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Saturday, September 13, 2014

I mapped out the circuit today. I discovered lots of free online circuit drawers, which was cool. I used the digikey one this time.

Stock Circuit. Note: there are two crossovers that look like 4-ways.
The only 4-ways that are actually 4-ways have dots.
For comparison. I have L1 and L2 flipped on my circuit diagram.
I attempted to lay it out in a similar manner to how the components are on the board (single layer PCB). I also attempted to understand what was going on. I am no EE though, so please correct me if I got it wrong. Starting from the red "+AC" line (left), there is a 150Ohm resistor, followed by a 250V film cap in parallel with a 1MOhm resistor (hidden under cap), followed by a shottky diode (forward) and a zener diode (reversed). These diodes are copied on the -AC line side. I think those components all make up the AC to DC converter, guessing 120VAC to approx. 5VDC. The DC side of the shottky diodes are linked and the DC side of the zener diodes are linked, giving the +5V and 0V references respectively. Following the +5V reference through a 33Ohm resistor (now about 3V), we get to the LED that makes the ball glow (the one labeled L1 on the board and L2 in my diagram). In series with that is a 1Ohm resistor and the projector LED, which then terminates into the 0V reference.

Now for the control part: Photoresistor resistance decreases with light and increases in dark. So if the room is lit up, power flows from the 5v reference node to through the low-resistance photoresistor, into the base of the transistor, which means it is on. I'm guessing the 20KOhm resistor in series to 0V with the photoresistor (with the transistor base tapped off between them) acts as a discharger or voltage divider, but I'm unsure. With the transistor on, the path that branches off between the 33Ohm resistor and the L2 (schematic) diode is now a low resistance path to ground, so all of the power (~0.25W?) goes that way instead of through the LEDs. If it is dark, the photoresistor resistance is high, causing the transistor to be off, preventing power to flow through it, allowing the LEDs to be powered. It's probably a fairly efficient circuit when it is on, but when it is off it is constantly bleeding some power (guessing about 0.25W) through the 33Ohm resistor and the transistor. Probably a good idea to unplug these when not in use. (see next post for update on this).

Modification plans: I want to add an option to power this using a Joule Thief circuit.

The "JT" voltage source represents the Joule Thief circuit. The switch will allow me to switch between AC and DC mode. It should be fairly safe this way. I'll probably put a diode in the JT circuit to prevent power from flowing across it in case I plug it in with the switch on. The good thing about this setup is that the photoresistor should still function normally in DC mode. However, the power drain when the transistor is on is may kill the already partially dead AA's I'll be using for the JT circuit in a single day. We'll see. If that ends up being a problem, I could probably cut a trace somewhere and install another switch.

Challenges moving forward:

  • I wasn't able to get exact values for many of the components, so I will have to use a scope/nice multimeter to figure out what voltages are where in the circuit. I'm not sure how I'll power it since everything will be exposed...I need some sort of low current 120VAC source.
  • Knowing the voltage difference across the LED's will give me the JT design voltage.
  • I want to figure out how much power the stock unit draws when plugged in (off and on).
  • Designing the JT circuit (lots of internet tutorials).

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