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From: rfwilson@intergate.nospam.bc.ca (Bob Wilson)
Newsgroups: sci.electronics.design
Subject: Re: Constant current source with low voltage drop
Date: Sun, 29 Sep 2002 23:05:29 -0000
Organization: Your Organization
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References:
In article ,
map.nospam@surfanytime.co.uk says...
>
>
>"Bob Wilson" wrote in message
>news:upct768hcck42c@corp.supernews.com...
>> In article ,
>> map.nospam@surfanytime.co.uk says...
>> >
>> >
>> >"Jim Thompson" wrote in message
>> >news:tfbcpu03pi46v2uust6f31164thsvuo76g@4ax.com...
>> >> 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
>> from
>> >a
>> >> |5V supply. This is to drive LEDs of an LCD backlight. The spec says a
>> max
>> >of
>> >> |4.4V drop @ 150mA (2 LEDs in series?) so I've got to do this with a
>> >maximum
>> >> |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
>transistor
>> >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
>solution
>> >is on the edge!
>>
>> Forget the FET; it is irrelivant. Using a PNP pass transistor, you can get
>a
>> 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
>is
>> 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.
No, I wasn't talking about a common base configuration, which (as far as I am
aware) has no possible application in this context. I was referring to the
normal topology where a PNP is used as the pass element. Its emitter is
connected to Vin, its collector to Vout, and current "pulled" out of its base
toward ground is controlled to regulate the output (i.e. collector) current.
This (in essence) is the core of most low dropout regulators, since when the
PNP is turned on "hard", the C-E voltage drop can be very small indeed.
With a more standard NPN pass stage, in order to saturate the NPN, you need
to drive the base voltage higher than Vin, which is obviously rather
inconvenient.
Bob.
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