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.

So I just realized the 75 ohm resistor above (R6)... is this going to mess with the 4.7 ohm (5w) resistor thats used in front of the darlington transistor?

 

So I just realized the 75 ohm resistor above (R6)... is this going to mess with the 4.7 ohm (5w) resistor thats used in front of the darlington transistor?

No, that resistor is only limiting current flow into the op amp. The OPA171 will need 0.5mA of quiescent current and a max load of 15 mA to the transistor for 15.5mA max. This will be a voltage drop of 1.16 volts to the op amp. (note, the 75 ohm resistor is not in series with the collector of the transistor - just the op amp). You will be fine. On the other hand, the op amp can handle up to 36 volts so there is no reason for the resistor.

 

No, that resistor is only limiting current flow into the op amp. The OPA171 will need 0.5mA of quiescent current and a max load of 15 mA to the transistor for 15.5mA max. This will be a voltage drop of 1.16 volts to the op amp. (note, the 75 ohm resistor is not in series with the collector of the transistor - just the op amp). You will be fine. On the other hand, the op amp can handle up to 36 volts so there is no reason for the resistor.

Wouldn't it also limit current going to the darlington transistor too though? For example, should it be like this instead? (see below)



I originally had problems with the opa171 failing randomly (there is a rediculous amount of EMF/EMI present), so I added in various forms of protective circuitry originally.

 

Or better yet..... (sorry for tossing in some of these trivial questions into the mix):

 

Or better yet..... (sorry for tossing in some of these trivial questions into the mix):
View attachment 95684

I don't see the difference in the pictures of posts 23 and 24. Both are fine from my point of view. I do assume that your input signal to the op amp is biased at 2.5 V, right? That is, the midpoint of the input signal (e.g. sine wave) is at 2.5v above ground.

 

I don't see the difference in the pictures of posts 23 and 24. Both are fine from my point of view. I do assume that your input signal to the op amp is biased at 2.5 V, right? That is, the midpoint of the input signal (e.g. sine wave) is at 2.5v above ground.

Oh sorry, the difference is that D6 is connected before or after R23.
The input signal to the op amp is 0 - 5 volts. You were saying -2.5 to 2.5 right?

 

Oh sorry, the difference is that D6 is connected before or after R23.
The input signal to the op amp is 0 - 5 volts. You were saying -2.5 to 2.5 right?

I was hoping that your input signal look like the top trace rather than the bottom trace...

 

Or better yet..... (sorry for tossing in some of these trivial questions into the mix):
View attachment 95684

I would return the diode to Vcc.
If you want 1 amp the 4.7 ohms may be to large. The tip102 can have a 2 volt drop from collector to emitter.
Move the 100pf cap back to the output of the op amp instead of the emitter or it may oscillate.

 

I was hoping that your input signal look like the top trace rather than the bottom trace...

View attachment 95700

Ok thanks for the clarification. Yes this is correct - the top one (0 to 5 volt signal)

I would return the diode to Vcc.
If you want 1 amp the 4.7 ohms may be to large. The tip102 can have a 2 volt drop from collector to emitter.
Move the 100pf cap back to the output of the op amp instead of the emitter or it may oscillate.

I am confused what you mean by moving the 100pf cap back to the output of the op amp. Are you saying it should be parallel with R22?

I will use a smaller resistor, such as a 3 ohm, and will move the diode to connect back up to Vcc instead.

 

Or are you saying that the 100pF cap should connect in-between R22 and TIP102.... and then to ground?

 

Or are you saying that the 100pF cap should connect in-between R22 and TIP102.... and then to ground?

From pin 6 to pin 2 of the op amp.

 

From pin 6 to pin 2 of the op amp.

Thanks for the extra info, I moved the capacitor and changed to a 3 ohm (5W resistor), so I think I should be good to go!:

 

The cap should be on the left hand side of R22.
Then your good..

 

The cap should be on the left hand side of R22.
Then your good..

Haha thanks!

 

So this has been working great by the way, thanks again for the insight!

I am wondering if its possible to take it 1 step further... to have protection if more than 1 amp is drawn from the output? Currently the darlington transistor and the 3.3 ohm (5 watt) resistor would suffer from this correct?

Would it be feasible to use an IC, such as a voltage regulator IC, so that if more than x amps is drawn then the IC shuts down and removes power?

Would it be better to use a PTC resettable fuse?

Is there a more elegant way of doing this?

What are your thoughts?

 

Hey guys,
Figured I would check back in on this. Is there more information I should be providing? I am still not entirely sure how I should approach this.

 

Hey guys,
Figured I would check back in on this. Is there more information I should be providing? I am still not entirely sure how I should approach this.

Here is an idea:
As the current goes up it will turn on the transistor .
PS. When I simulated it I saw an instability, so it might be a good idea to make the cap a 1 nf instead of 100pf.

 

Thanks a lot for the suggestion! Man I need to start playing around with that program you guys are using.

Would you mind elaborating a bit more on how this works?
So as the machine increases its current consumption, it begins to transfer the load into the other transistor? What happens if more than 1 amp is pulled from the machine? (Or a short circuit - drawing a ton of amps)

 

Thanks a lot for the suggestion! Man I need to start playing around with that program you guys are using.

Would you mind elaborating a bit more on how this works?
So as the machine increases its current consumption, it begins to transfer the load into the other transistor? What happens if more than 1 amp is pulled from the machine? (Or a short circuit - drawing a ton of amps)

It's free!
http://www.linear.com/designtools/software/
As the current increases the voltage drop across the 3 ohm resistor increases. The divider (1k, 330 ohm) causes the transistor to start to turn on at about .7 volts if the current is about 1 amp. This varies a bit, but I don't think you care if it is .8 or 1.2 amps. As the transistor turns on it pulls the - (minus) side of the op amp higher. Since it is the inverting input it turns the big transistor off holding the current steady around the 1 amp level.

 

How can I calculate the power dissipated from the transistor (T1)? I am trying to figure out if its necessary to use a heatsink or not. It generates a 0 to 10 volt analog signal, and needs to be able to provide up to 1 amp max, at which point there is a protection fuse.
The transistor is TIP102, and here is the schematic: