Sparkfun Weevil Eyes

weevil

Yesterday I ran the Learn to Solder workshop at the Tech Museum of Innovation.  We put together the light-sensitive Weevil Eye kits.  It was great fun, we had wonderful participants!  A few participants had questions about the circuit, which I will answer here. Sparkfun provides a circuit diagram on the product website. Here I will write out an explanation of the components.

Schematic

The LEDs light up when the photocell or light dependent resistor (LDR) at the bottom of the bug is covered. The LDR is at maximum resistance in the dark, and the resistance lowers when it receives light.

Why is there more current flowing through the LEDs in the dark? Because current only flows in that part of the circuit when the transistor gets a certain minimum voltage (at least 0.6V) and current. And more electricity goes to the transistor when the photocell resistance is high. So the darker it is, the more electricity flows through the transistor and so through the LEDs.

Why does more electricity go to the transistor base in the dark? Because the LDR and the 47kOhm resistor form a voltage divider whose output goes to the transistor.

What is a voltage divider? The simplest metaphor to describe a voltage divider is like a hose with a hole. Imagine water pouring down a hose with a hole in it. If the hose is empty, water flows freely, with a tiny trickle lost to the hole. But if the hose is pinched, the easier path is now out the hole. So when the resistance in the lower part of the voltage divider goes higher, more voltage is diverted out the hole instead of going directly to ground.

I wondered how bright the Weevil Eyes could possibly get in full dark – the LEDs are capable of glowing brighter. Most 5mm LEDs like these can handle around 20 mA. I measured total current in the circuit at full dark as 3 mA. I took a guess that the cadmium-sulfide photocell runs between about 10kOhm (light) and 100kOhm (dark), then ran a few numbers. The Weevil Eye is powered by a 3V battery so the voltage divider output at R1 = 47k and R2 = 100k should produce 3V*(100k/147k) = ~2V. 2V/147kOhm ~ 20 uA. How does that turn into 2 mA? The 2N3904 transistor has an amplification factor or “beta” (hfe) of about 100 under those conditions. So 100*20uA = 2 mA which is about what I found. So that is why they have a nice gentle glow in the dark instead of burning your retinas!

*If you assume 10kOhm is the lowest resistance, that leaves a forward bias of Vout = 0.68V, which is just over the bare minimum that could turn on the transistor – but if it is any lower, or there are any other fluctuations in the circuit, it falls short and the transistor stays off. Which is what we want! But I will measure one of their photocells later to see the real range.

**Wheee!

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