voltage dropping without stamp or stepper activated; can you look at my diagram?

Blake

Hi, I have a quick question. Here's my diagram (from tinycad, first time):


(where the X's are, wires are not joining, just to be clear)

Without my basic stamp powered on (power and ground wires removed) and with the isolating diode, battery and stepper removed from the circuit (by pulling one leg of the diode out of my breadboard, i get 13.75Volts after the regulator with my multimeter. If i connect the diode to the circuit, without even connecting the basic stamp, i get around 8-9 volts measured in the same place. Does this indicate my base pins on the tip120's are letting a little current through even without the stamp active? Or what else could be taking away my voltage? I just don't get it.

Any help would be awesome. Also I know i need reverse diodes on the transistors, but they are not even active at this point. The voltage drops without triggering the bases or turning on the stepper, so i don't understand.

Thanks,
Blake

Post Edited (Blake) : 3/10/2008 2:51:05 PM GMT

Phil Pilgrim (PhiPi)

I wouldn't expect enough leakage through the transistors to pull the voltage that low, since their bases are pulled low. With the diode in place, is the battery then connected again? If so, it may just be discharged enough to pull the voltage down on its own. What's the battery voltage with nothing connected to it?

-Phil

Blake

you're right the battery is at 7.6 volts. I guess i really need to make sure this circuit can charge that battery as well as set off the stepper. I've changed the circuit since I've started working on this project, so I think I'll charge the battery up and run it and see if the volts remain up at 13.75. Quick question: Is the battery drawing off that much and using it to charge? If so, if i ran the circuit long enough, would it build back up to the 12-13 volt range?

Thanks,
Blake

Post Edited (Blake) : 3/10/2008 3:21:53 AM GMT

Beau Schwabe

Are you sure that your schematic is correct?
The 5100 Ohm and the 510 Ohm resistor form a voltage divider with your 20Volt input.... the largest amount that your regulator would see in this configuration is only about 1.82V

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Beau Schwabe

IC Layout Engineer
Parallax, Inc.

Blake

you're right. thanks for pointing that out. the 510 ohm resistor should be between the ground and the Vout from the adjustable voltage regulator, not ground and Vin. And the 5100 ohm should be on the ground not the Vin. But, this is correct on my circuit, just wrong in my diagram. I fixed the diagram, refresh to see the new one.

Post Edited (Blake) : 3/10/2008 2:51:43 PM GMT

Beau Schwabe

Blake,

With the TIP120's make sure that the 22K resistor is between the B-E junction, and not the B-C junction.
Also.... make sure that pin 1 (Base)·and pin 3 (Emitter) on the transistor are not swapped.

Curious... Is the stepper energized? i.e. is it difficult to turn in the condition that you describe?

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Beau Schwabe

IC Layout Engineer
Parallax, Inc.

Post Edited (Beau Schwabe (Parallax)) : 3/10/2008 6:34:15 PM GMT

Blake

yes, stepper is energized and everything works. this circuit runs from a hand crank that turns a d/c motor with a toothed belt from an old scanner, generating 18-22 volts. I get a steady output of 13.25-13.75 volts which powers the bs1, the stepper, and charges the 12 volt cordless drill battery. The stepper moves freely and the generator is easy to keep going, not drag on the motor from the stepper's current draw.

Thanks,
Blake

http://www.blakefallconroy.com