I am a bit puzzled and ask for your help about the following:

A theoretical capacitor of 100 F is being charged with a constant power source of values V = 50V, I = 50A, ESR of capacitor = 5 mohm

To calculate the time to charge the cap:

Approach 1: [Calculate time using energy flow rate]

Capacitor capacity = 0.5xCxV^2 = 0.5x100x50^2 = 125 kJ

Charging power = VxI = 50x50 = 2500 W= J/s

Time to charge = Capacitor capacity / charging power = 125 kJ/2500 J/s = 50 s

Approach 2: [using standard capacitor charging formula]

V of capacitor = V(1-e^(-t/RC)) = 50(1-e^(-2.5/(0.005x100)) = 49.88 V

As one can see that after 5 time constants (2.5 s), the capacitor's voltage is 99% using approach 2.

Obviously, this is the correct approach using the established formula.

Why is approach 1 off this much? What am I missing in either case?

  • 5
    \$\begingroup\$ Because in the second case your current is not being limited to 50A. \$\endgroup\$
    – vir
    Commented May 31, 2022 at 22:33
  • 3
    \$\begingroup\$ What does a "constant power source of values V=50V, I=50A" mean? You can't have constant voltage and constant current (unless the thing being powered is a resistor) \$\endgroup\$ Commented Jun 1, 2022 at 9:26

4 Answers 4


Your time constant of 0.5 seconds clearly is derived from your capacitor ESR of 5 mohm. So, what you are effectively proposing as a charge circuit looks like


simulate this circuit – Schematic created using CircuitLab

So, start with the switch open and the capacitor discharged. Now close the switch. What is the charge current?

That is 50/.005, or 10,000 amps.

Compare this to the 50 amp limit of the constant power charger. You think that might have something to do with it?

  • \$\begingroup\$ All answers following the original post are good and carry much more technical information, but your reply is most to the point. The internal resistance of the power supply has to be included in the RC time constant. That said, the charging time will increase by many folds using the standard capacitor equation. New RC = 100.5 seconds. New calculations of Vcap = V(1-e^(-t/RC)) = 50(1-e^(-502.5/(1.005x100)) = 49.66 V. The more realistic time to charge the capacitor of this size is now about 8.5 minutes. Thanks to your very simple approach to help me stop pulling my hair. \$\endgroup\$
    – Raymond
    Commented Jun 1, 2022 at 18:08
  • \$\begingroup\$ @Raymond - I'n sorry, but you've confused me. Where does 100.5 seconds come from? The thing you need to realize is that the constant power charger does not have A resistance. Its effective resistance (voltage over current) changes with capacitor voltage level. A constant-power source will allow reaching 50 volts exactly in a finite time. An RC circuit will NEVER reach 50 volts. Not exactly. \$\endgroup\$ Commented Jun 2, 2022 at 18:40
  • \$\begingroup\$ 100.5 sec is the product of R (1-Ohm of charger + 0.005-Ohm ESR) and 100F. Even if the V/I ratio of charger is changing overtime, won't the average be 1? If it is not assumed to be 1, how would one approximate the charge time? Any suggestion on calculating the 100% (or 99% for that matter) charge time? \$\endgroup\$
    – Raymond
    Commented Jun 4, 2022 at 16:26
  • \$\begingroup\$ @Raymond - Ah. I see. "1-Ohm of charger" And you got that by dividing 50 volts by 50 ohms. A constant-power charger doesn't work that way, nor do current-limited power supplies in general. Presumably a constant-power supply will provide a constant load power regardless of impedance as long as the 50V/50A limits are observed. This means that "source resistance" doesn't really mean anything. For instance, a 50 volt supply with a 50 amp current limit will charge a cap at 50 amps regardless of voltage (as long as load voltage is less than 50). Such a setup will take 100 seconds for your cap. \$\endgroup\$ Commented Jun 6, 2022 at 13:41
  • you are neglecting ESR losses in your conservation of energy equations.

  • Charging a cap or battery is never "constant power" unless V*I= constant, which is not the case with CC for a cap with a low initial condition.

  • batteries have a 10% minimum charge voltage range, so CC is close to Constant Power but not.

For constant power , the current must decrease as the error to target voltage decreases = P=V*I out and that is an increasing output impedance. Yet for fastest charging time, CC is usually the best case to CV for <=10% of the time to charge the secondary charge to 5% CC for cutoff. All batteries and electrolytic caps have this memory feature , more or less depending on chemistry.

  • using a PV with constant Solar Power and efficiency is a special case.
  • the MPPT control MUST match the load impedance to the PV source impedance
  • then the cap or battery charger must use maximum efficiency at constant power to match the PV output, which is from a low Vcap initial condition, means I declines with rising V such that V(t)*I(t)out = effic. * Pin

The problem is already the following assumption:

A theoretical Capacitor of 100 F is being charged with a constant power source of values V = 50V, I = 50A, ESR of Cap = 5 mOhm

The voltage at the connectors of the capacitor is:

$$U_{\text{connectors}} = U_{\text{capacitor}} + I * ESR$$


$$U_{\text{connectors}}(t) = \frac{Q(t)}{C} + I(t) * ESR$$

"Charging the capacitor" means that \$Q(t)\$ increases over the time.

However, this means that \$U(t)=50V\$ and \$I(t)=50A\$ cannot both be constant.

You can use a constant power source; however, in this case, the product \$P(t)=U(t)*I(t)\$ is constant but neither \$U(t)\$ nor \$I(t)=\frac{P_{\text{const.}}}{U(t)}\$ is constant.


Your problem, as stated, is overconstrained. If you know V=50V and I=50A, the ESR has to be 1 ohm at t=0 and drop gradually over time as the capacitor stored voltage increases.

If we choose to assume that V and I can be 50 units or less, you can calculate the linear ramp easily...starting at 1/4V (due to ESR) and ramping up (remember, amps are coulombs per second and farads are coulombs per volt) at 0.5V/S all the way to 50V. At that point, however, 0.25V is due to ESR * current, so current will drop off exponentially as the cap finishes charging.


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