I made a simple circuit to take current from my dynamo to charge batteries and power lights in parallel. I'm having a problem that doesn't make any sense to me. For simple wires (not semi-conductors) I would expect that if A and B are connected (<2 ohms) and B and C are connected (<2 ohms) then A must be connected to C. In my situation A does not register as connected to C according to my multimeter. I'm really confused as to how this could be.

I made a schematic which shows where the points A, B and C are. The battery is actually disconnected at point C so it's voltage is not relevant.

 

When you measuring the wiring, the negative of 3.6V should be disconnected and the 6Vac should be power off, so the resistance of A, B, C should be very low as less than 50 mΩ.

 

When you measuring the wiring, the negative of 3.6V should be disconnected and the 6Vac should be power off, so the resistance of A, B, C should be very low as less than 50 mΩ.

The 6 Vac is actually a dynamo but it's only on if I spin the wheel (which I don't while testing continuity). Are you sure the negative battery terminal needs to be disconnected? I'm going to have to remove heatshrink and desolder it if that's the case. How does it interfere with the measurement?

 

The 6 Vac is actually a dynamo but it's only on if I spin the wheel (which I don't while testing continuity). Are you sure the negative battery terminal needs to be disconnected? I'm going to have to remove heatshrink and desolder it if that's the case. How does it interfere with the measurement?

As long as one end of the battery is disconnected that will be fine.
Actually, if the generator is stopped and the lights are off, both ends of the battery could stay connected without affecting the continuity reading.

 

As long as one end of the battery is disconnected that will be fine.
Actually, if the generator is stopped and the lights are off, both ends of the battery could stay connected without affecting the continuity reading.

I can only make the lights stay off if I disconnect the battery because the switch is closed for the continuity test. If I left the battery connected then I would have to open the switch to keep the lights off but then there would be no point in testing the continuity across the switch!

 

I can only make the lights stay off if I disconnect the battery because the switch is closed for the continuity test. If I left the battery connected then I would have to open the switch to keep the lights off but then there would be no point in testing the continuity across the switch!

OK, so one end of the battery should be disconnected, it doesn't matter which end.

 

OK, so one end of the battery should be disconnected, it doesn't matter which end.

Okay, that's fine.

 

Are you sure the negative battery terminal needs to be disconnected?

That is to avoid any accidently to damage the ee parts.
How is the led, does the led is some kind of package has resistor inside?
Normally the led should be in series with a current limits resistor.

 

That is to avoid any accidently to damage the ee parts.
How is the led, does the led is some kind of package has resistor inside?
Normally the led should be in series with a current limits resistor.

If the positive of the battery is disconnected and the generator is stopped, what damage is possible while measuring continuity?

 

If the positive of the battery is disconnected and the generator is stopped, what damage is possible while measuring continuity?

When he want to measure the node A, B, C as a shorted circuit, so disconnects the negative is the better way.

 

I agree with the you, there should be less than 1 ohm resistance from the battery positive terminal to the + end of C3. Are you sure the switch is wired into the circuit correctly? Can you post a photo? Besides that, here are a few comments.

1. I'm not familiar with Eagle, but - D1, D2, and D4 do not have connection dots for both ends of each diode. This won't affect the actual circuit, but it is good to make the schematic reflect what you have built.

2. Do the XP LEDs have current limiting built in?

3. It looks like you are relying on the internal resistance of the generator to protect the battery from over-charging, or over-current during charging. This is not a safe plan.

ak