battery load test

i need a simply circuit for load testing 7.4v lithium ion batteries with a one amp load. Have digital multimeters and thought to use that as measurement display but need help in building one amp load circuit.

The attached circuit should do it (depending on how precise you need that 1A load to be, you might want to replace the resistor with a variable resistor (perhaps one with a sliding screw tightened tap)). This will work down to a voltage around 3 Volts. If you need one that goes lower than that, then consider a constant current sink, such as depicted here:

http://www.eecs.tufts.edu/~dsculley/tutorial/opamps/opamps7.html

Just put the battery in place of "Current Out", and then set Vref to the lowest voltage you want the battery to drain down to (around .5V is probably the lower limit).

that looks like it should work. i am looking for something even more simple however. I am thnking simply a resistor in parallel between meter and battery with a momentary pushbutton contactor to load the circuit. i just need to know what ohm and wattage resistor to use.

Well, then you need two formulas:

Ohm's law: E = IR (where 'E' is the voltage, 'I' is the current, and 'R' is the resistance. Whip a little algebra on that puppy and you get: R = E/I

And the one for computing the power in a resistor with respect to current: P = I^2 * R

Consider that as the battery discharges, it's voltage is going to drop and thus the current through the resistor will drop (again according to ohm's law). That's why I suggested a constant current source/sink.

An even simplified circuit would be to get rid of the op amp. Replace R1 with two 200mA diodes in series (anode to positive battery terminal. Make R1 100 Ohms, The emitter resistor for 1A should be 0/7V/1A = 0.7 Ohms. This is not precise as with an op amp but is simple and with work down to about 2.5V or so depending on the gain of the PNP transistor.

A simplified op amp version would use a TL431, which does not need a bias supply and it has a built in reference, unlike the circuit cited previously. If you want a diagram for that I send it.

FTI, the reference voltage is needed so that when the voltage on the battery fluctuates the current stays constant. The circuit with the op amp has no reference so it will fluctuate, although that could be fixed.

The PNP transistor @1A dissipates 6.15W so it will need a heat sink with a thermal resistance to air of about 8deg/Watt, or you could use a smaller heat sink with a small fan, although the fan would need a power source,

Sorry for the prior diagram. It is incorrect because the feedback is positive and would result in a latch. Here is the corrected version. The 120Ohm resistor could be changed to a 330 Ohm probably but I don't know the gain of the NPN when Vce is say 2 - 3 volts, I could not find that data, but then I really didn't spend lots of time trying. The 330 Ohm would allow a 1/4W resistor and provide 17mA base bias current.

Again sorry for the prior diagram being incorrect.

But, consider that this circuit doesn't regulate the current as well, because the Base-Emitter voltage will change slightly with temperature. But, if you set the value of the emitter resistor to cause 1Amp to flow when the transistor is fully hot, then it should be fairly stable. Depends on how low the battery voltage will go before you cease the test. Because, as the battery voltage drops, so will the voltage across the Emitter-Collector junction which will allow the transistor to cool (and thus will the Base-Emitter voltage drop, causing the load current to drop slightly).

With the OpAmp, the current will not deviate appreciably.

It's still far better regulation than a mere resistor, though.

I don't know which circuit you are referring to? The circuit with the TL432 will regulate the current very well. It maintains 1.25V across the emitter (current sense) resistor and only runs out of steam when the loop gain is less than one, which is when Vce is close to saturation say 0.7V.

Yeah, I just noticed that. I'm completely wrong. This circuit will regulate the current precisely. I actually learned something new from this!

Earl,

After studying the datasheet for the TL432, it seems to me the emitter resistor should be 2.5 ohms, not 1.25 ohms. What do you think? And at that resistance, the power dissipated in that resistor will be 2.5W, so probably make that a 5W resistor.

Yes you are right. That is the case. I thought it was 1.25V. Anyway the current sense resistor should be 2.5 Ohms and that would make it a 5W (derating) resistor. Alternately use the part that I have a data sheet for, ZXRE060, which has a 0.6V reference which means the current sense resistor would be a 0.6 Ohm and 1W (derating).

In any event I though I had a 1.25V adjustable shunt regulator part number at hand, I'll have to look it up tomorrow. The lower voltage reference just allows the battery to discharge lower if that is a goal, which isn't stated, just assuming a wider dynamic range to be optimum. With the 2.5V reference the constant current would fade at around 3V or so. If that isn't a problem then go with the TL432.

With the ZRE060, constant current would fade at about 1.2V or so. I don't know why someone would discharge their batter that low unless they have lots f money to buy new ones.

Better choice than the ZRE060 part. Data sheet enclosed.

