# Calculate Charge time of capacitor with current limiting power supply

If I have a 10VDC supply current limited to 1AMP to charge a capacitor through a 1 ohm resistor, what will happen and how long does it take to charge? I need a capacitor that can "eat" this power for 2ms.

1A*1ohm = 1VDC

When I don't have a capacitor connected, the voltage out of the supply reads 1V, because the resistance is only 1 ohm. But when I add the capacitor, the voltage accross it reads 10VDC. I know that the capacitor builds up in voltage as time passes. Is this voltage buildup linear with a constant current source?

Given a certain Capacitance, how do I calculate the voltage at any time T?

If I use a resistor higher than 10 ohms, will my current drop below 1A? Does the resistor still have the same operation at ALL resistances under 1 ohm? (yes the current regulator shown here is set for 1.25A) That's what I'm designing for.

I expect a 3-4V drop across my current regulator+resistor. What happens if my load is only 2.2 ohms? Will the gate of the MOSFET still reach ~6-7V? I'm wanting to delay voltage to the MOSFET gate reaching 5V by more than 2ms.

Some of the things you say are conflicting, so I want to be clear about the question I am answering. If this is not the question you intended, then you need to write a better question. The question is what is the voltage accross C1 over time when it initially starts at 0. You didn't specify a capacitance, so we'll leave that as the variable C.

There are two separate regimes to the C1 voltage over time. The first is when the supply is in current limit mode. In that case, the capacitor is being charged up linearly. When the capacitor reaches 9 V, the supply switches over to constant voltage operation. After that, there is a exponential decay from 9 V to 10 V governed by the RC time constant.

The voltage rise of a capacitor is dV = I dT / C. We know the current (I) is 1 Amp and that dV will be 9 V in the first part of the function. The time to reach 9 V is dT = dV C / I. For example, if C = 470 µF, then the time to charge from 0 to 9 V is 4.2 ms. The capacitor voltage will rise linearly during that time.

From 9V on, the supply will be at a constant voltage of 10 V. This remaining 1 V rise will occur as a exponential according to the time constant RC. Again using 470 µF as example for C, that time constant would be 470 µs. That means, for example, that 470 µs after the capacitor has reached 9 V, it will have gained another 630 mV, which would put it at 9.63 V in absolute terms.

## Added to clarify why the crossover point is 9 V:

Work backwards and assume the supply is always putting out 10 V. At what capacitor voltage does that require 1 A or more? Since R1 is 1 Ω, 1 A thru it causes a 1 V drop. If the supply is 10 V and R1 drops 1 V, then there must be 9 V on the capacitor when the current is 1 A. If the capacitor voltage is lower, then the voltage on R1 must be higher, which means the supply has to source more than 1 A. However, we know the supply puts out the lesser of 10 V or 1 A, so delivering more than 1 A is not possible. This means for capacitor voltages below 9 V, the supply voltage will be the capacitor voltage plus 1 V, since 1 V will be accross R1 when the supply is putting out 1 A.

• 1) where are you guys getting 9V as the voltage where it stops being linear? Your answer helps a lot. Sorry for the vagueness. I didn't specify a C because I need to figure out what size cap and resistor to use. I merely used stock easy values to figure out the time vs. voltage/current characteristics are. – brett s Aug 30 '13 at 17:12
• @brett: See addition to the answer. – Olin Lathrop Aug 30 '13 at 17:57
• Thanks!!! Makes perfect sense. I scrapped some parts together in a quick RC circuit to verify the equations. If you ONLY include the capacitor in my current regulated circuit, the voltage across the capacitor will rise linearly until it reaches the supply voltage correct? So... in an already current limited state, there is not much reason for the resistor correct? For my needs, I don't think it's going to be needed. – brett s Aug 30 '13 at 19:53
• I take that back. I still need something to drop voltage across before the capacitor, or else it won't accomplish what I need. – brett s Aug 30 '13 at 20:11
• @brett: Without a resistor, a constant-current supply will charge up a capacitor linearly with time until the supply gets to its voltage regulation threshold. As for "drop voltage accross before the capacitor", I have no idea what that's supposed to mean. Else is won't accomplish what exactly? You haven't said what you need this thing to do. – Olin Lathrop Aug 30 '13 at 20:58

If I have a 10VDC supply current limited to 1AMP to charge a capacitor through a 1 ohm resistor, what will happen and how long does it take to charge?

The charge current will be limited to 1A until the voltage on the capacitor exceeds 9V.

So, the capacitor voltage will increase linearly with time from 0V to 9V. The time required is determined by the capacitance C:

$\Delta t = \dfrac{\Delta V}{I}C = \dfrac{9V}{1A}C = 9C$

Once the capacitor voltage reaches 9V, the power supply comes out of current limit so the capacitor voltage increases from 9V to 10V in approximately 5 time constants where the time constant is:

$\tau = RC = 1C$

So, the total time to charge to about 10V is:

$\Delta t + 5 \tau = 9C + 5C = 14C$

If the desire charge time is 2ms, the required capacitance is:

$C = \dfrac{.002}{14} = 143 \mu F$

• How are you getting 9V as when it stops linear rise? – brett s Aug 30 '13 at 17:14
• @bretts, the power supply voltage is limited to 10V. If the capacitor voltage is greater than 9V, the voltage across the resistor must be less than 1V and so, the current through it must be less than 1A. – Alfred Centauri Aug 30 '13 at 17:40