I'd use a P Channel MOSFET with appropriately low Rdson at the max required current and Vth low enough that it is adequately exceeded at all voltage conditions where operation is required.

The MOSFET is capable of a lower loss circuit in most cases both because gate current is ~=0 and because Rdson can be so low on even a quite cheap device as to make it hard to match with a bipolar.

  • BC327-40 / BC337-40 TO92, BC807-40 / BC817-40 SOT23 are about the best "Jelly bean" transistors going outside Asia. (Zetex make better but cost is usually MUCHG higher). Current is up to 500 mA+ (varies slightly with manufacturer), Beta is 250 - 600 and price is usually about as cheap as any. Buy 100 when on special. Should cost under 10 cents each. (About 1 cent in manufacturing volumes).
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As far as I can see the Beta is going to be arount 10 when it's properly saturated, right? –  mFeinstein Jan 10 '13 at 6:30

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