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A DAQ board is outputting 0-10VDC analog voltage via its DAC. The DAC cannot supply more than 5mA so I'm planning to use the following emitter follower for more current(x-axis is the DAC outpout voltage incresing 0 to 10V, y-axis is the base emitter voltage difference):

enter image description here

I thought emitter follower keeps the difference between the base voltage and emitter voltage constant around 700mV.

But if I use the following resistor 1ohm, the emitter follower cannot function anymore and will draw hundreds of mA current from DAC which may cause its damage. Here is the 1ohm case:

enter image description here

If I setup with the right resistor as in the first figure I'm safe but if someone accidentally shorts the resistor or ect. I dont want the DAC to be damaged.

I have two questions:

1-) Why emitter follower cannot keep the base emitter voltage constant in 1ohm case?

2-) Is there a way to totally be safe for DAC will never get loaded?

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  • \$\begingroup\$ You have a diode (B-E) and a 1 ohm resistor connected to ground and you wonder why it draws large currents from your DAC. Why no resistor to limit base current? Why use a small signal transistor with only 1 ohm load? \$\endgroup\$ – JIm Dearden Apr 26 '16 at 9:29
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A BJT has current gain - this might be 100 for instance. 10V across 1 ohm implies a current of 10 A and this would require (approximately) 10A/100 into the base. That's a current of 100mA to be sourced by the DAC and clearly the DAC isn't geared-up for that. This fact stops a BJT being a true voltage follower i.e. there are limits to how far things can be pushed.

Try using an op-amp and BJT (or MOSFET) configured as a voltage follower: -

enter image description here

You get the added benefit that the 0.7V base-emitter voltage error is "fixed" by the op-amp due to the action of negative feedback. The op-amp also keeps the output voltage stabler under a wider set of load conditions again, die to negative feedback. If you need 10A on the output then look for a power op-amp that can deliver up to an amp to the BJT base because, under worst case conditions, the current gain of a BJT can fall to as low as maybe 10.

Just so the BJT police are happy, I'm aware that current gain is strictly speaking collector/base rather than emitter/base current but it makes little difference to this answer.

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  • \$\begingroup\$ Will this also prevent loading in case of Rload shorted? And would any opamp work? Any suggestion? \$\endgroup\$ – user16307 Apr 26 '16 at 9:39
  • \$\begingroup\$ No of course not. If the load is shorted you cannot produce any voltage across it. I've edited my answer to give some more information. Regards an op-amp that can source 1A into the BJT I'm afraid nothing springs to mind but, there are devices out there I'm sure. You could also use a Darlington BJT and give yourself more op-amp options because base current requirement will be much less. \$\endgroup\$ – Andy aka Apr 26 '16 at 9:43
  • \$\begingroup\$ No no I dont need 1 ohm , I will use 470 ohm in my application. It is for safety if someone the load I just dont want DAC to be damaged \$\endgroup\$ – user16307 Apr 26 '16 at 9:46
  • \$\begingroup\$ I'd just use an op-amp unity gain buffer and rely on the internal inherent protection that op-amps have to current limit. Maybe add a small output capacitor of ten ohms too. No need for a BJT. \$\endgroup\$ – Andy aka Apr 26 '16 at 9:51
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    \$\begingroup\$ A 741 is horrible but depending on what power rails you choose it may be acceptable. An LM324 is better. \$\endgroup\$ – Andy aka Apr 26 '16 at 10:51
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As mentioned in a comment, transistor base draws it's own current, it's not a mosfet! You can in fact push quite a bit of current through this.

If your first case, 5mA would be about 470*5mA = 2.35V so you'd be pulling 5mA from DAC with approximate output voltage of 3V.

On the second one you'd see 1*5mA = 5mV so anything over 0.7V on the base would saturate the transistor and allow as much current through as possible and pull as much current from the DAC as possible too.

Use a series resistor on base as others have pointed out, to prevent the DAC from overloading you'd want at least (DAC max output - 0.7V) / 5ma. 560R for 3.3V DAC would work but since the output current is multiplied by a 100 or so and 10V/470R = 21mA, you might as well as use 4k7 or something. Then adjust the emitter resistor based on the voltage drop on the current range you're looking for.

If someone shorts the emitter resistor, the base current will be limited by the base resistor but your transistor will fry.

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  • \$\begingroup\$ Any suggestion for base resistor? \$\endgroup\$ – user16307 Apr 26 '16 at 9:55
  • \$\begingroup\$ @user16307 Amended that. Transistor amplifies current by about 100 (check datasheet) so anything that allows 21mA / 100 = 0.21mA through would work. However this way your transistor would work as a current controlled device meaning your output would wary according to current gain value and make setting the output harder. I'd aim for something like 1/10th of the output current to keep the base voltage dominating. \$\endgroup\$ – Barleyman Apr 26 '16 at 9:59
  • \$\begingroup\$ 4.7k i will use \$\endgroup\$ – user16307 Apr 26 '16 at 10:12
  • \$\begingroup\$ Can you explain what you mean when you say "If your first case, 5mA would be about 470*5mA = 2.35V so you'd be pulling 5mA from DAC with approximate output voltage of 3V". I can't make sense of this at all where does the 5mA come from because 5 mA through the 470R doesn't mean 5mA from the DAC. \$\endgroup\$ – Andy aka Apr 26 '16 at 10:28
  • \$\begingroup\$ @Andyaka Just cutting some corners to keep the calculations simple(r). Of course you'll have the current divided between the base current and the collector current but if you're driving the base without a base resistor, calculating the base current is not that straightforward. Add a series resistor and it becomes much more reasonable. If I had to shoot from the hip I'd say the actual current division would be by a ratio of the voltages (minus VCE and VBE). And if you remove the 10V then it indeed would mean 5mA ;-) \$\endgroup\$ – Barleyman Apr 26 '16 at 11:39

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