Turning on and off current source

Hi all

I am thinking which way is the best to turn this on/off?

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I need to turn this on/off, and the easiest way would be simply to turn power off at the main transistor. But since I have a few amps there, it is not that easy.

Therefore I came up with these solutions:

Q3 can turn my reference off, that should make it all fall to ground and turn off. I can also pull down pin 13 on the opamp.

Q4 will also do the trick, but will cause the current when turning on to be very high, therefore there is a condensator to give a soft start (which might be a good idea no matter how I turn it on or off)

Finally I can add a transistor to Q2 in series, which will also control it.

Please dont discuss specific values as I ask for a general functionality.

WBR Sonnich

Reply to
Sonnich Jensen
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"Sonnich Jensen"

** Google Groups Groper Alert !!!

** You want to turn the load and off remotely ??

FFS - say what you mean.

** You want to turn the * load * and off ??

FFS - say what you mean.

** Looks dead simple to me.

** FFS - say what the load is.

Such details are crucial.

therefore there is a condensator to give a soft start

** Yawnnnnnnn ....
** WTF is it you want to know - pal ???

YOU have NOT asked US an ACTUAL question !!!!

.... Phil

Reply to
Phil Allison

On a sunny day (Mon, 23 May 2011 23:57:00 -0700 (PDT)) it happened Sonnich Jensen wrote in :

It is attractive to change the reference, but would decrease the quality of reference perhaps. Q4 seems to make sense.

Reply to
Jan Panteltje

Are you sure this isn't homework? Do you have access to the left end of the resistor to Q4 base?

Thanks, Rich

Reply to
Rich Grise

h

of reference perhaps.

I agree. Also turning it off could cause funny problems.

Yes, I and I think I got a better idea - Q4 sources my R1. By that I still have the loop running. The capasitor was mostly for a soft start, which still might be a good idea.

Thanks Sonnich

Reply to
Sonnich Jensen

This is an idea I am working on, and the dead end is for control. I am talking ideas, not anything specific. So a high on the dead end would turn it off, the rest can be adjusted as I need it, so NPN or PNP is not (so) important as of now. Currently I dont work with specific values, more with ideas.

WBR Sonnich

Reply to
Sonnich Jensen

Not important, the dead in should give it - add what you want

None as of now. Not important.

It is, but the pricetag of yet another power transistor makes the Q4 solution better.

Not in the basic idea. Well if you want to know, a nuclear reactor.

Well the Dutch guy below got it.

Reply to
Sonnich Jensen

On a sunny day (Tue, 24 May 2011 03:59:31 -0700 (PDT)) it happened Sonnich Jensen wrote in :

I dunno what you mean by that, and I think you have + and - reversed n the opamp?

Reply to
Jan Panteltje

"Son of a Bitch Jensen"

This is an idea I am working on, and the dead end is for control. I am talking ideas, not anything specific. So a high on the dead end would turn it off, the rest can be adjusted as I need it, so NPN or PNP is not (so) important as of now. Currently I dont work with specific values, more with ideas.

** This SOB sure is one wackadoodle.

.... Phil

Reply to
Phil Allison

ich

y of reference perhaps.

You could make Q4 a MOSFET, and use it to short the op-amp input to GND. That speeds recovery by keeping the op-amp's output from railing while the current source is OFF.

Like Jan said, you have the + and - inputs reversed.

You might consider speeding up the feedback (for stability): op-amp output to capacitor, capacitor to "-" input AND to R1 through an isolating resistor.

HTH, James Arthur

Reply to
dagmargoodboat

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nic=3D

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e opamp?

You are right, it just seemed like an idea. And you are right about the opamp imputs.

Reply to
Sonnich Jensen

nnich

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ity of reference perhaps.

New schematic

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I guess I prefer Q4 as here, it will cause it to rail to max, but the capacitor will also be short circuited. When released, it will give a soft start. Also, there might be a change - the resistor in series with the pot might be removed, so I rather leave the inputs untouched.

Jepp, corrected already.

I got the capacitor, and I agree. But where do you exactly want to put the isolation resistor? I can not quite follow you. I agree, that the capacitor here could cause some negative voltage to the input, and a resistor to the - input would be in place. Is that what you mean?

Otherwise I think it is a proper solutuon by now.

WBR Sonnich

(and for those wondering, the part below is an option of a LM3914 - to display the level. It also has a reference of 1,25V which could replace my LM385. Though as it is optional I will probably not do that)

Reply to
Sonnich Jensen

e:

Sonnich

:

nd

on

tart

ality of reference perhaps.

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Between C4B and R1B.

-- Cheers, James Arthur

Reply to
dagmargoodboat

e:

Sonnich

:

nd

on

tart

ality of reference perhaps.

g

You want to put something like 10K between IN(+) and the wiper of VR1, then move that Q4/C3 pulldown junction to IN(+). Also put a 10K between C4 and R1.

Reply to
Fred Bloggs
** Funny how nobody has yet noticed the OP's schem is NOT a current source.

..... Phil

Reply to
Phil Allison

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