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Help with voltage follower

Discussion in 'General Electronics Discussion' started by igorcarajo, Oct 30, 2014.

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  1. igorcarajo

    igorcarajo

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    Oct 30, 2014
    Hello, this is my first post on this forum. I'm a mechanical engineer. I've built some simple circuits before, but I'm really an "outsider" in the world of electronics. This is my problem: I have a circuit with an output signal that is 0-10 VDC, with a maximum current of 1 mA. I need to feed this signal to another circuit that requires a much higher current, maybe as high as 500 mA. I've been researching, and what I think I need is called a unity gain voltage follower, or voltage buffer. It is not terribly important that the output voltage exactly match the input voltage, so long as they are with 2-3% of each other. I saw some articles showing a very simple schematic using an op amp. The part where I'm stumped is finding an op amp that can handle 500 mA. So, am I looking at the wrong type of circuit? Any suggestions? Also, is there any commercially available product that can do this and that comes in an enclosure with electrical connections, so I don't have to build a bunch of circuits and mount them in a box and all that? (I might need 20). Thanks in advance.
     
  2. Gryd3

    Gryd3

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    Can you share some additional info for us?
    What device are you wanting to feed the signal to?
    500mA for a signal wire seems kind of high, and because current is a function of voltage and resistance, it would help to know a little more about the devices.
     
  3. Supercap2F

    Supercap2F

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    Mar 22, 2014
    Is it a analog signal or digital? If it's digital you can use a transistor. If it's analog your going to need something more complex.
    Dan
     
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  4. Arouse1973

    Arouse1973 Adam

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    You could use an OP-amp with an NPN transistor on the output as a current amplifier. Choose a rail to rail opamp powered from the 12 Volt supply. The transistor should be rated for at least 1A if not 2A and you might need a small heat sink. You could look at using a Darlington transistor instead which will reduce the output current needed from the opamp . You will need a bit of supply decoupling also. A 100nF ceramic and a 22uF Tant or electrolytic capacitor across the supply close to the opamp, these need to be rated higher than 12 V, 16 V should be fine.
    Adam

    CONTV.JPG
     
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  5. igorcarajo

    igorcarajo

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    Oct 30, 2014
    Thanks for the responses. To answer a couple of the questions above, the device being controlled is a driver for a dimmable LED light. The analog 0-10 V signal determines the intensity of the light (doesn't provide the power for the light, it's just a signal). The thing is, I need to control a bunch of these LED drivers with a single signal, that's why the current gets high.
     
  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    I recommend Adam's circuit in post #4. But you will need a higher voltage power supply, especially if you use a Darlington (which I recommend). I think your power source should be around 15V DC because most op-amps can't drive their outputs closer than about 1.5~2.0V below the positive rail, and the Darlington could need up to 2V base-emitter voltage.

    The op-amp needs to be a "single supply" type whose input range includes the 0V rail. There are many suitable op-amps - LM358 (two op-amps per package) and LM324 (four per package) are widely available and cheap.
     
  7. Colin Mitchell

    Colin Mitchell

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    In circuit #4 you are not gaining anything by adding the op-amp. Just use the emitter-follower transistor.
     
  8. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    The reason for using the op-amp is that the output voltage will then follow the input voltage accurately.

    If you use just an emitter follower, the output voltage will be about 0.6~0.8V lower than the emitter due to the base-emitter voltage drop (twice that, if you use a Darlington transistor). Also the base-emitter voltage will vary somewhat with load current, which will normally vary with output voltage.

    With the op-amp, the input-to-output voltage difference is reduced to roughly the input offset voltage of th op-amp, which is typically a few millivolts for older designs, and less for modern devices.
     
    Last edited: Oct 31, 2014
  9. Colin Mitchell

    Colin Mitchell

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  10. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    In post #1 he mentioned an accuracy of 2~3%. That may just have been a guessimate. A 0.7V base-emitter drop is a 7% error (offset), or a 14% offset if it's a Darlington.
    Yes. I pointed that out in post #6 where I suggested a power supply voltage of 15V.
    That will still have the 0.7V offset, but it would be better than a Darlington, and a 12V supply would work.

    Depending on the characteristics of the drive circuit, removing most of the base-emitter offset could be as simple as adding a diode between the input and the base (anode to the base) and a pullup resistor from the base to the positive power supply.

    But I think given the requirements (undefined, low-current source; 2~3% accuracy) an op-amp feeding a Darlington is probably his best option.
     
    Supercap2F likes this.
  11. BobK

    BobK

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    500mA sounds really high for a dimming signal, even if you had 100 of them. Can you post a link to the LED driver specs?

    Bob
     
    Last edited: Oct 31, 2014
    Gryd3 likes this.
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