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morizabal (Electrical)
6 Feb 06 14:56
What is the best method of amplifying output current of op-amp.  I need a cheap solution without using power amps.

Thanks
IRstuff (Aerospace)
6 Feb 06 15:28
Not sure I understand.  You need more power, you need to have a power amp stage.  

Depending on the accuracy and other characteristics required, you could get away with driving a push-pull stage directly from the op amp output and wrapping the feedback to the output of the push-pull stage.

TTFN



itsmoked (Electrical)
6 Feb 06 15:53
IRstuff is correct...  Unless you have other circumstances like a single quadrant output.   Does the output only swing one way?

There are single chip power op-amps..
morizabal (Electrical)
6 Feb 06 16:03
Thanks for the input,

Basically I am using a 5 or 10 volt reference to excite a resistive sensor with sense (feedback) lines.  Since the voltage references provides only small currents I am using an opamp to pass the reference voltage.  I need to be able to source about 150 mA.  The op-amps I am working with generally output about 12-20 mA.  I tried using a single NPN transistor (C = Power, B = Op amp output, E = Load) E feedbacks to (-) input of op-amp.  This approach only seems to work with certain conditions.  For instance it would work with 10 VDC, but not 5 VDC.  If I change the transistor to another type it would work for 5 but not 10.


 
itsmoked (Electrical)
6 Feb 06 16:23
morizabal (Electrical)
6 Feb 06 16:35
Thanks for the links itsmoked,

These seem to be reasonably priced, cost was the main issue with not getting a power amp.  These amps say they have a "minimum" output.  Does that mean they will not operate at less than the specified output current?

Helpful Member!  OperaHouse (Electrical)
6 Feb 06 16:36
The basic circuit should work except for three problems.
1. The transistor may not have enough gain for the current you want and a darlington output may be appropriate.

2. Your supply does not have enough voltage for 10V considering the op-amp will not go up to the rail voltage.

3. You could make a circuit change, but I'm not here to do your homework!
itsmoked (Electrical)
6 Feb 06 16:57
Some might need a minimum, but the several I looked at didn't mention that limitation.
cbarn24050 (Industrial)
6 Feb 06 18:32
A complementry emmiter follower is the simple way to increase output current, don't forget to keep it inside the amplifiers feedback loop.
logbook (Electrical)
11 Feb 06 7:47
cbarn

I'm not sure what you mean by a "complementary emitter follower". Did you mean an emitter follower using a PNP transistor, and then resistively sourcing say 200mA from the positive power rail? Ugh!

I could see one using a PNP common emitter hung from the positive rail, but the designer would have to take care of the increased loop gain to prevent it oscillating.
Skogsgurra (Electrical)
11 Feb 06 7:59
A complementary emitter follower uses a PNP and an NPN with the NPN emitter connected to the PNP emitter. Bases tied together (if zero crossing distortion is of no concern - otherwise a biasing network or diodes).

Gunnar Englund
www.gke.org

logbook (Electrical)
11 Feb 06 13:54
skogs,

ok a standard Class B output stage.

Well that won't help the problem of the output range of the opamp. The load is static so the complementary output stage won't help at all.
cbarn24050 (Industrial)
13 Feb 06 7:17
A complementry emitter follower is a general way of boosing the current from an op amp allowing both source and sink capabilities. Biasing is not required if a small resistor (470) is placed between the emitters and the bases and the feedback is taken from the emitters. If you only need unidirectional current then you can remove the unused transistor.
IRstuff (Aerospace)
13 Feb 06 10:20
This is nearly identical to the LH0032 buffer amp.  OBviously, rail to rail is not possible with this configuration.

There is a pull-push using a common emitter configuration using 741's powerpins to drive the bases, which was published in Popular Electronics about 30 yrs ago.  Pretty cool concept; claimed 30 watts from complementary MJE2955/3055, a 741, and a handful of resistors.

TTFN



logbook (Electrical)
13 Feb 06 14:10
Yes I have used them, but the point here is that the NPN emitter follower is doing all the work; it doesn't work because of the headroom issues. Adding the PNP, which will not be turned on at any point, does nothing.

To solve the headroom issue we need either a PNP emitter follower biassed by a resistive pullup, or we need a common-emiiter PNP hung from the positive power rail, or we need a higher power rail.
OperaHouse (Electrical)
13 Feb 06 15:15
And with that common-emiiter PNP hung from the positive power rail, you will probably need a diode or two in addition to the resistor because the op amp output won't go to the rail.  This will cause the transistor to start to conduct, an emitter base resistor would also work.  Good reason to use an open collector op amp and avoid all that.
Warpspeed (Automotive)
13 Feb 06 18:08
If headroom is a problem, how about replacing the transistor with a P type mosfet ?

Source to +ve, drain to load, and the gate voltage should operate well within the output range of most op amps.

That will obviously require inversion of the inverting and non inverting inputs of the op amp. High frequency stability may be an issue, but that will depend on the particular components used, and is not going to be a difficult problem to solve if it occurs.
cbarn24050 (Industrial)
13 Feb 06 18:50
I cant see any headroom issue if a rail to rail op amp is used. Adding a resistor from the output to the rail, ie making the circiut sink current, would help.
logbook (Electrical)
14 Feb 06 14:14
Agreed, but isn't a 12V rail to rail opamp going to be pricey?
Helpful Member!  RJ44 (Industrial)
17 Apr 06 3:55
Greetings all, I just joined up.

I realize this is an old topic, but thought it might still be useful.

One simple way to boost output current is to place a resistor in series with the op-amp's +V supply pin. A PNP transistor is added with the emitter on the +V rail, the base on the op-amp's +V pin, and the collector on the op-amp's output. The resistor is sized to turn on the PNP at an output current somewhat below the maximum available from the op-amp's output. The +V supply current, which is through the resistor in the PNP's base circuit, includes the output current plus the op-amp's operating current.

This also allows the output voltage to swing as positive as with the unassisted output.

Works only with single amp IC's, of course.

Same circuit can be used to boost current from regulators.
RJ44 (Industrial)
17 Apr 06 6:12
Need to correct that. The resistor reduces the + supply to the op-amp by about .7V and the output swing by the same amount.

Anyway, it works if you can tolerate the reduced swing.
OperaHouse (Electrical)
17 Apr 06 8:54
I'd think if you wanted to get closer to the rail, the op amp output would drive an appropriately sized resistor and just be wasted (not added to the output).  Otherwise the output is limited to the op amp max output - 0.7 volts.

I still give you a star for a creative idea.  I love 3 pin regulator circuits - they can even be turned into oscillators (not just by forgetting the cap!).

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