Simple 1A Load Circuit for 7.4V Li-ion Battery Load Testing Using Multimeter

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#1 21664821 20 Nov 2012 16:15

Knico Knico

Anonymous

Post #1
21664821 20 Nov 2012 16:15

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.
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#2 21664822 20 Nov 2012 16:37

Steve Lawson Steve Lawson

Anonymous

Post #2
21664822 20 Nov 2012 16:37

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).
#3 21664823 20 Nov 2012 18:00

Knico Knico

Anonymous

Post #3
21664823 20 Nov 2012 18:00

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.
#4 21664824 20 Nov 2012 18:14

Steve Lawson Steve Lawson

Anonymous

Post #4
21664824 20 Nov 2012 18:14

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.
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#5 21664825 20 Nov 2012 18:25

Earl Albin Earl Albin

Anonymous

Post #5
21664825 20 Nov 2012 18:25

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.
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#6 21664826 20 Nov 2012 18:48

Earl Albin Earl Albin

Anonymous

Post #6
21664826 20 Nov 2012 18:48

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,
#7 21664827 20 Nov 2012 19:47

Earl Albin Earl Albin

Anonymous

Post #7
21664827 20 Nov 2012 19:47

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.
#8 21664828 20 Nov 2012 20:18

Steve Lawson Steve Lawson

Anonymous

Post #8
21664828 20 Nov 2012 20:18

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.
#9 21664829 20 Nov 2012 20:20

Steve Lawson Steve Lawson

Anonymous

Post #9
21664829 20 Nov 2012 20:20

It's still far better regulation than a mere resistor, though.
#10 21664830 20 Nov 2012 20:25

Earl Albin Earl Albin

Anonymous

Post #10
21664830 20 Nov 2012 20:25

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.
#11 21664831 20 Nov 2012 20:27

Steve Lawson Steve Lawson

Anonymous

Post #11
21664831 20 Nov 2012 20:27

Yeah, I just noticed that. I'm completely wrong. This circuit will regulate the current precisely. I actually learned something new from this!
#12 21664832 20 Nov 2012 20:53

Steve Lawson Steve Lawson

Anonymous

Post #12
21664832 20 Nov 2012 20:53

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.
#13 21664833 20 Nov 2012 21:15

Earl Albin Earl Albin

Anonymous

Post #13
21664833 20 Nov 2012 21:15

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.
#14 21664834 20 Nov 2012 21:43

Earl Albin Earl Albin

Anonymous

Post #14
21664834 20 Nov 2012 21:43

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
#15 21664835 20 Nov 2012 22:04

Frank Bushnell Frank Bushnell

Anonymous

Post #15
21664835 20 Nov 2012 22:04

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.
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#16 21664836 20 Nov 2012 22:08

Earl Albin Earl Albin

Anonymous

Post #16
21664836 20 Nov 2012 22:08

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.
#17 21664837 20 Nov 2012 23:43

Knico Knico

Anonymous

Post #17
21664837 20 Nov 2012 23:43

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.
#18 21664838 20 Nov 2012 23:54

Frank Bushnell Frank Bushnell

Anonymous

Post #18
21664838 20 Nov 2012 23:54

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 . . . !-)
#19 21664839 21 Nov 2012 00:04

Frank Bushnell Frank Bushnell

Anonymous

Post #19
21664839 21 Nov 2012 00:04

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http://www.allaboutcircuits.com - a series of online textbooks covering electricity and electronics

http://electronics-for-beginners.com/links.php

http://www.hobby-hour.com/electronics/ - Teaching from basic principles, resistor code calculators, projects

http://www.radio-electronics.com - useful information and tutorials
#20 21664840 21 Nov 2012 00:24

Knico Knico

Anonymous

Post #20
21664840 21 Nov 2012 00:24

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.
#21 21664841 21 Nov 2012 11:07

Steve Lawson Steve Lawson

Anonymous

Post #21
21664841 21 Nov 2012 11:07

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]
#22 21664842 21 Nov 2012 11:29

Steve Lawson Steve Lawson

Anonymous

Post #22
21664842 21 Nov 2012 11:29

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!?!
#23 21664843 21 Nov 2012 11:49

Steve Lawson Steve Lawson

Anonymous

Post #23
21664843 21 Nov 2012 11:49

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.
#24 21664844 21 Nov 2012 11:51

Steve Lawson Steve Lawson

Anonymous

Post #24
21664844 21 Nov 2012 11:51

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.
#25 21664845 21 Nov 2012 23:09

