# Automatic Capacitor Charge/Discharge Circuit

I am still somewhat of a beginner in electronics, so please go easy on me, but here is a situation I have encountered, and hopefully some of you have as well.

I have a voltage source of 5volts, but extremely low current. This power source is connected to a large capacitor (in my case 3F 5.5V). After a period of time, the capacitor does eventually charge up to 5V as expected, this is all fine and dandy, but now I need to use this energy. The capacitor's load is a small, 500ma DC-DC boost converter, to utilize most of the capacitors charge, but it only works down to a minimum of 0.7 Volts.

So here is what I am looking to do, build a circuit that allows the capacitor to discharge only once it's reached 5V (or 4.5V to leave some wiggle room), and stops discharging once it drops below 1V or so. Only to start discharging again once the cap reaches its 5V threshold.

It should also be said that efficiency is very important here, as power is extremely limited for this circuit. The power source is a nuclear device capable of producing small amounts of power, unfortunately I am unable to provide exact specifications, other that that it is capable of powering the circuit enough to slowly recharge the storage capacitors.

I apologize if I am not being clear, feel free to ask for more details. :)

EDIT: Here is a schematic of my circuit, built only by memory so lets hope its accurate xD

MAX756 Datasheet

simulate this circuit – Schematic created using CircuitLab

• You need to be more specific. How low is "extremely low current"? Give us some numbers. – Dave Tweed Mar 23 '14 at 15:40
• It's likely that the boost regulator circuit connected to the cap can offer some assistance here - do you have a circuit for this section? – Andy aka Mar 23 '14 at 18:33
• You really should post some info about the boost circuit. Almost all boost controllers have a shutdown or enable pin (like the one in Andy's answer). Some that are meant for battery use will also have a low battery sensor. Use of the SD with LBS could do what you are talking about with not many parts added. – gsills Mar 23 '14 at 20:47
• Added schematic. – Zoidsfan77 Mar 24 '14 at 12:28

The shutdown (SD/) pin of the MAX756 could be used to turn on the boost when the super cap voltage ($V_c$) is charged to ~4.5V. Then, the Low Battery Indicator (LBI) could be used to turn the boost off when $V_c$ has decayed to 1.25V. A comparator with added hysteresis would be lowest current, lowest part count, and most repeatable way to control the SD/ pin for turn on. Since the MAX756 will start at input voltage as low as ~1V, any comparator used will need to be operational down to that voltage to maintain control of device turn on. The comparator will also need to be open drain so tha it can be wire-Or'ed with the LBO output for turn off control. Comparators that fit this requirement are rare, but the NCS2202 would work.

Using the LBI function for boost turn off makes things a lot easier. When the LBI pin is lower than 1.25V, the LBO pin is pulled down through an open drain N channel FET. It is a built in function, so requires no extra parts. More important, the low transition of the comparator doesn't have to be precise, it doesn't even need to be possible to reset by decay of $V_c$ as LBO pull down of the SD/ pin handles that. This makes it a lot easier to calculate resistor values. It isn't desirable to operate the MAX756 at $V_c$ lower than 1.25V because output current there is only about 100mA, and dropping quickly as $V_c$ goes lower. Here is a schematic of how this could be done:

Boundary cases are what is important for defining the hysteresis. First case is $V_c$ approaching 4.5V with U3-1 still low. Bias current in the hysteresis circuit ($I_b$) is from $V_c$ through $R_{\text{in}}$,$R_h$, and U3-1 and from $V_c$ through D2 (whose forward drop is $V_{\text{sch}}$), $R_{\text{pu}}$, and U3-1. With the U3-3 sense point voltage being set by the voltage divider of $R_{\text{in}}$ and$R_h$ :

$\frac{V_{\text{ch}} R_h}{R_h+R_{\text{in}}}$ = $V_{\text{ref}}$

where $V_{\text{ch}}$ is $V_c$ at its high point. Bias current can be written (with 100 $\mu$ A as target) :

$\frac{V_{\text{ch}}-V_{\text{sch}}}{R_{\text{pu}}}$ + $\frac{V_{\text{ch}}}{R_h+R_{\text{in}}}$ = $I_b$ = 100 $\mu$ A

Second case is $V_ {\text {out}}$ is 5V with $V_c$ decaying to its low value $V_ {\text {cL}}$. U3-3 sense point in this case is set by the voltage divider of $R_ {\text {pu}}$ + $R_h$ and $R_ {\text {in}}$. Since LBO controls boost turn off, the only use of the second case equation is to help determine hysteresis resistors.

