# How long to charge a super-capacitor with LM317

I am trying to charge a super-capacitor to 2.5V with LM317. According to datasheet the maximum output current of LM317 is 1.5A and there is internal current regulator in side it. So what is the charging time now if there is no resister in series between LM317 output and a 600F super-capacitor? I don't think using t=RC and see R=2.5V/1.5A is still useful here. Can anyone tell me what to do in this condition?

The relationship you're looking for is $$\frac{dV}{dt} = \frac{i}{C}$$ or, more usefully here, $$\frac{\Delta V}{\Delta t} = \frac{i}{C}$$ and plugging in your numbers gives $$\Delta t = \frac{{\Delta v} \times C}{i} = \frac{2.5 \times 600}{1.5} = 1000 seconds$$

Now for potential problems.

First, and most important, is the matter of cutoff voltage. You'll need to keep at least 3 volts across the LM317 (Section 8.4.2), so let's say you use a 6-volt source on the regulator input. When you get to 2.5 volts, the regulator will not simply cut off. Instead, depending on the unit you've got, it may well keep on supplying current, although probably at a lower level, and may well drive your cap to 5 volts or more. The 3-volt headroom requirement is in no way guaranteed - it is the worst-case number.

Second, you cannot count on the internal regulator to limit your current to 1.5 amps. From Section 7.5 (Electrical characteristics) 1.5 amps is the guaranteed minimum, while typical is 2.2, and there is no stated maximum. You can use a resistor to set the current output, and as long as you set it for 1.5 amps or less you'll be OK, but the cutoff problem remains. Also,

Third, you must worry about thermal effects. If you set the current to 1.5 amps, and use a 6-volt input, at the beginning of the capacitor charge the LM317 will dissipate 9 watts (6 x 1.5), dropping to 5.25 watts when the cap is at full charge. Unless you provide a very good heat sink (capable of keeping your package cool with 10 watts on it) the regulator will go into thermal shutdown, cool off, start conducting and heat up, cool down, etc.

• Thanks a lot. Still confused about the over charge. My output is set to 2.5 volt, how can the cap still be charged when it reaches 2.5 volt? Do you mean my output voltage increase the same time? And does a output resistance help with this? – tendo Sep 16 '15 at 4:56
• If you use a 317 set for 2.5 volts, you cannot count on the resulting current level. If you use the 317 set as a current source, you cannot control the output voltage. From Section 8.3.3 "easy output voltage or current (not both) programming". If you set the output for 2.5 volts and it happens to current limit at 1.5 amps, per the answer it will take 1000 seconds to charge. If the current limit is 2.2 amps, it will take 682 seconds. If the current limit is 3 amps, 500 seconds. – WhatRoughBeast Sep 16 '15 at 21:00
• Does it mean when setting LM317 as voltage mode (2.5V) and charging a capacitor with no resistor, the capacitor voltage will be no more than 2.5V? From ti document SNVA558 (page 16) if output current is too high it will change to current limiting mode with unknown (usually lower) output voltage. And when it goes back to voltage mode the output voltage may not be restored to nominal(page 16) – tendo Sep 16 '15 at 21:44
• Good find on SVA558. That is correct. If you draw too much current, the output voltage will never recover to the nominal value. You will have to disconnect the cap, then reconnect. (I had not realized the LM317 uses foldback limiting), but then I've never tried to depend on the onboard limiter. – WhatRoughBeast Sep 16 '15 at 22:08
• The reason I don't want a current limiting resister is because the longer charging time (t=4RC). Do you think a trickle charging mode circuit like this circuitstoday.com/battery-charger-circuit-using-lm317 would give me constant current (1.5A) and stable output voltage? Thanks so much for your help! – tendo Sep 16 '15 at 23:39

The definition of capacitance is:

\begin{align} C = \frac{Q}{V} \end{align}

Where C is capacitance, Q is charge stored, and V is the voltage across the capacitor.

Re-arranging and plugging in your voltage and capacitance gives $Q = 1500~ \textrm{coulomb}$.

The first approximation is assume you have an ideal capacitor and ideal wires. The only thing limiting how fast the capacitor charges is the current limit on the supply, so the equivalent circuit is:

simulate this circuit – Schematic created using CircuitLab

Then using the definition of current, we can derive the time required to charge to 2.5V as:

\begin{align} \boxed{ t = \frac{Q}{I_1} = 1000 s } \end{align}

This gives a pretty good first order approximation. If the capacitor has a significant ESR (equivalent series resistance), then the regulator will first charge at a constant current, then switch into a constant voltage charge. In order to solve this, you need to define how close to 2.5V is "fully charged". Then you can use the basic RC low-pass filter result to figure out how long it will take to reach "fully charged".

For example, suppose the capacitor has an ESR of $1\Omega$ (pretty high, but you can substitute your own values in). Then the voltage you can charge to in constant-current mode before the regulator switches to constant voltage mode is $V_C = 2.5V - 1\Omega \cdot 1.5A = 1V$. The time it takes for this to happen is $t_1 = 400s$.

The RC time constant is $\tau = 600 s$. Defining 99% of 2.5V as an acceptable "fully charged" voltage, the time spent in constant voltage mode is

\begin{align} t_2 = -\ln\left(\frac{1.5V-2.5V \cdot 0.01 }{1.5V}\right) \tau \approx 2457 s \end{align}

Thus, the total charge time is $t = t_1 + t_2 = \boxed{2857 s}$.

These are theoretical answers and assumes a model for the LM317 and supercap which probably don't match reality (i.e. reality may give wildly different results, including setting your board on fire). See WhatRoughBeast's answers for a few of the practicalities and problems with this setup.