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LM317 constant current charger are quite popular as it seems. I made one for my 12v 7Ah Lead Acid battery. I used a 2.2 ohm 10 watt resistor and calculated that it would give a decent amount of constant current to charge my battery. To my surprise, I noticed that the battery is drawing only 20 - 30mA current when charging. Currently the battery is little over-discharged to 11.8v.

I had an impression that LM317 Constant current mode pumps constant amount of current, no matter what the load is. Was my understanding wrong? If so, should the constant current value, i.e 1.25/R is actually the max current value? Because when I sorted the output terminals of the power supply, it reached upto desired current level, i.e (1.25/2.2) * 1000 mA

And if my understanding of LM317 constant current supply is correct, why is the battery drawing only 20mA current?

My circuit is as simple as following.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Please provide a schematic diagram that shows the exact circuit you created. \$\endgroup\$
    – Jim Fischer
    Aug 10, 2016 at 22:40
  • \$\begingroup\$ Let me guess. You are using a voltage source of something like 12 volts, with an LM317 current source, right? Your problem is that the LM317 itself should have at least 2 volts across it, plus another 1.25 across the current set resistor. Since 12-volt SLAs will charge to about 13.5 volts, you need a voltage source of at least 17 to 18 volts. 20 would be better. \$\endgroup\$
    – WhatRoughBeast
    Aug 10, 2016 at 22:47
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    \$\begingroup\$ What is the charger's output voltage during the CC charging phase (with the charger attached to the battery)? Also, measure with a DMM and report back to us the LM317's unregulated input voltage (the "15V input voltage") when the battery charger circuit is connected to and charging the battery. \$\endgroup\$
    – Jim Fischer
    Aug 10, 2016 at 22:59
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    \$\begingroup\$ Charging a 7ah battery is really beyond the capabilities of an LM317 unless you figure a way to use an external pass transistor to dissipate all the heat. Honestly you would be better off using an op-amp to control a power FET or BJT. \$\endgroup\$
    – user57037
    Aug 11, 2016 at 3:49
  • \$\begingroup\$ I measured the open corcuit Input and Output voltage across the regulator. It was 15.2v and 14v respectively. \$\endgroup\$
    – sribasu
    Aug 11, 2016 at 5:23

1 Answer 1

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enter image description here

Figure 1. Line regulation for the TI LM317 indicates that it could require 3 V headroom to operate properly.

The LM317 has a relatively high dropout voltage compared with modern LDO (low drop-out) regulators. Add to this the 1.25 V across the current sense resistor and you can see that your supply voltage is too low.

The big question is why you are charging lead acid batteries using constant current. You should be using constant voltage.

From the datasheet:

9.3.8 50-mA Constant-Current Battery-Charger Circuit

The current limit operation mode can be used to trickle charge a battery at a fixed current. \$ I_{CHG} = 1.25V ÷ 24Ω \$. \$ V_I \$ should be greater than \$ V_{BAT} + 4.25 V \$. (1.25V [\$ V_{REF} \$]+ 3V [headroom]).

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    \$\begingroup\$ Well, the first stage is usually constant current until a specific Voltage target is reached. Then hold that Voltage (stage 2) then drop to a long-term float Voltage (stage 3). \$\endgroup\$
    – user57037
    Aug 11, 2016 at 6:38
  • \$\begingroup\$ As mkeith said, I learned first constant current charging then constant voltage for float voltage. Did I learn wrong? Sorry, I am new in Electronics! \$\endgroup\$
    – sribasu
    Aug 11, 2016 at 15:45
  • \$\begingroup\$ Yes my question gets a classic example here. See the contradicting facts in the LM317 Datasheet shared by @Transistor \$\endgroup\$
    – sribasu
    Aug 11, 2016 at 16:01
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    \$\begingroup\$ There are no contradicting facts in the datasheet. (The LM317 is more than 40 years old so they know what they're talking about.) Section 9.3.8 clearly shows that you need 4.25 V more on the input than you want on the output. See my update. \$\endgroup\$
    – Transistor
    Aug 11, 2016 at 16:19
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    \$\begingroup\$ The output will rise as it tries to get 1.25 V between the output and the ref. pin. If charging a battery you will get constant 50 mA until the voltage of the battery gets within 4.25 V of Vin. At that stage the regulation will start to fail. \$\endgroup\$
    – Transistor
    Aug 11, 2016 at 20:46

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