Now, of course, this is not how this circuit is supposed to work. Um, because there's an extra bit, which is this little voltage detector I see and a potential divider here. There's also a little capacitor in there for stabilization, but yeah, a potential divider between positive and negative and a voltage detector I see looking at the midpoint of that potential divider. But let's take a look at the part number of this voltage detector I see and I don't know whether you can see this I've got to get the light in the right place.

Um, it says on it CNP 3 I Think that is right? Well, we're burning power unnecessarily in these resistors. so I'll wind the current of this thing back down to about 500 milliamps where it was a little happier and we'll go to the PC and we'll take a look at that Cnp3 thing. So we look up Cmp3 sot 23 and we get this: Reddit Uh article I Bought some super capacitors on the balancer. There is a yes, there is a component missing U1 and the component number is Cnp3.

Well, there's there's a thing. There's a surprise: the Cnp3 component is missing on his dual supercapacitor module. Does someone maybe know and uh, the very helpful Lab 9049 says it's an Ln 61c voltage detector from Nat Linear N specifies its output configuration is open drain. Uh, the P specifies a detection voltage of 2.2 volts and there's a link to the data sheet.

so there's an ultra small package. High Precision Voltage detector. Um, here we've got C means it's an Ln61c N means it's open drain with a voltage range of 0.1 volts to 3 volts. uh P in as the middle letter would be a higher voltage range and then the p is this 2.2 volts.

So they're using a 2.2 volt voltage detector um to protect a 2.7 volt super capacitor. And when you read more about this thing, you realize that its main use is to detect that VCC is falling and signal to a microcontroller. So here microcomputer reset circuits signal to the microcomputer that VCC is falling and that the micro computers should either reset or respond in some way. Perhaps go into a subroutine to back some data up, or do something because the powers disappearing.

That's it's intended or that was its design purpose. So if you look on this table here, electrical characteristics, you've got the detection voltage. Now Remember this is a 2.2 volt device. It can be a minimum of 0.98 times that voltage.

or it can be the detection voltage in itself. or it can be as high as 1.02 times that voltage. So let's assume that it's bang on accurate and it is the detection voltage of 2.2 volts. But there's also the hysteresis voltage.

now. this thing is designed to trigger to change its output State when the voltage is falling. so it will trigger at 2.2 volts when falling. But the hysteresis means that um on its Rising Edge It's going to be higher than that detection voltage because you don't want it to Jitter at 2.2 volts and the higher voltage is typically Vdf times 0.05 Um, minimum is 0.02 and the maximum is 0.08.

But let's assume this is absolutely designed to Perfection and the hysteresis is Vdf times 0.05 So adding to the schematic now we've got this voltage detection device. Let's draw it here. Um, it's hooked up to no it, it's VCC. Yeah, it's hooked up to ground there, but it's VCC is on a potential divider which I'll draw here that's also or across the super cap.

Oops, that's slightly off camera that goes to the top of this device and then this device is hooked up. Do a little jump there and a little jump there to the gate of the mosfet. So the idea is that as the voltage on the supercapacitor Rises When this voltage detector device sees 2.2 volts at this midpoint, it switches on it's open drain. remember, so it pulls its output low, which turns off this device.

this mosfet now Rebecca Pardon, it's the other way around. This device is designed when the voltage at this point drops to 2.2 volts that it pulls low on its output. so it will be pulling low on its output when there's a low voltage here. thus holding this mosfet off.

When this point here reaches 2.2 volts, which we assume is when the capacitor voltage is 2.7 volts, this will stop pulling low on its open drain output and allow this mosfet's gate to pull High via the 5k1 resistor turning on and the therefore allowing an amp of current to flow down here, protecting the super capacitor from right, it's voltage Rising Any higher, as long as the current coming into the Super capacitor is no more than one amp and of course, at the same time, this Led will turn on. So now we need to know what these resistors are in the potential divider. Well, there's also a capacity here. actually.

where's the capacitor? It's across the lower resistor. That's only for stabilization. So these are 1k8 and 6K 8. So we can work out what the voltage is.

Say, we have 2.7 volts and these wires are right in the way. Aren't they 2.7 volts across this super capacitor. We can work out what this midpoint voltage is here. So it will be the 68 part divided by the addition of these two which is 8k6.

Is it So 86 times 2.7 volts? That will be the voltage at this potential divider point. So that is 2.13 Now the problem there is that that's not 2.2 So we'd actually need a higher voltage across the capacitor up here to get 2.2 at this potential divider midpoint. Let's work out what that voltage is. Yeah, the calculation is 86 over 68 times 2.2 volts, which is 2.78 So it's nearly 2.8 volts that that capacitor has to get up to to put 2.2 volts on this.

Ln 61 voltage detector chip. Now here's the thing. Remember this voltage detector chip is designed to see a voltage falling and to trigger at the threshold voltage which is 2.2 volts when it's coming up the other way, which is not what this is intended to detect. There is hysteresis and it was what was it? 1.05 times the threshold voltage.

So assuming the chip is all accurate, that's actually 2.31 volts when you include the hysteresis and that gives you a super capacitor voltage at which this thing will detect the trigger voltage, but with the hysteresis added in of and that is 2.92 volts. So assuming this chip is absolutely accurate, um, it will actually allow the mosfet to pull current down through these resistors. which means that its open drain output turns off at 2.92 volts. that's really too high for a 2.7 volt super capacitor.

I have just bought actually from AliExpress some 2.85 volt super capacitors, but even those really shouldn't have 2.92 volts across them.