I'm planning to charge auxiliary battery from car's alternator, by using voltage sensitive relay to connect positive poles of car's battery.

The only obstacle now is that auxiliary battery (12V) allows maximum charging current of 20A. Therefore I need current limiter in series with relay.

Which is the most simple circuit that I can use?

I've read that people use light bulbs due to their non-linear behaviour. In that case how can I calculate the needed wattage of (12V) lightbulb, to limit current to 20A?

EDIT: In other words I need 12V lead-acid battery charger that gets power from another 12V lead-acid battery with charging limit of 20A

EDIT: System info:

  • Car battery: 100Ah 760A start current - regular lead-acid car battery
  • Auxiliary battery: 100Ah (C20), max charging current: 20A, 500A/ 5s start current - cyclical solar battery
  • Wires: 2x 10mm^2, cca. 2x 3m, copper
  • \$\begingroup\$ What type of lead-acid battery? What's the capacity of each battery? What's specific about 20A limiting current? \$\endgroup\$
    – dirac16
    Commented Jul 10, 2017 at 16:02
  • \$\begingroup\$ I add it to the original post \$\endgroup\$
    – Delphi4U
    Commented Jul 10, 2017 at 16:43
  • \$\begingroup\$ How "good" do you need it? Light bulb is your by far cheapest and easiest solution. \$\endgroup\$
    – winny
    Commented Jul 10, 2017 at 17:08
  • \$\begingroup\$ Well, just that it doesn't go over 20A... and still charges somewhat quick. \$\endgroup\$
    – Delphi4U
    Commented Jul 10, 2017 at 17:51

2 Answers 2


This is not easy as a linear limiter and better with a PWM limiter and LC filter.

If the 12Vaux is fully discharged and requires (14.2-11.2)V*20A=60W series load dump . That is a lot of heat before the Imax reduces to a CV equal voltage to the alternator.

If one considers a light bulb a good solution , this is effectively a PTC almost constant current source when used in the 0 to 10% voltage range defined by DCR which is 10% of rated voltage R. e.g. 12V/20A where filament resistance drops to 10% of DCR at rated voltage at 3200'K Thus 3V*20A=60W , with a DCR of 0.15 ohms then at 10V and 10% DCR the bulb may be something like 30V 1.5Ohms or 45W which is really not a practical value. It might be something like to 12V 70W car headlamps in series/parallel unless you want an AUX light when charging.

So ok in lower currents, but not 20A. This is just a ballpark estimate, not a rigorous calc.

A better solution is PWM with a series choke rated for 20A such as those air coils found in ATX PSU's and a fast switch rate with a MOSFET switch rated for 50A ( hi side or low side. ) THe battery acts as a 10kfarad capacitor with some ESR from 5 mOhm to 1 Ohm when dead. but an RF cap will reduce EMI.

Another way is to just use a fixed 150 mOhm 60W heater wire (NiChrome) and use that to keep your coffee cup warm ;) Or use a Cap Pulse discharger rated for high RMS ripple current ( several large plastic caps in parallel) This is essentially an active SMPS current limiter with a 0.1V max drop at CV mode at 14.2V or 0.1V/20A = 5 milliohm (MOSFET + choke DCR)

I said to do this right is not easy and stay within limits.

Ultimately the simplest solution is get a bigger battery rated for the current of the alternator, but then you may end up blowing your Alternator diode bridge with both batteries are weak and drawing max current of alternator for long periods at 180'C junction temp.

You can look at NTC disc surge limiters, but they cannot protect a battery as they are designed to protect perhaps only 0.5Farad and not a battery with 10k~100K Farads. To understand read , https://www.digikey.com/en/ptm/a/amphenol-advanced-sensors/cl-series-inrush-current-limiters/tutorial. To dump 40 Watts of heat it either has to run extremely hot or be big like headlamp bulbs ( which will get hot and must be sealed from moisture).

  • \$\begingroup\$ I guess I will use heater wire then... I would need schematics otherwise, and I worry that I might produce unwanted RF... But if I understand the principle is that you power inductor (choke) that then charges battery? \$\endgroup\$
    – Delphi4U
    Commented Jul 10, 2017 at 17:49
  • 1
    \$\begingroup\$ Yes choke raises low DCR impedance of series L pulsed current and battery acts as LPF such that L(f) acts as variable Z with PWM control loop. So 5 mohm copper square coil (SMD) and 100KHz switch with current sense and hysteretic control of on off switch. Shielding on tiny choke reduces noise with RF caps on input and output. This is how its done on your PC MOBO. \$\endgroup\$ Commented Jul 10, 2017 at 18:05
  • 1
    \$\begingroup\$ e.g. 150nH at 200kHz switch ~ 150 mΩ the required current limit R @ 20A and thus with low DCR copper dissipating only 250mW must have a DCR of 0.6mΩ something like these digikey.com/products/en/inductors-coils-chokes/fixed-inductors/… \$\endgroup\$ Commented Jul 10, 2017 at 18:19

The simplest current limiter I have come across is 2 transistors and a resistor..


simulate this circuit – Schematic created using CircuitLab

If you ignore any values then you can imagine it like this: As the current increases, the voltage drop over Rsense increases. When that drop becomes more than approx 0.7V, Q3 is turned on and the base voltage of Q2 is pulled down to prevent the current increasing.

Rsense is calculated by Vbe/I

I have used this circuit myself a few times and it has worked great at limiting current for me. I am still learning electronics myself so if you want a more detailed description of this circuit, I can't provide any more than I just have, but I am sure someone will be able to!

Hope this is useful to you

  • 3
    \$\begingroup\$ There are Q2, Q3 in the schematic Q1, Q2 in the description. :) \$\endgroup\$ Commented Jul 10, 2017 at 15:08
  • \$\begingroup\$ Ha! So it is! I was copying my hand written notes when writing the description and didn't look! I'll change that now! \$\endgroup\$
    – MCG
    Commented Jul 10, 2017 at 15:18
  • 1
    \$\begingroup\$ I'm away from the computer, on my phone so had to amend the description rather than the schematic! \$\endgroup\$
    – MCG
    Commented Jul 10, 2017 at 15:19
  • \$\begingroup\$ Q3 is in the wrong orientation, typically a pullup resistor is used from input to control on your schematic. \$\endgroup\$
    – sstobbe
    Commented Jul 10, 2017 at 15:27
  • 1
    \$\begingroup\$ Current limiter must dump 60W at reasonable temp rise of 60'C , this design would FAIL \$\endgroup\$ Commented Jul 10, 2017 at 17:32

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