Increasing Current Capabilities For Op Amp Output - OPA171

Discussion in 'General Electronics Chat' started by Mahonroy, Oct 9, 2015.

  1. Mahonroy

    Thread Starter Member

    Oct 21, 2014
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    Hey guys,
    Currently I am using a DAC to generate a 0-5V signal, which then goes through the Op-amp circuit and becomes a 0-10V signal. The maximum frequency of this signal is 50Hz and is an analog range - meaning its not just 0V or 10V on off pulse... its anything in-between as well. This works great for loads under 25mA (this is the limit on the OPA171).

    Here is my current circuit (the zener diode is 16V, the op-amp is "OPA171AIDR", and the DAC is "MCP4921-E/SN"). All of the resistors/capacitors are small 1206 package parts):
    opa171_increase_current.jpg

    I now have a device that I want to use with this circuit, but it draws around 500mA instead of 25mA like the previous one. I figured to be on the safe side, my circuit should be good up to 1A.

    I have been looking at a few circuits that show how to boost the op-amp current using a transistor. I am confused how this would work for voltages in-between 0 and 10 volts... its looks like it would only be an on/off type of a function? I may not be understanding it correctly.

    I have also seen people mention to just use a different op-amp. Is one method preferred over the other? I figured since everything is working reliably as-is, that adding a transistor around an already working design would be the better option?

    I found this circuit, is this all I would need to do?
    [​IMG]

    Can you recommend a specific transistor and protection resistance I should use around this? Thanks and any help is greatly appreciated!
     
  2. dannyf

    Well-Known Member

    Sep 13, 2015
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    Put a pnp+npn (+ optional resistor) on the output.
     
  3. Mahonroy

    Thread Starter Member

    Oct 21, 2014
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    Would you mind being more specific please? In my original post, I was looking for an explanation on why the transistor and how this is not just on/off. I was also asking what specific transistor. Did you take a look at either of the 2x circuits I posted?
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    Assuming that your dotted box is just a wire or a resistor, the inverting input of the opamp is sensing the actual output voltage. If that voltage is too low, then the opamp has a positive differential signal and the output voltage of the opamp goes up and so too does the actual output since the transistor is acting as an emitter-follower. If the voltage is too high, the opamp has a negative differential signal and the output voltage of the opamp goes down. As long as the load is drawing some current from the positive rail through the transistor, the voltage will then drop until the actual output is just barely below the input signal, at which point the output of the opamp will come up to just be that voltage that need to be on the base of the transistor.

    If your circuit needs to sink current as well as source it, then you need a push-pull output stage.
     
  5. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Based on your existing circuit and
    1) ASSUMING THE LOAD IS CONNECTED TO GROUND,
    2) ASSUMING A SINGLE POWER SUPPLY (0 - 15 volts).
    You DO NOT NEED A TWO- transistor circuit. Keep it simple.

    I would recommend that you keep the similar op amp with the following changes (including a ONE NPN pass transistor). It must be gain of 250 or more and can handle one amp (a Darlington type is ok).

    Change R22 to a bigger value, your op amp cannot handle such a small load (1k should work).

    Move the feedback transistor (R9) and capacitor (C20) to the emitter of the new pass transistor. R9 and R8 should keep your gain of 2 and similar to the same circuit as your existing circuit.

    Use a transistor with 1 amp or more current carrying and gain of 250 or more.

    NOTE: the 10 ohm resistor (highlighted in yellow) is your LOAD. It is there to demonstrate that the circuit handles 1 amp. Do not add a 10 ohm resistor and your load!

    image.jpg
     
    Last edited: Oct 10, 2015
  6. crutschow

    Expert

    Mar 14, 2008
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    If you only need a positive output then the single NPN you show should work.
    The transistor is operating as an emitter follower so the emitter voltage is essentially equal to the base voltage minus the base-emitter drop of about 0.7V.
    Since that's located inside the feedback loop this offset is not seen at the circuit output (the feedback causes the op amp output to be 0.7V above the transistor emitter) so that the output follows the plus op amp input.

    To help protect the transistor against momentary short circuit outputs you can add a resistor in series with the collector of Q1. An 8Ω resistor would limit the short circuit current to a couple of amps.
    Since the resistor's peak power dissipation with a 500mA load would be about 2W, the resistor's power rating should be 4 or 5W for proper derating.

    The transistor will also dissipate about 2W peak with a 500mA load and 8Ω resistor in the collector.
    A good transistor would be the old 2N3055 in a TO-3 case, which can absorb a lot of abuse and dissipate the 2W of power without a heat sink.
     
