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I found this circuit, which supposedly limits the charging current to 2A max. This is for acid/lead batteries.

enter image description here

Then, I would like to NOT use a power transistor like 2N3055, but instead use a relay as, perhaps, follows (ignore the 1A relay rating in diagram):

enter image description here

I couldn't simulate it with LTSpice because I don't have the relay component. I do have Electronics Workbench but it won't run the simulator with Win10. In any case, I would like to know if it should work using a low power transistor plus relay. Also, perhaps the BD135 + 2N3055, ideally using a low power transistor instead the 3055, could be simplified using a Darlington like TIP122?

UPDATE

This is from the site that has the circuit:

How does the current limitation work? If the load current is 2 amps, there is a voltage of approximately 0.94 V at the terminals of the R7 resistor. This voltage charge the C2 capacitor through the R3 resistor. This voltage is applied between the base and the emitter of transistor Q1. (0.65V approximately) and activate Q1.

The transistor Q1 start to conduct and remove some of the base current from the Q3 transistor, which in turn removes a bit of base current from the Q4 transistor, which determines the amount of charge current.

So why use a 15A transitor? Why not one slightly greater than 2A?

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  • \$\begingroup\$ Why do you want to use a relay? \$\endgroup\$ Commented May 2, 2023 at 1:41
  • \$\begingroup\$ @KevinWhite where I live, we no access to such power transistors like 2N3055 \$\endgroup\$ Commented May 2, 2023 at 1:50
  • \$\begingroup\$ Please simulate above original circuit with a voltage sweep for the battery from 12 V to 14.4 V, instrumenting current through the PNP as well as charge current. Rectified 15 V sine isn't anything like 15 V DC: compare. \$\endgroup\$
    – greybeard
    Commented May 2, 2023 at 5:02
  • \$\begingroup\$ A relay is not a suitable replacement. I'm surprised that power transistors are not available if other components are. Almost any power transistor could work even ones scavenged from old equipment. The 2n3055 in particular is over 50 years old. \$\endgroup\$ Commented May 2, 2023 at 16:56
  • \$\begingroup\$ @KevinWhite 2N3055 are quite scarce here, more easy to find are those used in CRT horizontal control, but the ones I see are <15A, basically 8A is all the power I can find. Ah oh...and maybe some Mosfet like IRF630 but nevertherless are 9A as well. \$\endgroup\$ Commented May 2, 2023 at 19:47

2 Answers 2

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Be aware that R7 should be 0.47 ohms, not 0.47k.

With the relay as drawn, there's nothing limiting the current through the battery when the relay is on, not even R7. And when it's off, there's no charging at all.

The circuit is now sensing the current through the relay coil, which doesn't tell you anything about the state of the battery.

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  • \$\begingroup\$ that's what i was afraid of \$\endgroup\$ Commented May 1, 2023 at 22:49
  • \$\begingroup\$ it's almost impossible to do active current limiting without a power transistor, so this is maybe not a helpful approach. The nearest you can get is a series power resistor of 7R which of course limits current to 2A but gives half the voltage you want at 1A. In the end you will need power transistors to do this properly. Where do you live? \$\endgroup\$
    – danmcb
    Commented May 2, 2023 at 6:04
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If you can get a simple and very common LM317 regulator, they're easy to configure as a battery charger. Look in the datasheet for example circuits, or google a bit. This one has a battery charger example on the first page! Texas Instruments version of the LM317

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