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

The circuit achieves stable current above 0.7 V Vbat in a PSpice simulation, also stable transient response (with and without the below shown compensation circuit).

Sajjadkhan, I simply don't understand your comments in post #17 and #19. They sound misterious to me, because you describe the restrictions related to Vth correctly in post #13.

I'll show the intended compensation circuit for completeness, but it's apparently not required with LM324/IRF520.

alexan_e said:

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

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Ok may be you are getting it wrong.By Vgs i mean the voltage provided by op-amp output. Little Vgs or small voltage on op-amp make sense. Look its a feed back loop so op-amp is not operating in saturating and cutoff levels. Now how the output of the op-amp that is Vgs is limited by V+? We know that V- = V+ in an ideal op-amp. So if you look at the general inverting configuration of op-amp then you will see that op-amp will output the amount of voltage as required to balance or to equal V- and that value of output voltage depends on the resistors Rf and Ri and V+ value. Now if V+ is increased and Since Rf and Ri is a voltage division network so more output voltage is need to make same amount of voltage on V- as V+ and vice verse.
same principle goes for this circuit.

---------- Post added at 21:18 ---------- Previous post was at 20:37 ----------

FvM said:

The circuit achieves stable current above 0.7 V Vbat in a PSpice simulation, also stable transient response (with and without the below shown compensation circuit).

Sajjadkhan, I simply don't understand your comments in post #17 and #19. They sound misterious to me, because you describe the restrictions related to Vth correctly in post #13.

I'll show the intended compensation circuit for completeness, but it's apparently not required with LM324/IRF520.

Click to expand...

I will still say that small Vgs can be the problem. Why don't you guys look at the fig in datasheet (Vgs Vs Id graph).
Post 17 and 19 (addressing them separately)was for his experiment only.
All i wanted to say is in the fig i posted. varying V+ obviously effect the Vgs and interns it effect the current.

I know that voltage on Rd is directly related to V+ and Vice verse.

My simulation is also giving me the correct result but i don't know if it will be true on ground realities. Thats why i am saying that if your circuit is failing then just lower the RD value and increase the V+ value. Just try it by lowering Rd and having a POT on V+. You will believe.

Hello Saj.. and Alex..
can u tell me the circuits have constant current ability ?
i dont see how using a comparator or the constant current option is out .
I know it is very important to discharge nimh properly without internal overheating (too large discharge current)that may degrade the batt .
thanks dselec

Ok saj for some reason did not see your drawing ... its truly a good constant current Amplifier
well done.
still i think it can be done with a simple current controlled supply 1 transistor 1 diode 2 resistors.
u must be knowing what i am talking about.

dselec said:

still i think it can be done with a simple current controlled supply 1 transistor 1 diode 2 resistors.
u must be knowing what i am talking about.

Click to expand...

The Transistor dose not compensate unlike the op-amp. Transistor's Beta or hfe changes due to temperature and due to the flow of current through it. look at the beta graph vs temperature and current.

So if you look at the general inverting configuration of op-amp then you will see that op-amp will output the amount of voltage as required to balance or to equal V- and that value of output voltage depends on the resistors Rf and Ri and V+ value. Now if V+ is increased and Since Rf and Ri is a voltage division network so more output voltage is need to make same amount of voltage on V- as V+ and vice verse.

Click to expand...

I still don't understand what you exactly want to say here. The feedback operation of the constant current circuits hasn't to do with the gain of an inverting amplifier. The OP is bringing it's full open loop gain to increase Vgs to the required value. OP saturation is the only restriction in this regard, as you already stated in post #13.

With IRF610, that has been initially used, a current of 0.9 A cant be achieved with a single battery. But IRF520 should do. According to the datasheet, it achieves Rds of 0.4 Ohm typically with a Vgs value of 5 V. So with maximum Vth, 6.5 or 7V would be sufficient in any case. LM358 with 9.5 V supply can output 8 V which results in Vgs of 7.5 V. The results of post #14 can't be explained this way, there's apparently and additional problem not visible in the basic circuit.

It's still reasonable to use a logic level FET, that's a different thing.

still i think it can be done with a simple current controlled supply 1 transistor 1 diode 2 resistors.

Click to expand...

Yes, but with reduced accuracy. We would need to talk about specifications. Also, the output current will be affected by transistor and diode type variations, that are neglectable with the OP cicrcuit.

FvM said:

The OP is bringing it's full open loop gain to increase Vgs to the required value.

Click to expand...

I don't mean any disrespect FVM but every feedback circuit is a close loop circuit.

basic idea here is r4 and cr1 or q4 control current(hfe of transistor q1) q3 can be replaced by diode or omitted ,r2/3 calculated for 0.8volts and required discharge current cr3 omited.

Turns out the problem is the resistances of my cheap Chinese jumper wires.
Even the contacts in my battery holder are causing a huge drop in current.

(the following was done with a battery I bought yesterday)
Short circuit battery directly with multimeter leads = 2.46Amps

Multimeter shorting output terminals on battery holder = 1.96Amps.
Same circuit, but with total of 3 jumper wires between battery holder and multimeter = 0.96Amps.

Sorry for making such a rudimentary error, guys.

A really big thanks to you guys for all your very helpful, informative advice! I will use it!!
Thank you!

Some thing funny about the lm324 input.
this chip usually works on +ve=4.8v and -ve=4.8v for getting a balanced ov on output.
thats why the both inputs should be + 4.8 volts to start with.
now the voltage from sense to _input should be a smaller than +4.8 volts for that rsense should be calculated for 4.8v divided by discharge current needed .
so i think this circuit cannot work.
ill try it out on ltspice .has anybody done that please se3nd me the asm file.

Obviously it wont work. 4V is the Vth of Mosfet i think. Just attach a higher supply or switch to LM358.

Simple answer
read the batt voltage with the analog output of pic and then with soft ware decide how much volts to give transistor to discharge and how much discharge current u need each step .