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I'm an electrochemical engineer. I saw a video showing although the output voltage from transformer is about 16.5V without connection, the voltage immediately drops to 9.86V, slightly higher than the open circuit voltage of a battery to be charged.

Video link: https://www.youtube.com/watch?v=k_kEtElESvw // please see from 3:20

Could someone clarify the following points:

  1. Why the output voltage cannot maintain the voltage value seen with this simple “charger” when unloaded?
  2. When I charge battery with a potentiostat equipment, that also can charge the battery with controlled voltage and current, this is not expected to happen, if the potentiostat is properly adjusted. Why does this not happen?
  3. What are the differences between this simple circuit and a more sophisticated circuit, as a potentiostat?
  4. Can a potentiostat be used as a battery charger?
  5. Are there problems to use potentiostat as a battery charger?
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    \$\begingroup\$ Series resistance and/or core saturation. \$\endgroup\$
    – DKNguyen
    Commented Dec 12, 2021 at 3:08
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    \$\begingroup\$ Or loose coupling between primary and secondary; leakage inductance. \$\endgroup\$
    – jonk
    Commented Dec 12, 2021 at 3:47
  • \$\begingroup\$ I'm sorry, could you give more details? I could not understand. \$\endgroup\$
    – Donghoon LEE
    Commented Dec 12, 2021 at 4:09
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    \$\begingroup\$ @DKNguyen - The flux in the core of a transformer is lowered when there is a load on the secondary as the current is such that it opposes the flux from the primary. So it is less likely to saturate under load. \$\endgroup\$
    – Kevin White
    Commented Dec 12, 2021 at 4:20
  • \$\begingroup\$ @DonghoonLEE It's basically why you run slower when you're pushing or carrying something than when you run without pushing anything. Everything slows down or goes down under load. Even batteries don't maintain their voltage when supplying current. \$\endgroup\$
    – DKNguyen
    Commented Dec 12, 2021 at 4:42

1 Answer 1

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The 'battery charger' in question is an unregulated, full wave DC power supply that utilises a transformer having a centre-tapped secondary, a pair of diodes and a filter capacitor.

At no load, the filter capacitor holds the DC voltage close to the peak secondary voltage (16.5V).

The fully-discharged battery draws a high charging current from the power supply and overloads it, causing its output voltage to dip to the battery terminal voltage (close to 9 V) and rise as the battery charges.

The sealed lead-acid battery, used for the demonstration, would already be as good as dead, with its terminal voltage far below the safe lower limit of 10.8 V.

Constant-Current, Constant Voltage (CCCV) charging is good for lead-acid batteries to maintain their life span.

enter image description here

Charging is at a constant current, till the battery terminal voltage reaches 14V, after which charging is continued at a constant voltage of 14 V till the charging current becomes zero.

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  • \$\begingroup\$ As I understand, this is because the output voltage is not so 'strong' to maintain its value from high drawing current. What determines to the output voltage's retentivity ? \$\endgroup\$
    – Donghoon LEE
    Commented Dec 12, 2021 at 10:28
  • \$\begingroup\$ I do not understand the term 'retentivity' with respect to voltage. Suffice to say that the high uncontrolled charging current drops a higher voltage across the internal impedance of the secondary. \$\endgroup\$
    – vu2nan
    Commented Dec 12, 2021 at 12:42
  • \$\begingroup\$ @DonghoonLEE "Retentivity" has a very specific meaning in EE and it is related to magnetic core materials. What you are talking about is just referred to as low output impedance. \$\endgroup\$
    – DKNguyen
    Commented Dec 12, 2021 at 21:34

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