  7. #12

    Expert

    Nov 30, 2010
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    AnalogKid and GopherT like this.
  8. dannyf

    Well-Known Member

    Sep 13, 2015
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    Tie the pnp/npn's bases to the opamp's output;

    Tie the emiters together, and to the load - this is the output of the combined amplifiers;

    Tie the npn's collector to positive rail and the pnp's collector to the negative rail.

    Put the optional resistor between the bases and the emiters: its value is determined by the opamp's output current level above which you want the npn/pnp to kick in. Typically from 110 to 1k.
     
  9. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    @Mahonroy
    The push-pull (NPN/PNP output) is not needed if your load is connected to ground and you are using a single rail power supply (as shown in your circuit in post 1). The single transistor is fine.
     
  10. Mahonroy

    Thread Starter Member

    Oct 21, 2014
    186
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    Hey guys, thanks a lot for all of these responses.

    So originally I was feeding 15 volts into an op amp, then the op amp configuration was generating the 0-10 volt signal. Here is that circuit for reference:
    opa171_increase_current.jpg

    So now I will want to create a strong 10 volt power rail instead and run that into the transistor?
     
  11. Mahonroy

    Thread Starter Member

    Oct 21, 2014
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    Sorry let me be more specific...

    Should I just use something like a LM7810 for this 10V rail? Or is there a switching regulator circuit you can recommend for this? As you can see in my previous circuit, I never had a 10V rail and used it like that. Thanks again and any help is greatly appreciated!
     
    Last edited: Nov 30, 2015
  12. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    The circuit in my post above (post 5) will work best with a power supply of 12 to 15 volts. Just be sure to add a heat sink to the transistor. The output voltage of that circuit depends on the input signal and the R8 and R9 resistors. The power supply voltage does not determine the voltage of the signal.
     
  13. Mahonroy

    Thread Starter Member

    Oct 21, 2014
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    Ok that makes sense now, thanks again! In that case I can just leave my current power rail as is.
    I'll give this a shot.
     
  14. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Back to your second question, APEX and Burr-Brown (now TI) make opamps with output current capabilities 1 A and more.

    ak
     
  15. Mahonroy

    Thread Starter Member

    Oct 21, 2014
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    One more quick question for ya. It was mentioned that the 2n3055 would be a good candidate. Looking at the datasheet, it does not seem to fulfill the minimum gain of 250... will this be a problem?
    Here is the part I was looking at:
    http://www.mouser.com/ds/2/308/2N3055-D-105416.pdf

    Also, instead of generating a 0-5 volt signal and amplifying it to 0-10 volts... I'm going to change the circuitry to run off 3.3 to simplify everything. So will generate a 0-3.3 volt signal, then amplify it up to 0-10 volts. All I need to do is change the ratio of the resistors for the voltage divider and gain and should be good to go. The transistor setup will still function the same correct? Thanks again!
     
  16. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    @Mahonroy

    The op amp doesn't like to send out more than 0.01 A (10 mA) if you want to keep distortion low (25 to 30 mA max - depending on op amp). The 1000 ohm current limiting resistor on the output does this current limiting. Now you ask it to put out 1 amp. That is a current gain of 100. If the gain of your transistor exactly matches your needs, then you are always behind. It is best to shoot for 150 to 250 of gain.

    Since you are running very low frequencies (50 Hz), you can get away with a fat darlington (TIP 102) or similar. Put it on a big heat sink. On the other hand, you can make your own darlington with a 2N2222 or similar combined with your 2n3055. Then you will have plenty of current gain.
     
  17. Mahonroy

    Thread Starter Member

    Oct 21, 2014
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    Thanks again for the info.
    Here was my attempt at it:
    opa171_increase_current_2.jpg

    I am planning on using the TIP102 with one of these heatsinks:
    [​IMG]
    Do I still need an 8 ohm (5W) resistor directly before the TIP102 to protect it from momentary shorts if I have the diodes configured this way for protection? Thanks again!
     
  18. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    The diodes only help if you connect it backwards. I think it is a good idea to add a small value resistor to take some of the heat from the transistor and minimize the damage if you have point-to-point wiring supplying power to something like a motor or speaker.
     
  19. Mahonroy

    Thread Starter Member

    Oct 21, 2014
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    Ok cool, I will put that in then. So I think this should do the trick!

    opa171_increase_current_3.jpg
     
  20. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    A 3.3, 4.7 or 5 ohm would be better (in that order as fat as I'm concerned). The darlington transistor will drop a minimum of 1 volt, so your 15 V supply is down to 14. 8 ohms from your fixed resistor and an 8 ohm load mean you only get (14V/16 ohms) of current through the load.
     
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