: 50mA-500mA constant Dc current source

[lkari]

URGENT: 50mA-500mA constant Dc current source

Hi everyone,

I am trying to design a 50mA-500mA constant current source for my project. I found the following link. But I think it would give slightly less than the required or designed value.
http://www.ecircuitcenter.com/Circuits/curr_src1/curr_src1.htm
SO I change the transistor to a 2N3055 and the opamp to a LM348. I want to know what can i use to simulate this circuit before i can order nay parts.
Also please suggest if I can use any better circuit.
I will have to supply two different voltages for this. So I will have to use voltage regulators for the same I understand.

Its very urgent for me and Can someoen please help me.

Thanks in advance
Kari

 

[pantaz]

Try LTspice (free)
http://www.linear.com/designtools/software/

 

[lkari]

Current amplifier

Hi I am trying to build an adjustable current supply using an opa547, so the thing is the output current is dependent on the load that i am attaching. I was wondering as to what I could do to get rid of this dependence?

Thanks in Advance
Kari

 

[wywong]

It shouldn't. An op-amp circuit doesn't work if you exceed its current or voltage range, in which case there will be a measurable voltage difference across its two inputs. Could that be the case? Is the circuit design a proven one or one of your own? Current supply circuits should be easy to find, e.g. http://ecow.engr.wisc.edu/cgi-bin/get/ece/340/schowalter/opampckts.pdf .

 

[berkeman]

(Merged the two threads on the same subject...)

 

[lkari]

So the output of the opamp should be independent of the load you mean right? please correct me if i am wrong.

Thanks in advance
Kari

<< txt speak edited out by berkeman >>

 

[Redbelly98]

In theory, yes. In practice, yes as long as the load current is kept below the maximum current for the op-amp.

Its very urgent for me ...

When is your deadline?

 

[berkeman]

So the output of the opamp should be independent of the load you mean right? please correct me if i am wrong.

Thanks in advance
Kari

<< txt speak edited out by berkeman >>

No, the output voltage will not be independent of load. Opamps have an output impedance and a maximum current drive specification. Check the datasheet.

To get more current, you need to include some power amplification circuit after the opamp, either an open-loop amp, or include the power amp in the feedback loop of the opamp (being careful about your phase margin at unity-gain crossover...!).

 

[lkari]

Hi the maximum rating on the opamp is 500 mA and I am not exceeding it. I am worried that I am not able to work properly with the inputs and supplie sto the opamp. I am using an OPA547 . I am reffereing to teh datasheet and there is also a application note on current source design. I am not able to understand it i guess. Please someone help me
Kari

 

[Proton Soup]

do you have a basic sense of how the circuit works?

you won't come anywhere near the current limits of the opamp using that circuit you referenced in your first post. all the current to your load is coming directly from VCC. the opamp is basically just outputting a small signal to control that transistor. the transistor does the heavy lifting in the design, so you need to make sure it meets specs.

 

[berkeman]

As Proton says, it's the external pass transistor that is passing the load current, not the opamp. There is no reason to use a 500mA opamp in a current source/sink circuit.

Have you read the explanation of the circuit that you linked to in Post #1? It's pretty clear how this kind of circuit works. Post specific questions about that explanation if it still doesn't make sense.

 

[lkari]

So let me start from the begining. I am using an OPA 547 high precesion high current opamp to build a power supply. I used the application note for the opamp 547 in its datasheet. SO now the problem is that I am not able to control the output current. I am using TINA Ti simulator for simulating the same. All my issues are with the load. it has to operate over a load of 10-100OHMS and current of 50-500mA.
1. I used a V+ of 30V
2. V- of 0V
3. RCL calculations as shown in the circuits in the datasheet.

Problems:
1. Current is not remaining constant
2. I am worried that i am not understanding teh datasheet specifications like the input signals to be given.

 

[Averagesupernova]

Without knowing anything about the circuit other than what you have posted I don't think there is any way to help you. I would like to know resistor values, power supply voltages, etc.

 

[Proton Soup]

So let me start from the begining. I am using an OPA 547 high precesion high current opamp to build a power supply. I used the application note for the opamp 547 in its datasheet. SO now the problem is that I am not able to control the output current. I am using TINA Ti simulator for simulating the same. All my issues are with the load. it has to operate over a load of 10-100OHMS and current of 50-500mA.
1. I used a V+ of 30V
2. V- of 0V
3. RCL calculations as shown in the circuits in the datasheet.

Problems:
1. Current is not remaining constant
2. I am worried that i am not understanding teh datasheet specifications like the input signals to be given.

well, yeah, that's your problem. think about what the beta of your transistor is and i bet you need a lot less than 50mA to drive the base. like others are telling you, you need to choose another opamp. a less optimal solution might be to put a dummy load from the output of the opamp to ground of 100 ohms. and then it might be able to drive the output at the correct voltage. but you're wasting a bunch of power by kludging it that way.

