[SOLVED] Constant current from AA nimh (down to 0.9V)

Status
Not open for further replies.
L

lee321987

Member level 5
Joined
Apr 3, 2011
Messages
86
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,288
Activity points
1,987
Constant current from AA nimh (down to 0.8V)

Hello.
I'm trying to design a circuit to drain a single AA nimh battery at a constant current down to 0.8V.
I was going to try an LM317 or an LM431, but they're voltage references are 1.25V and 2.5V, so since the battery needs be drained down to 0.8V those two ICs won't work -- right?
I'm trying to build this as cheaply and with as few components as possible. Any ideas?
 
Last edited:

You didn't tell two things:
- acceptable regulator voltage drop
- intended accuracy
- current order of magnitude

A JFET or depletion MOSFET current source may be sufficient for your purposes.
 
Okay, I'm a bit of a noob...
Regulator drop: I don't think that really matters? I just want to drain the battery, a PIC microcontroller will be monitoring the battery voltage, and when the battery gets to 0.8V the PIC will turn of the drain circuit.

Intended accuracy: Hopefully it won't vary much. Guessing... 1%?

Current order of magnitude: Don't know what that means... I want to be able to drain at several different currents, e.g. 200mA, 500mA, 1A.

Also, I have an LM741, and an LM324 op amp.
 

O.K., in other words, you have a higher supply voltage available in addition to the battery. Then it's rather easy. Use an OP based current regulator comprised of a ground referred sense resistor, an OP with a common mode range down to zero, e.g. a LM358/LM324 and a transistor as control element.
 
**broken link removed**Build a simple transistor current source with a diode transistor and 2 resistors.
the batt will discharge thru current source and once the batt voltage goes lower than 0.7 the circuit will not conduct.

checkout this link
**broken link removed**
see fig 6a

---------- Post added at 19:37 ---------- Previous post was at 19:25 ----------

**broken link removed**
Current Sources, Sinks and Mirrors in Audio
see fig 6a
 
Last edited:
@FvM
Thanks.
I tried one (pic attached), but the current drops as the battery voltage drops.

---------- Post added at 13:00 ---------- Previous post was at 12:58 ----------

@dselec
Thanks, but I don't want to go below 0.9V (I said 0.8V just to give me some room)
 

What is the current that you have tried to keep?
with a 10 ohm load there is no way to maintain 1A for example and in addition you have the Vce saturation drop which will decrease the voltage output even more (and the current on the load)
 
I tried one (pic attached), but the current drops as the battery voltage drops.
Click to expand...
The circuit is exactly implementing what I meant, with two necessary corrections:

- The reference voltage (at positive OP input) and the voltage drop at the resistor respectively must be below about 0.5 V, so that you have still room for transistor current control operation. Now the voltage is 2.5 V, which can be never achieved by the battery.

- A LM741 can't work with 0.5 V input in single supply operation. You either need an OP like LM358, or a dual supply.
 
 
@FvM
Thank you.
Okay, I changed the voltage divider to give me 0.5V a OP input.
When you say the "voltage drop at the resistor", do you mean the voltage measured between the emitter-R3 junction and ground? That would limit how low I can go on the battery drain current...?

About the op amp... Are you saying there is no way to do this with an LM741, or LM324?
What characteristic on the datasheet tells me that it won't work with 0.5V input?

Again... thank you.

---------- Post added at 14:23 ---------- Previous post was at 13:51 ----------

@dselec
Thanks, I'm gonna breadboard the circuit on the right in the link (fig 6a).
 

Reference voltage + transistor saturation voltage = minimum regulated battery voltage

OP input voltage range or common mode range is the specification you are looking for. With LM324 or LM358 it's ranging down to zero. Many other OPs dedicated for single supply operation can be used as well. LM741 or TL08x in contrast don't work.
 
