ante

I just remembered ::) that they are called magnetic reed switch or reed relay. And there is one of these inside the coil:

Hi Guys, I am fairly sure this is a relay. The kind with a glass tube in the middle!

You can

Hi Audioguru, This is not entirely true, I have charged many batterys in this way and never caused any damage. If you just do it right there is no problems as I see it. For example to charge a car battery, set the voltage to 13.75Volts then connect the battery and set any current up to 5% of the capacity. For a 60Ah battery this will be 3A, and leave it. There will be no destruction or gassing or explosion, what will happen is the voltage is now lower (current foldback) than the 13.75V set but the current will stay at 3A. Slowly the voltage will rise to 13.75V but not more, then the current will start to fall. Gassing will only occur in a lead acid cell if the voltage exceeds 2.38Volts (14.28V) so there is no gassing here. The voltage set at the start is the maximum for the charge that

Hi Dazza, It

Hi Kain, Just make wire connections for R7 on the pcb and bolt the resistor to the chassis instead. Some of these resistors can only take one third of the wattage without any heatsink. At 5A the resistor will dissipate almost 7W and you don

Hi Rhonn, I think it will work fine with this transformer if it

John, I tried to see it but all I got was this:

You have to explain what you mean, it

Audioguru, You know the difference between the 486 and a Pentium 4 is enormous! It

Hi Audioguru, Amazing how cheap computers are now compared to the early 90:s. A 386 with 20Mb HD and 256kb memory was almost $4000 US and now the latest with extra everything is less than $800 US.

Hi John, Have you considered this circuit http://www.electronics-lab.com/projects/sensors/004/index.html By changing the 270k resistor to a potentiometer you can vary the size of the

Yea, or a wise ass! ;D ;D ;D

Hi Dazza, I can

Dazza, I think this is a very commendable thing you want to accomplish. However the shipping costs might be the biggest obstacle to overcome. Or perhaps if you let the receiver pay for the shipping it will work out fine.

Hi Yevgeni, Why don

Hi Audioguru, Maybe it

Well, I couldn

Sasha, Remember if you go for IR (infra red) there must be a clear path between transmitter and receiver (no obstacles) for it to work properly.

Matthew, Ok, the 17.5Volts reading is on the battery connectors I assume? If so it means one of two possible reasons, first a faulty battery (most likely) and second an enormous charge current. Have you measured the current (ADC) going to the battery? Is the motor running nicely, no misfiring or uneven rpm:s at the presence of the 17.5V at the battery? The statement

Hi Stuee, Yes it will work, but you don

Hi Matthew, Welcome to this forum. I am not following you here, is it possible for you to post a schematic of your circuit?

OK, this comes down to 10Watts and I

Hi Tubby, How much current (mA) do you need?

Why not, a PWM in slow motion can do this?