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Subject: Re: Constant current source with low voltage drop
Date: Sun, 29 Sep 2002 12:46:44 +0100
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"Bob Wilson" wrote in message
> In article ,
> firstname.lastname@example.org says...
> >"Jim Thompson" wrote in message
> >> On Sat, 28 Sep 2002 23:18:35 +0100,
> >> "markp" ,
> >> In Newsgroup: sci.electronics.design,
> >> Article: ,
> >> Entitled: "Constant current source with low voltage drop",
> >> Wrote the following:
> >> |Hi All,
> >> |
> >> |I need a constant current circuit adjustable from 0 to 150mA powered
> >> |5V supply. This is to drive LEDs of an LCD backlight. The spec says a
> >> |4.4V drop @ 150mA (2 LEDs in series?) so I've got to do this with a
> >> |of 0.6V voltage drop. Any suggestions?
> >> |
> >> |Thanks!
> >> |
> >> |Mark.
> >> |
> >> P-Channel power MOSFET controlled by OpAmp with common-mode range that
> >> includes positive rail; small sense resistor in source of P-Channel.
> >> ...Jim Thompson
> >Yes, that's my favourite so far. However I was considering a PNP
> >as a switch followed by a P channel FET. The PNP would turn on at around
> >0.6V or so and shut off the FET. Current is then controlled by a resistor
> >from +ve to base of PNP. Your solution is better though as the PNP
> >is on the edge!
> Forget the FET; it is irrelivant. Using a PNP pass transistor, you can get
> very low dropout voltage. You are incorrect that the PNP will have 0.6V
> across it. When fully saturated, you it will get downt to mere tens of mV.
> Now a word of caution with ANY low dropout regulator (Built from discretes
> OR integrated): The PNP (or P-channel FET) pass element has gain (i.e. it
> not an emitter or source follower). Thus you may run into self oscillation
> unless you use the correct value of output capacitance. This is unlike
> normal regulators using NPN or N-FETS what are stable with no output caps.
I was thinking of using the PNP as a switch that would turn on when the
current was 150mA and produced a 0.6V drop across the emitter-base. I'm
assuming you are talking about common base configuration, but I'd still need
to measure the current and turn it off somehow.
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