Here is the part I was looking for from Diodes Inc. AP432, which is a 1.25V part which should use the 1.25 Ohm resistor, 2W

These engineers are all aiming to provide a very high degree of precision in their circuits.
However, a technician like myself would propose something much simpler, which might be adequate for your purposes.

With the battery fully charged at 7.4 volts, it will put 1 Amp into a 7.4 Ohm resistance. (Amps = Volts/Ohms)
7.5 Ohm resistors are easily available if that is close enough. (7.4Volts / 7.5 Ohms = 0.99 Amps)
The power rating of the resistor MUST be at least 7.4 Watts ( = 1 Amp x 7.4 Volts). Any higher power rating is fine.

(I see on Ebay "2 Pcs 7.5 ohm 20W Wire Wound Ceramic Cement Resistor" for £2.37 UK pounds. But that's from China, which can take a month to arrive.)

(By the way, the resistor will get VERY hot, so put it on something fireproof or clamp it (gently) in a vice.)

An ordinary digital multimeter on its highest current range (usually 10 or 20 Amps) has only a small fraction of an Ohm resistance, virtually a dead short, so its resistance can be ignored.

Put the first DMM as a voltmeter across the battery.

Put the second DMM as an ammeter (make sure the positive lead is in the 10 or 20 Amp socket) in series with your resistor.

You can then watch the battery voltage gradually drop as it discharges (at slightly less than 1 Amp).

Circuit diagram attached.

yes that would work as well. I would think (I could be way off base on this one), that the Knico would have though about that on his/her own. Again could be way off base on that assumption.

I do like the simplicity of the resistor, you can't get that wrong, but could be off base on that too.

i thank you. this is more what i was looking for. this is a simple hobby application where i need to periodically check the 7.4v battery with a 1 amp load to make sure the voltage is not down to 7.0 volts which is the safe cutoff point for the application. It will be a momentary test. I will most likely breadboard the resister witha momentary push contactor to apply the load. I believe your diagram will do the job. I thank you. I have a very basic understanding at best of electronics. once you tell me what i need i can do the job but am not good at figuring the circuit.

P.S. After completing the test, remember to put the ammeter DMM's positive lead back in the volts socket.
(The number of times even us experts forget that . . . !-)

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yep good advice. i will most likely use only one meter as i am looking for the voltage under load. just need to make sure to not use after reaching 7 volts.

No, actually, I'm wrong again. I took a closer look and discovered that the TL432 has a ref voltage of 1.24V. It's the TL431 that is 2.49V! So, you had it right!

Gosh! I look like such an IDIOT!! [aka Napoleon Dynamite]

Son-of-a-b...oy am I screwing up left and right! I just noticed that it's NOT the TL431 I was looking at, it's the AP432/AP431!! The TL431 and TL432 are both 2.5V (the only thing different are the pinouts). The AP432 has a Vref of 1.24V and the AP431 is at 2.495V.

Was that a conspiracy to make guys like me look like idiots!?!

How about a circuit that constantly monitors the battery voltage and turns off an LED when the battery voltage drops below 7V? It's below.

It will put an approx 5ma load on the battery when it's voltage is above 7V and a 35uA load on it otherwise. I added the switch just to show which side of the switch to place this circuit on if there IS a switch.

interesting. i like the idea. what lamp to use and what is the AP432 shown in the circuit? What wattage would the resistors need to be? reminder i know nothing about circuit design but with instructions i can build anything.

Simple comparator, should work although a little flicker right at transition perhaps because no hysteresis. No biggie.

Yeah, then the fact that the AP432 is in a tiny little package might be a deal breaker. The lamp is simply an LED.

As Earl stated, a simple comparator should do it. The resisters can all be 1/4 or 1/8 Watt.

And, hysteresis can be added with a few more resistors.

The trouble with using a comparator is the need for a voltage reference, but as you pointed out, earlier, these adjustable shunt references contain both a reference and an opamp and the bypass transistor.

So, there are two possibilities:

1. Jameco has a TO92 version of the TL431 (p/n: 33347). That would be much easier to work with. This one requires a bit more current to run so the circuit puts a load of around 50uA when the LED is off.
2. Jameco also has a TO92 version LM385Z-1.2 (p/n: 24117). Also easy to work with, and because it only needs, worse case, 35nA on the FB input, and 10uA for the voltage reference, it's possible to design a circuit that puts much less of a load on the battery (around 16uA when the LED is off -- it could be made to draw even less, but I'm concerned about noise immunity and current leakage as the board gets old and dusty).

BTW, I haven't tested these circuits, so I'm not positive they will work.