 

[Averagesupernova]

Just because the OP doesn't need the current that the op-amp is able to source doesn't mean that the circuit won't work. A 2N3055 doesn't have a very good beta to start with. I want to know resistor values.

 

[lkari]

Hi

Please c the following circuit. how can I decide the specifications of the opamp?
I will be choosing a BC 328-40 pnp transistor for the output stage. It will be bale to take a maximum collector current of 800mA, a little more than my requirement. the input to the opamp is through a Zener.

Will the zener essentially be doing in this circuit? how do i calculate the resistances at the input terminal?

Please help me. I tried to replace the previous circuit with this now to source more current.

Thank You
Kari

 

[berkeman]

Why is there an alpha in your Iout equation?

 

[lkari]

that was how the circuit was described. it means the transistor parameters change teh load current rt? would that be a bad design?

 

[berkeman]

that was how the circuit was described. it means the transistor parameters change teh load current rt? would that be a bad design?

The opamp is there to take out the variabilities of the transistor's parameters. The only thing you need to worry about is that the transistor can handle the heat being generated as it drops a voltage and passes the current.

 

[lkari]

but what abt the alpha paramenter? that changes with the transistor currents rt? which is teh load current? so not just the value of thr emiter resistance but the alpha value is also varying while we try to vary the load current. rather is it that the alpha value is stablized by the opamp feedback for the comman emitter configuration?

 

[Proton Soup]

it looks like the zener and bottom resistor create a voltage reference below +Vs, then the two resistors either side of Vin form a voltage divider of that voltage. those two resistors will probably need to be a bit larger than the bottom one. depending on the size of R, your amp may need to get reasonably close to +Vs.

 

[berkeman]

but what abt the alpha paramenter? that changes with the transistor currents rt? which is teh load current? so not just the value of thr emiter resistance but the alpha value is also varying while we try to vary the load current. rather is it that the alpha value is stablized by the opamp feedback for the comman emitter configuration?

The opamp is going to do all it can, within the limitations of its (very large) open loop gain, to keep the bottom of the resistor at Vin. So Iout = (Vs - Vin) / R (not the equation you have written).

 

[lkari]

should the opamp input operate very close to the supply? I thought the bottom resistor was to maintain teh minimum current through the zener. then the other two resistors i thought are as voltage divider. please correct me if i am wrong

 

[lkari]

hmmm yea. but what would limit the open loop gain here? then what criteria determines the input to th eopamp? is that what we want as the emitter voltage for the pass transistor?

 

[Proton Soup]

yes, it operates fairly close to the supply. this circuit is a lot like your last one, but referenced to +Vs instead of ground.

 

[lkari]

but then in using a pass transistor of vbe approximately 0.9 volts, the base and the emitter will be almost at the same voltage and not operate teh transistor in the required region of operation rt? it need to be in the active region. so the vbe should be slightly less than -2 V for all ranges of current? Please find the attahced BC328-40 datasheet.

 

[berkeman]

but then in using a pass transistor of vbe approximately 0.9 volts, the base and the emitter will be almost at the same voltage and not operate teh transistor in the required region of operation rt? it need to be in the active region. so the vbe should be slightly less than -2 V for all ranges of current? Please find the attahced BC328-40 datasheet.

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

Maybe now I see the error term that you're trying to account for, the fact that Ie does not equal Ic. Yes, to the extent that there is some base current, that is subtracted from the Ie that the opamp is holding steady. That's the only error term I see here.

 

[lkari]

yea makes sense to me now. I will try simulating the circuit and write back what happens so other may benefit. Also by tweaking the resistance between the diode anode and the input positive to the opamp, we will be able to get the input close to teh supply. i could get it as close as around 95%

 

[lkari]

I had one question, the opamp stage is totally not the load carrying stage and so only needs to operate at the supply voltage and nothing else rt? and also if the negative terminal of the opamp maintains the supply voltage, then how would there be a current in the emitter resistance? that too upto 800mA?

 

[Proton Soup]

the resistor labelled "R" in that circuit is there for current measurement. it's a decently small value. if you consider the base current for the transistor to be negligible, then Ir is pretty much the current output of your current source.

the output of the opamp is just a couple of voltage drops below +Vs. this is only important if you're also using +Vs as a supply for the opamp. you'll notice that some opamps are "rail-to-rail" amps. that means the output can go really close to the supplies. it's probably not a big deal, just something to remember. your output must be in the range necessary to drive the transistor.

yes, the opamp is not carrying the load here. if you think of the base as the control on a valve, the opamp is just there to adjust the valve.


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?

 

[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.