Thread: Your Basic Coil
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Old 02-15-2016, 04:05 PM
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wantomake wantomake is offline
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Too many questions

Originally Posted by gyula View Post
Hi Wantomake,

Okay on the 17.36 V battery and 16.56 V cap bank voltages, the difference is 0.8 V. On leaking current via the LED I mean the following: the cap bank from its 16.56 V level slowly self discharges when left alone (disconnected) and the battery is is able to supply this current via the LED when the cap bank is in the circuit (this is just some microAmper maximum) while the voltage drop in the forward direction across the LED is just 0.8 V, and this small forward bias is normally not enough to emit visible light. This tiny microAmper is what I meant on leaking current via the LED to supply the self discharging loss of the cap bank.

When you discharge the cap bank, the 0.8 V difference invariably increases because the cap bank voltage reduces as your voltmeter showed, this must be the 3 - 5 V or so you measured in the process. (The battery voltage nearly remains the same.) The difference increases as the cap bank loses charge due to the motor or other load and the difference decreases when the load is removed from the cap bank and the battery starts recharging it.

So what do you mean by extra voltage? If you mean the recovery of the unplugged batteries (some ten to some hundred mV) I think that is normal for both the rechargable and the alkaline batteries, depending mainly on their age / usage. The capacitor bank is also able to recover some hundred mV when they were charged up earlier but got discharged from say 16.5 to 14 V and you disconnect them completely (the dielectric material in the capacitors has a 'memory' effect, they have been stressed by a higher voltage and then this voltage disappeared by the discharge). When the cap bank is included in the circuit, the batteries are able to charge them up via the LED, starting from the 3 to 5 V difference till this difference settles at 0.8V and the LED becomes dark. If you still have questions, ask.

What I mean by extra voltage is this.

Why aren't the batteries being slowly drained by the load of the led?

I understand the 'memory' of the batteries and the dialectic of the caps. But if connected in series a battery, led, and small resistor would it not discharge the battery? I thought it would? So the resistance of the combined coil and resistor plus the slow leakage of the led will keep the battery and caps from discharging??

So the led turns off after the caps are charged up to max memory and the batteries nor the caps will never drain or loose charge??

Thanks for the help,

Last edited by wantomake; 02-15-2016 at 04:21 PM.
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