$\frac{V_{\text{cL}} R_h+V_{\text{cL}} R_{\text{pu}}+R_{\text{in}} V_{\text{out}}}{R_h+R_{\text{in}}+R_{\text{pu}}}$ = $V_{\text{ref}}$

The three equations can be solved to find values for the hysteresis resistors. The set of values shown in the schematic result in $V_c$ topping out at 4.5V.

Edit:

It should also be mentioned that straight boosts like these don't isolate the input source from the load. There is always a leakage path through the inductor and diode. It's not clear what your load is like, but if there is no separate isolating element between the super caps and load, a load switch may need to be added to prevent leakage during charge up.

Edit:

As to the load switch, it could be controlled by the hysteresis comparator (U3). Something like this could work:

When U3-1 goes high to turn on the boost it will also connect the load by turning on Q2 (Si1304B) and Q1 (NTR4101P). And of course, load is disconnected when LBO and U3-1 go low. An N channel FET is used for Q2, instead of a BJT, to keep bias current low. Zener D3 (BZT52C5V6T) is there because a boost at light load can peak detect and over voltage the load. It may not be needed.

• Thank you for the informative answer, this is exactly what I needed. :) I appreciate that you explained how the circuit works, including equations, thank you. Unfortunately, I am new to stack exchange, and don't quite have enough reputation to upvote, but as soon as I get enough, I will upvote this answer. – Zoidsfan77 Mar 26 '14 at 12:09
• @Zoidsfan77 please see edit for info about boost leakage, that may apply here. – gsills Mar 26 '14 at 16:43
• Thank you for the edit, I would have very easily overlooked that, it would indeed be a problem without a switch. – Zoidsfan77 Mar 27 '14 at 0:00

Seems like you could use just a simple Schmitt trigger which could control MOSFET's for charging and discharging like this:

simulate this circuit – Schematic created using CircuitLab You can simulate it (Time domain 0-60s, 1s step).

Problem with this solution would be that it requires an independent power supply (You could alter the schematic to work with say 2 lithium coin cells (~6V)) for the opamp, although it shouldn't be too power consuming if you choose the right opamp (it'll leech off Ucap/16k from the cap and the rated supply current from the autonomous power supply). You could also increase both resistor values by a factor of 10, to get even less current draw from the cap, but going higher might get sketchy.

Whether it's feasible depends on your current draw though.

Hope it helps or gives material for new ideas.

• I will upvote as soon as I have enough rep, I am new to stack exchange so I apologize. – Zoidsfan77 Mar 24 '14 at 12:27
• Oh. No worries. :) – Linards Mar 24 '14 at 16:28

In the absence of the boost regulator circuit the OP used I have picked one to show how this might work: -

I've chosen this circuit because it has an application with supercaps - note the two 50 farad caps in series to the left and note that this boost chip will work from 0.5V to 5V. Note also that it has a shutdown pin (SD) and ideally this would be an input that has a tight spec on its threshold but unfortunately it doesn't and this slightly complicates things. 1.6V (or above) on this pin turns the switcher on and 0.25V (or below) turns it off.

The trick is to ensure that as the supercap voltage got to about 5V, the SD pin acquired a voltage of 1.6V. This would then turn on the booster and its output (via a resistor) could lift the SD significantly higher. This ensures that as the supercap voltage dropped, it would have to drop significantly below 1.6V before the switcher turned off.

That's the idea and this could be achieved by 3 resistors but it will be device specific because the real levels at which SD turned on and off are unknown between 1.6V and 0.25V. If it were a one-off design I'd suggest using a pot on the switcher to see where its shut-down and power-up points were and then pick resistors. If it's a production job then use a comparator as suggested by Linards in his answer BUT the comparator output would feed the SD pin.