: 50mA-500mA constant Dc current source

[Averagesupernova]

If you are worried that collector current is not matching emitter current why don't you just ground the collector and put the load resistor in series with the emitter between the emitter resistor that is already there and the emitter? Keep the connection that goes back to the inverting input on the op-amp tied to the same place on the existing emitter resistor. BTW, 2N3906 won't do 500 mA. Also, I think your equation should read: Iout = (+Vs - Vin) / R

 

[lkari]

hi guys, i understood everything of what you r saying proton soup. I am an electrical engineer. With background in mixed signal circuit design as well as IC Packaging. I am just having problems reading out datasheets and choosing rt components.

yea I replaced the 2N3906 with BC328-40 transistor.

i just have too stringent specs to meet for these supplies as these have to run constantly for 6 months.

Thanks

 

[Corneo]

also, I'm not sure how much of this you're taking in. something tells me you've never studied any electronics. why are you being tasked with building current supplies from components?

I was getting that feeling as well. If I remember correctly the sketch of the circuit that the OP posted is from a design article from ED magazine (I think the author was Jim Williams). I first came across the article and could not see how the author arrived for the equation for the current. What I derived matches up with what berkemen arrived at as well.

The base will be *wherever* the opamp sets it to make the emitter current be (Vs - Vin)/R

Absolutely! We do not know what the base voltage is required to get the emitter current we desire. But the beauty of this circuit is that the op amp will decide for us! This is one of the most important things I have learned from working with analog designers from work.

Ikari,

There is really nothing to simulate here! I am willing to bet you that what your SPICE program tells you what the base voltage is will not be close to what you measure in the lab! The transistors will vary all over the map in terms of base-emitter voltage vs. collector/emitter current. But what is so great about this circuit is that your op amp will decide for you. Please don't take offense to this, but have you been a designer very long?

 

[Proton Soup]

heheh, i should probably not post this, but here is Bob Pease blowing your mind on precision current sources...

http://www.truveo.com/Precision-Current-Source/id/3143447996but maybe important for thinking about heat sinking and other problems you may encounter like oscillations.

edit: and here's another on not-precision current sources. i haven't watched it.

http://www.youtube.com/watch?v=411f0DvXu18&feature=channel_page

 

[berkeman]

Yikes the volume is loud! Holy cow, Bob Pease. Man, what a Mentor to thousands of EEs, including me. Thanks, Proton.

 

[Bob S]

Maybe I misread the specs, but if the output range is 500 mA max into a 1 to 100 ohm resistor, then you need to use a power supply capable of 60 volts at 500 mA, and output transistor ratings of Vceo of 60 volts and 50 watts. I suggest a PNP darlington TIP125, which has an hfe of 1000. Also I suggest that the output load be ground referenced, so with the suggested PNP darlington, the sense resistor (about 10 ohms) is between the +60 volts and the emitter, with the load connected between the collector and ground.

 

[ksac]

Hey, I know this thread is about a year old, but I was working with the same thing, so thought I'd post here itself..
Even I derived the output current as I= (+Vs-Vin)/R
but I found a book that says I=Vin/R
and what's more! I tried the circuit out in lab and found the latter to be the current.
I am confused..
Any thoughts anyone?

 

[berkeman]

Hey, I know this thread is about a year old, but I was working with the same thing, so thought I'd post here itself..
Even I derived the output current as I= (+Vs-Vin)/R
but I found a book that says I=Vin/R
and what's more! I tried the circuit out in lab and found the latter to be the current.
I am confused..
Any thoughts anyone?

If you mean for the circuit shown in post #16, your first equation is correct. Your 2nd equation can apply to other topology current source/sink circuits. For example, if you are sensing the current on a low-side resistor, with the load on top of the resistor, and the bottom of the resistor grounded. Do you see how that equation would then apply?

 

[ksac]

Well my circuits exactly the same, except there's no voltage divider at V_in (I've used a zener for reference voltage) and a resistor between the op amp output and pnp base. My sensing resistance is between V_s and emitter (so no low-side current sensing) and the load between collector and ground.

How would low-side sensing work here? Emitter to V_s, load to collector and the other end to R_sense, which is grounded, and the inverting terminal of op-amp connected to the junction of R_load and R_sense? Is that how?

Thank you for replying!