Knico Knico

Anonymous

Post #25
21664845 21 Nov 2012 23:09

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.
#26 21664846 21 Nov 2012 23:15

Earl Albin Earl Albin

Anonymous

Post #26
21664846 21 Nov 2012 23:15

Simple comparator, should work although a little flicker right at transition perhaps because no hysteresis. No biggie.
#27 21664847 21 Nov 2012 23:41

Steve Lawson Steve Lawson

Anonymous

Post #27
21664847 21 Nov 2012 23:41

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.
#28 21664848 22 Nov 2012 04:49

Steve Lawson Steve Lawson

Anonymous

Post #28
21664848 22 Nov 2012 04:49

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).
#29 21664849 22 Nov 2012 04:50

Steve Lawson Steve Lawson

Anonymous

Post #29
21664849 22 Nov 2012 04:50

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

✨ The discussion focuses on designing a simple 1A load circuit for testing 7.4V lithium-ion batteries using a digital multimeter for voltage measurement under load. Several approaches are considered: a basic fixed resistor load, a variable resistor for adjustable current, and more precise constant current sink circuits using op-amps or shunt voltage references. Ohm’s law (R = V/I) and power dissipation (P = I²R) calculations are emphasized for selecting resistor values and wattage. A straightforward solution involves a 7.5Ω, 7.5W (or higher) wire-wound resistor combined with a momentary pushbutton to apply the load, suitable for brief testing to ensure battery voltage does not drop below a safe cutoff (7.0V). More complex circuits using PNP transistors with emitter resistors, TL431/TL432 adjustable shunt regulators, or AP432 voltage references provide better current regulation and stability but require heat sinking and more components. The TL431 family and AP432 are discussed for their reference voltages (1.24V to 2.5V) affecting current sense resistor values and discharge cutoff voltages. Additional suggestions include simple comparator circuits with LEDs to indicate low battery voltage, using low-power voltage references like LM385Z-1.2 for minimal battery drain. Practical advice includes ensuring proper resistor wattage, heat dissipation, and correct multimeter lead placement after testing. Several online electronics tutorial resources are recommended for further learning.

FAQ

TL;DR: For a 7.4 V pack, a 7.5 Ω resistor draws ~0.99 A and must dissipate ≥7.4 W; “The resistor will get VERY hot.” [Elektroda, Frank Bushnell, post #21664835]

Why it matters: This FAQ shows how to do safe, repeatable 1 A load checks with parts you can buy today, using one or two DMMs; it’s for hobbyists validating 2‑cell (7.4 V) Li‑ion battery health during quick, momentary tests.

Quick Facts

Simple load: 7.4 V ÷ 7.5 Ω ≈ 0.99 A; choose ≥10–20 W wire‑wound to handle heat safely. [Elektroda, Frank Bushnell, post #21664835]
Constant‑current option: AP432 (≈1.24 V), AP431/TL431/TL432 (≈2.5 V) references set sense‑resistor value. [Elektroda, Steve Lawson, post #21664842]
Example sense resistor: 2.5 V ref → 2.5 Ω at 1 A; rate it ~5 W for margin. [Elektroda, Steve Lawson, post #21664832]
Pass‑device heat: ~6.15 W at 1 A; use ~8 °C/W heatsink or fan. [Elektroda, Earl Albin, post #21664826]
Metering: Put ammeter in series on 10/20 A jack; voltmeter across battery. [Elektroda, Frank Bushnell, post #21664835]

What resistor do I need for a simple 1 A load at 7.4 V?

Use a 7.5 Ω wire‑wound resistor. At 7.4 V it draws about 0.99 A (I = V/R). This meets the 1 A target closely enough for hobby checks. It’s inexpensive and easy to wire with a momentary pushbutton. Expect significant heat, so mount it on a fire‑safe surface. For visibility, clamp the ceramic body lightly in a small vise. [Elektroda, Frank Bushnell, post #21664835]

What wattage should the resistor be for that load?

Minimum dissipation is P = V × I = 7.4 W. Choose a resistor rated well above this, such as 10–20 W, to stay cool enough and survive repeated tests. Wire‑wound cement resistors are robust and widely available. “The resistor will get VERY hot,” so avoid plastic breadboards and keep flammables away. [Elektroda, Frank Bushnell, post #21664835]

How do I connect two multimeters to measure voltage and current correctly?

