3
\$\begingroup\$

I'm looking into modding an old battery charger, but I want to understand exactly how it works before I start. What is the purpose of the choke in this circuit?

Circuit Diagram

Some people at my work have suggested the choke should be on the output of the bridge rectifier not the input - but it was obviously manufactured this way for a reason. Why would you want a choke on the AC side of a bridge rectifier?

\$\endgroup\$
  • 1
    \$\begingroup\$ Is a battery really the only thing on the output of that bridge? Because most of the time the reason you put differential or common mode chokes on the AC part of any line is to prevent your noisy switching circuit from injecting noise back onto the mains and using those wires as a radiator. \$\endgroup\$ – DKNguyen May 12 at 5:48
  • 1
    \$\begingroup\$ @Toor it is just a simple battery charger, there is a current shunt and a circuit breaker in between the bridge rectifier and the battery, but nothing doing any switching. \$\endgroup\$ – James May 12 at 6:21
  • 1
    \$\begingroup\$ It acts like a current limiter. \$\endgroup\$ – Marko Buršič May 12 at 13:36
2
\$\begingroup\$

Pretty sure that is there to limit the current into a largely discharged battery.

Basically it adds reactance in series with the (almost) voltage source that is the transformer so as to ensure the current does not rise to silly levels if the battery is very flat.

Another approach to the same thing that you also see in battery chargers is a transformer designed with a magnetic shunt thereby raising the leakage inductance (to largely the same effect).

Somewhat crude, not great for the power factor, but cheap and VERY robust and reliable.

\$\endgroup\$
  • \$\begingroup\$ Thanks for your concise answer. Could you tell me if the circuit would behave differently if the choke was on the output of the bridge rectifier rather than the input? \$\endgroup\$ – James May 31 at 8:22
1
\$\begingroup\$

Thus the purpose of L (depends on value) could be to limit current in the diodes by ΔI=VΔt/L in time domain. The diodes are often driven very hot on a heatsink.

In the frequency domain I(f)=V/(Rs+DCR+ESR+jXL(f)) but only for the duty cycle when the diodes are conducting.

Rs= Diode resistance above 0.8V
DCR= inductor DC resistance
ESR= battery ~= 5V/CCA

Transformer output impedance is low limited by mutual coupling and winding resistance.

L1 impedance Z(f)= DCR + jωL

When diode bridge BR1 is conducting just before peak AC voltage, they are also very low Resistance, Rs (which inverse to diode power rating).

The battery ESR can be even lower in the mΩ (which is inverse with Ah capacity when charged)

The battery CCA rating determines its ESR.

An automotive high current charger shown without the transformer but 16Vp enter image description here

\$\endgroup\$
  • \$\begingroup\$ You have exaggerated the difference because you aren't putting the same amount of energy into the battery on each cycle. But, you have illustrated the concept, the conduction period is wider, so the peak will be less even if the energy is the same. \$\endgroup\$ – Mattman944 May 12 at 9:23
  • \$\begingroup\$ The OP put the inductor on the other side of the bridge. Can you re-run this way? I don't think that it matters, unless somehow you were to get the voltage and current far enough out of phase to upset the diodes. I am trying to decide if this is a real possibility. \$\endgroup\$ – Mattman944 May 12 at 9:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.