Hello, Uh, this feels a bit wrong really. but uh, needs must So let's drill this out and there. Okay, let's fix the volt meters to the Dad's Office sign. Okay, the first one.

Yeah, I Think that's gonna work fine. So temporarily I've set up these two self-powered voltmeters which run down to half a volt which is quite handy because I've set them up on this super capacitor bank and I'm putting 500 milliamps 500 milliamps into this super capacitor. Bank But it's only gone up to 2.4 volts and now it's kind of sitting there and won't go any higher. and you can see that the 2.4 volts is split equally between the two capacitors 1.2 volts each.

Now, if you remember from my post bag video, this super oh, these are quite warm. Actually, this super capacitor module turned up with a chip missing. There it is. it's missing from that position there.

The chip is in place on the other side, but I've actually lifted a leg on it so that it is effectively out of circuit. Um, because I wanted to just show both sides what the circuit does if that chip isn't present now. I Set up an AliExpress dispute resolution thing and the seller gave me a partial discount for that missing chip. So they're happy I'm happy and we now move on.

So what is actually happening in this circuit with that chip missing? That gives rise to this 2.4 volts upper limit: 1.2 volts per capacitor. Now remember, these are 2.7 volt capacitors, so they should go a lot higher. And when they get up to their maximum voltage, this Led is supposed to turn on. which means that the protection circuit is on and the capacitor is being discharged through those four R7 resistors and this mosfet here.

Well, the fact is, without that voltage oops voltage detector chip there and with this one having its leg lifted in this corner here, Um, these mosfets are ever so slightly turned on. And that means that some current is flowing through these four R7 resistors. and we've reached an equilibrium where the 500 millivolts of incoming current is being sorry. 500 milliamps of incurring current is being offset by current flowing through these resistors.

Now at the full 2.7 or let's call it 2.5 volts, these resistors, which are about 2.5 ohms would dissipate one amp or allow one amp to flow so it could offset an amp of incoming current. Here, it's offsetting 500 milliamps of incoming current at 2.4 volts across the whole capacitor. 1.2 volts per individual capacitor. So let's draw the schematic.

We can see that the two resistors are connected to the positive capacitor terminal, so I'll draw these two resistors here. they're in parallel. Like So then we have this mosfet here. Now it's marked as I think it's 2N 4.

Now if you do a search on Google for 2N for Sot 23 and actually this is one of the results. It's this Mosfet N Channel 30 volts 2.1 amps. Nowhere on this data sheet does it actually say the marking is 2f4. not that I can see anyway.

Uh, there's this specific device code, but it doesn't say 2n4. But I Do think this is a pretty good candidate for the mosfet that's on there. and what makes me think that is in the characteristics. The on characteristics we have here that uh V GS the the gate threshold voltage, voltage gate Source threshold in bracket.

There's a minimum of one volt typically 1.7 volts and a maximum of 2.4 volts. Uh, trying to remember how to draw? Oh, why is it not focused on that? uh A N Channel Mosfair. Actually I could look at that date sheet can't I Yes. So in here we have the arrow and then we've got the body diode running up there.

Okay, that'll do so. that's the N-channel mosfet. Um, the uh source of that. This is the source is tied to zero volts or the negative of the capacitor.

So I should draw the capacitor in. Really, these are the two big resistors. so it's for R7 for R7 Um, about two and a half ohms with those in parallel. And then the Gate of this mosfet which we draw as an insulated gate like so um is pulled High by a resistor.

Let's take a look at that. Yeah, so the gate is this connection here. it's pulled High through that resistor which I'll have to get a magnifying glass to have a look at. Um held down by the output pin on this voltage detector chip.

But that's the pin that I've lifted. so that's not doing anything at the moment. So actually all that's happening is the Gate of the mosfet is pulled right up to positive through that resistor. and it's a 5 101.

So it's a 5k one. I Think isn't it Five One? Oh Oh yeah. Five K one up to positive of the capacitor and that goes to the gate of the mosfet. So you can see here that when the capacitor voltage reaches a point where the voltage on this gate and that resistor is fairly irrelevant.

Actually, you could probably tie that directly up to positive positive negative. When it reaches the gate threshold of this mosfet, this mosfet turns on and a current will flow down through these resistors, down through the mosfet to the negative of the capacitor and the voltage at which this happens will be when the current flowing into this capacitor reaches an equilibrium with the current flowing down here and they're balanced out. The voltage can thus go no higher and it is at 2.4 volts. Um, 1.2 volts per capacitor.

If you can see that I took my light away. I'll bring it back. Yes, this voltage here. 1.21 volts per capacitor.

Now there's also a resistor coming down from positive. I'll extend that out. a resistor here that has an LED on it. Let's put some arrows on there and that goes to this point here, which is the drain of the mosfet.

So when this mosfet turns on, it also allows a current path down here through this red LED and down to ground. Now the problem is at the moment, with only Uh 1.2 volts across this capacitor that simply isn't enough voltage to illuminate that Led, so it's not illuminated. So now what happens if I increase the current here? We've got 500 milliamps, so let's take that up to one amp, a nine, one amp. That's why these buttons wear out.

Okay So we've got one amp, four volts. Let's turn that on. The voltage on these capacitors will now rise further and I'll wait until they get to their equilibrium point because of course we've got more current coming into the capacitors. So what will happen now is the voltage on the capacitors will rise up a little bit higher.

Turn this mosfet on a little bit more fully. More current will flow down here through these four R7s and we'll reach another equilibrium point, but at a higher capacitor voltage. I've just noticed that the LED resistor is one double one, so that's one K Ohms. So not a huge amount of current going through that Led, but they're very efficient these days, so typically you'll find quite a high value resistor on LEDs So now we've got 3.4 volts across the pair of capacitors that's oh, I'll slightly move the camera 1.7 volts on each of them.

oh, and it is visible here. You can see that with that additional voltage across the big super capacitor which I've drawn there, there is now enough voltage across this led to allow it to conduct and the current's flowing. So you can see that this mosfet is on partially or fully, not quite fully probably and you can see the red. LEDs here are lit and I can tell you that these resistors are now really quite hot because they're dissipating a fair amount of current.

You can work it out if you want. It's a 1.75 volts across whatever two four R7s are in parallel. that's the amount of and then multiply volts by amps to get watts and that's how much power there is being dissipated in those resistors minus the on resistance of this fet of course. And I can see from the data sheet that the on resistance of this set is going to be something like a hundred milliohms so it's quite small in comparison to these resistors.