Fast, high current switching

[(*steve*)]

One option is to read the battery voltage. I would assume that the battery voltage is higher when the engine is running.

Another option is to pick up the signal from a tachometer.

Also, it's good to hear you've got a programmed chip and a cool mosfet.

Remember that any incoming signal will need to pass through a circuit to protect the chip from voltage spikes.

The AC signal you're talking about might have a relatively high peak voltage, but low average voltage (I don't know)

I'll draw you a quick circuit so you can test that.

 

[dilbert9]

Thanks Steve! There is currently no tach connected to the bike so that is really not an option. I do have another option though. The voltage from the stator (AC) runs into the rectifier for the charging system. The only time the rectifier is outputting the charge voltage (up to if not higher than 15v) is when the engine is running. So I have the option to step that down and into the arduino for reading.

So I have another option if I need to. I just figured it would be easier to use the .05ish vAC over the 15ish VDC.

 

[(*steve*)]

Connect the signal source to the left, your meter to the right (voltage range) and this will allow you to see the maximum DC value.

The diode can be almost anything, the cap anything from 1uF to 10uF should be fine.

Connect that up to the pickup wire and see what you get.

 

[(*steve*)]

That didn't show for some reason.

 

[dilbert9]

Ok. I will connect this up tomorrow and see what I get. I have a bunch of 1n4007 diodes sitting around. I will use one of those. I will report back when I get the reading. It is 8:30 pm here is PA so it wont be until tomorrow sometime.

Quick question though. What happens to the electrolytic cap when the AC cycles through the negative voltage? Will that still be ok?

 

[(*steve*)]

The diode stops that from happening.

 

[dilbert9]

I have not had time to run out and test the pickup voltage, but I did setup the power supply circuit on the breadboard here. The problem I am having is I am putting 12v into the power circuit but am only getting 3v out. I have tested throughout the entire circuit. There is 10v going into the regulator but still only 3v coming out.

I followed the schematic we have above there.

 

[(*steve*)]

Measure the voltage between pins 1 & 2, and the voltage between pins 2 & 3, and between 1 & 3

The first one should be 7V or higher, the second one should be 5V (you're getting 3) and the last should be 2V or more.

Alsop measure the current the circuit is drawing.

I anything getting hot?

 

[dilbert9]

Pins 1 & 2 is 10.7v
Pins 2 & 3 is 3.1v
Pins 1 & 3 is 7.63v

The current is reading 10.5mA

The zener diode is getting pretty hot. I have it connected up just like the schematic. The cathode (black line) is connected after the inductor and the anode is going to ground.

 

[dilbert9]

I tried a different regulator and still the same results. I am unsure what is causing it to be only 3v. I removed the zener and still same result.

 

[kpatz]

If the zener is getting hot, it's shunting a lot of current. This means the zener's voltage is lower than your supply voltage. If the pin 1&2 voltage is 10.7V, that's probably a 10-11V zener diode. You want one that's higher than the peak output voltage of your supply voltage, since it only exists to protect from spikes/transients.

As for the low output voltage coming out of the regulator, you don't have pins 1 and 3 reversed, do you? Are you using a 7805 or something else?

 

[dilbert9]

The zener is a BZX85C18-ND (http://www.fairchildsemi.com/ds/BZ/BZX85C10.pdf) which is a 1W 18V zener. I am using a true 7805 regulator (NCP7805TGOS-ND -- http://www.onsemi.com/pub_link/Collateral/NCP7800-D.PDF ). I have it wired up as the datasheet says. Pin 1 is input, pin 2 is ground and pin 3 is output. Same as the schematic above. I even rewired it completely with the same result.