Thanks a lot for all the help, guys.
@FvM
You said that the voltage drop at the resistor must be below about 0.5V. Are you saying there is a limit how how small I can make the sense resistor?
I seem to have a working circuit now. Using an LM324, and an IRF610 N-channel enhancement mode MOSFET (pic attached), but with a 0.5 ohm resistor for Rsense, i'm only getting 360mA -- it should be about 1000mA...?
 
Last edited:


Yep its fine. the only thing in this circuit u should be caring of is that your op-amp voltage must be greater than your mosfet's Vth i.e. Vth +2 and that you did. since op-amp output never equals to power supply.If current is not that big e.g. (Constant current x [Rds + Rs]) <0.8 then its fine else use mosfet with a low Rds and try to lower your resistance Rd. Set these values considering the max constant current you need to draw.
 
Okay I made Rsense 0.5 ohm.
Switched MOSFET to an IRF520 ::: Rds=0.27 ohm // Vgs(th)=2-to-4V.
Made the voltage divider feed 0.45V to the non-inverting op amp input.
But the battery drain is only ~580mA (should be 900mA). Two batteries in series gives me only 810mA.

 

The reason for the unexpected behaviour should be revealed by a measurement. With a faster OP, I would suspect an unstable current control loop due to the high MOSFET input capacitance, which can be isolated by a compensation circuit.
 

Can you please tell the basics of a compensation circuit? What kind of parts on what pins?
 

lee321987 said:
Can you please tell the basics of a compensation circuit? What kind of parts on what pins?
Click to expand...

I won't go into compensation circuit. the datasheet revels the answer. Look at the Vgs in fig 5. Vgs must be between 4 to 5 to have your required current. and you are offering only 0.45v on V+. 580mA(0.5 x 0.27) =0.4466V which is approx equal to the voltage on V+ pin. The voltage on V- pin reaches very soon and op-amp is not able to supply more voltage to mosfet.try to increase your V+ of the op-amp. The op-amp will try to put the same voltage on the negative V- by supplying more voltage to mosfet. in turn mosfet will allow more current Id to have that voltage drop on Rd. try to keep Voltage on V+ pin a little less than the SUM of voltage on Rd and Rds. in addition try to lower your Rd a little less if u still face problem.. use irf3710 which i used.

I will suggest to add a pot instead to Voltage divider and vary the voltage on V+ pin.
 

I don't see the problem with Vgs, LM324 can output V+ −1.5V (max output 1.5v lower than the positive supply of the opamp), this would lead to a max Vgs of about 7.5v which should be more than enough (assuming 0.5v drop on the source resistor).

Why do you expect a Vgs of only 0.5v, the voltage will be as high as needed to balance the plus and minus opamp inputs.

Alex
 


I am only saying that he is supplying low Vgs. The op-amp do what ever to balance voltage on V+ and V- pins. as V+ is small i.e. only o.45 the op-amp don't need to strive that much. It only produce a little Vgs in order to balance V+. So a little Vgs is the problem. increasing V+ will force op-amp to produce more Vgs to balance the V-.
The Vgs will be low in two cases.
1. low V+ reference.
2. High Rd.

The first case i have described. second obviously if high Rd the a little current will make the voltage on V-to be equal to V+.

He is changing both the cases at the same time. there are 2^2 = 4 combination of 2 cases.
1. Low V+, Low Rd.
2. Low V+, high Rd.
3.High V+, low Rd.
4. High V+, high Rd.

He need to address them separately.
 

I don't understand the definition of "a little Vgs in order to balance V+" , the opamp will drive the mosfet as needed to maintain the balance of the inputs, the inputs are either balanced or not, there is not little balanced or more balanced as long as the needed output value is in the range that the opoamp can provide.

If the Vgs needs to be 6v in order to have the opamp inputs balanced then that will be the actual output of the opamp.
I don't understand how 0.5v in the inputs will drive the output differently compared to 2v at the inputs (as both input voltage level), the only difference I see is in the final balanced difference between the inputs which will be more accurate if you compare the input offset as a percentage of the 2v or 0.5v , the higher voltage at the inputs will balance the resulting current in a more strict area becasue the offset will represent a lower output current error.

Alex
 
Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…