Suppose I have a 20 AmpHour battery at 4V (as in, the actual capacity of the battery is 80 Watt hours). And I'm using a 10V 2A charger to charge it (this is a 20 Watt charger). How can the charge time be calculated?

Is it 4 hours?

Note: The charge time is definitely not 20Ah/2A = 20h as a lot of people state, since these numbers are close to the actual values for most fast-charging smartphones today, and the charge time should be of the order of 1 hr).

Note 2: Please explain using the fundamentals of physics, instead of obscure formulae.

Since the charger is at 10V and the battery is at 4V, how does this affect the charging?

  • 6
    \$\begingroup\$ Your charger is the wrong voltage. You can't use it to charge the battery without additional circuitry. If you can't handle obscure formulae you'd better stay away from "the fundementals of physics". Ask for "basic electrical formulae" instead. If you just drop the voltage from 10 to 4 V then charge into battery is 4 V x 2 A = 8 W. \$\endgroup\$
    – Transistor
    Jul 30, 2016 at 12:02
  • 4
    \$\begingroup\$ electronics.stackexchange.com/questions/24160/… \$\endgroup\$
    – user57037
    Jul 30, 2016 at 16:34
  • \$\begingroup\$ The charge rate is usually specified as a fraction of "C" by the manufacture of the battery. As for capacity, the slower you draw off the charge the longer the battery will last. For example, a 20Ah battery will probably not last an hour if you continuously pull 20A from it. So, not to frustrate you, but there are hardly any canned / clear-cut answers with respect to secondary battery maintenance. On top of that, a battery pack may contain it's own electronics which will further abstract basic assumptions about its secondary battery chemistry / behavior. \$\endgroup\$
    – st2000
    Jul 30, 2016 at 17:22
  • \$\begingroup\$ C/I is true during the first part of the charge cycle, the constant current(CC) part. ('C' is capacity, and 'I' is charge current.) Later the charge mode transitions to constant voltage (CV). During the CV stage, charge is replenished more slowly. So the overall charge takes longer than C/I. \$\endgroup\$
    – user57037
    Jul 30, 2016 at 18:23

1 Answer 1


You don't tell us what voltage is required to charge your 4 V battery so let's assume it's 5 V.

Linear regulator / charger

  • If we use a linear voltage regulator or charger we can drop the 10 V down to 5V and can draw 2 A from the power supply. Power into the battery will be 5V x 2A = 10 W. Power dissipated in the regulator will be 5V x 2A = 10 W. Efficiency will be 50%.
  • Charge time would be \$ \frac {Ah~capacity}{current} = \frac {20Ah}{2A} = 10~h \$ if the battery could store all the energy. Since the battery gets warm during charging we know that some of the input power is being lost so 12 h may be a better estimate.

Buck charger

Using a buck voltage converter is more efficient and when reducing the voltage the current can be increased. Let's assume we could get one with 85% efficiency.

$$ V_O \cdot I_O = \eta V_I \cdot I_I $$ where \$ \eta \$ is the efficiency. From this we can work out the maximum output current is

$$ I_O = \eta \frac {V_I \cdot I_I}{V_O} = 0.85 \frac {10 \cdot 2}{5} = 3.4~A $$

Ideal charge time will be \$ \frac {Ah~capacity}{current} = \frac {20Ah}{3.4A} = 5.9~h \$ so say 7 hours allowing for heat losses.

I hope the formulae weren't too obscure. ;^)

  • \$\begingroup\$ The issue is not heat losses. The issue is that cell phone batteries transition from CC to CV. The CV stage replenishes charge more slowly than the naive C/I calculation would lead you to believe. Hence the full charge period is longer than C/I (where I is nominal charge current). \$\endgroup\$
    – user57037
    Jul 30, 2016 at 18:21
  • \$\begingroup\$ I agree, of course, but I just wanted to give the OP something to go on. Please feel free to give a fuller or complementary answer. \$\endgroup\$
    – Transistor
    Jul 30, 2016 at 18:42
  • \$\begingroup\$ I may have actually misread the question. I think your answer is actually pretty good. The OP was hung up on power equivalence which is what you addressed. I was thinking it was about charging taking longer than C/I. \$\endgroup\$
    – user57037
    Jul 30, 2016 at 18:55

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