Let's Make Robots!

Putting voltage through a photo cell and to motors?

While I wait to get some real parts, I've been tinkering around a bit. I made sure that one of my photocells could endure 9 volts without exploding (out of voltage regulators), and connected it to a buzzer and battery to see what would happen. It worked pretty well, and I found that the more light the photocell took in the more power the buzzer got (and thus the louder it got). However, when I replaced the buzzer with a small motor I got nothing. I tried a few other motors and still wound up with zip. How to I make the motor spin faster as more light is taken into the photocell?

 

Thanks. 

Comment viewing options

Select your preferred way to display the comments and click "Save settings" to activate your changes.

Just an FYI, in case you're as anal-retentive as I.

Voltage doesn't go "through" something - current does. Voltage goes "across"

Spam?
Heh, yeah, the circuit drawing software I'm using was a freebie. It has a decent library of component icons, but the voltage source and ground icons are a bit different from what I'm used to. Fortunately LTspice lets you modify and draw your own icons, I've just gotta get around to doing it...

Thank you very much for the schematic and the great explanation. I'm not very good at reading schemas, but I'll do my best.

 

Thanks again, that was very helpful.

 

1 quick question. What is the part at the bottom right of the schematic that is in between Q1 and D1?

 

Thanks 

The arrow pointing down? That should be ground (earth, 0V, common)
AKA the negative terminal of your battery.

A photocell is just a resistor that becomes less resistive as more light hits it. The buzzer you were using probably draws a very small amount of current from your 9V source, which is the case for piezo-type (little golden disc) speaker units.

By applying the V=I/R rule to the photocell we can see that as long as the resistance isn't huge, a small current causes only a small voltage drop across the photocell, which means there is enough voltage left over to run the buzzer. If R (the photocell resistance) gets smaller because more light hits it, then the photocell will eat up even less voltage, making the buzzer louder.

DC motors on the other hand have very low resistance, especially when they're not already running, which means they consume a lot of current. In this case the voltage drop across the photocell will be much larger than before which effectively starves the motor of voltage, so much so that it won't even turn at all.

A more versatile way to control a motor with a photocell is by using a transistor as a current amplifier:Photocell controlled motor

Q1 is any regular NPN transistor that can handle enough current to drive the motor. D1 is a diode that protects the transistor from back EMF should the motor be stopped suddenly. R1 is there to limit the amount of current going into the base of Q1 when the photocell resistance is at a minimum, but it might not be needed if the photocell resistance is always enough.

Current flows into Q1's base via R1 and the photocell. More light on the photocell allows more current to flow into Q1. The current flowing through the other half of the transistor, and thus the motor, is a magnified version of the base current. So, to sum up, the more light on the photocell the more current the motor gets fed, but without the drawbacks of having a high current going through the photocell and wasting all the power that you want going to the motor.