Connect the voltmeter directly across the battery. Put the ammeter in series with the 7.5 Ω resistor and the battery. Ensure the red lead is in the 10 A (or 20 A) jack on the meter before testing. This lets you watch the voltage sag under an ~1 A load as the pack discharges. [Elektroda, Frank Bushnell, post #21664835]

Can I test with only one meter if I only care about voltage under load?

Yes. Wire the 7.5 Ω load with a momentary pushbutton, then measure the battery voltage during the brief press. Stop using the battery once it reaches your 7.0 V cutoff. This quick check verifies usable capacity for the application without continuous instrumentation. [Elektroda, Knico, post #21664837]

Why is a constant‑current sink better than just a resistor?

Battery voltage drops during discharge, so a fixed resistor’s current falls too. A constant‑current sink keeps current near 1 A despite voltage and temperature changes, improving test repeatability. “With the OpAmp, the current will not deviate appreciably.” Use it when precision matters beyond a quick go/no‑go check. [Elektroda, Steve Lawson, post #21664828]

Which reference IC should I use (TL431/TL432/AP431/AP432), and what are their reference voltages?

AP432 is about 1.24 V. AP431, TL431, and TL432 are about 2.5 V (TL431 vs. TL432 mainly differ by pinout). Choose lower reference for lower dropout in the current sink, or 2.5 V types for simplicity and availability. This choice sets the sense‑resistor value. [Elektroda, Steve Lawson, post #21664842]

What value and wattage for the current‑sense resistor at 1 A with a 2.5 V reference?

Use R = Vref / I = 2.5 V / 1 A = 2.5 Ω. Dissipation is I²R = 1² × 2.5 = 2.5 W, so pick a 5 W resistor for margin. This keeps it within safe temperature rise during a sustained 1 A test. [Elektroda, Steve Lawson, post #21664832]

How low will a TL431/TL432 current sink regulate as the battery discharges?

It maintains constant current until the pass transistor’s VCE nears saturation. Expect regulation to fade when VCE is roughly 0.7 V. Below that, current falls because the control loop loses gain. Plan your cutoff above this region to keep results consistent. [Elektroda, Earl Albin, post #21664830]

How much heat does the pass transistor dissipate at 1 A, and what heatsink do I need?

Plan for about 6.15 W in the PNP pass device at 1 A. Use a heatsink near 8 °C/W to keep junction temperature reasonable. A smaller heatsink with a small fan also works but needs a power source. This prevents thermal runaway and protects the device. [Elektroda, Earl Albin, post #21664826]

Is there a simple 7.0 V indicator I can add so I stop in time?

Yes. A reference‑based LED indicator can watch pack voltage continuously. One implementation draws about 5 mA above 7 V and ~35 µA when below. Place it on the load’s battery side. It turns the LED off when the pack dips under your 7.0 V threshold. [Elektroda, Steve Lawson, post #21664844]

Will that indicator flicker near 7.0 V? How do I avoid it?

Without hysteresis, small noise or ripple can cause LED flicker exactly at the threshold. Add a little positive feedback (a resistor network) to introduce hysteresis and create a clean on/off transition. It simplifies interpretation during slow voltage changes. [Elektroda, Earl Albin, post #21664846]

I can’t solder tiny packages—what parts are easier to handle?

Use through‑hole TO‑92 parts. Options include a TL431 in TO‑92 and an LM385‑1.2 V in TO‑92. They are easy to wire, and the LM385 solution can draw ~16 µA when the LED is off. That minimizes idle drain on your pack. [Elektroda, Steve Lawson, post #21664848]

Any gotchas with the ammeter connections after testing?

Yes. Move the DMM’s red lead back from the 10/20 A socket to the volts socket after the test. Forgetting this is common and can short or blow the meter fuse the next time you try to measure voltage. Make it a habit. [Elektroda, Frank Bushnell, post #21664838]

Will a fixed resistor method still be accurate enough?

It’s adequate for quick health checks. Current will sag as voltage drops, but it’s far better regulated than no load and needs no calibration. For precise discharge profiling, use a constant‑current sink instead. Choose based on your accuracy needs. [Elektroda, Steve Lawson, post #21664829]

Quick how‑to: How do I perform a momentary 1 A load test safely?

Wire a 7.5 Ω, 10–20 W resistor and a momentary pushbutton in series with the battery.
Press the button briefly while reading pack voltage; stop if it reaches 7.0 V.
Release, then let the resistor cool on a fire‑safe surface. [Elektroda, Frank Bushnell, post #21664835]