Experiment 4 Part 1 (Chapter 4) -- Fun With The Phototransistor
I'm now up to Experiment 4, which actually consists of a few simple circuits you'll need to wire up. The chapter starts out with a very simple circuit -- a phototransistor and a 3.3k resistor. Doesn't get much easier than that to wire up!
This little circuit is perfect for helping understand voltage division. In the first test, the phototransistor's Collector is connected straight to 5V (regulated). The 3.3k resistor connects the PT's Emitter to Ground. We know that light lowers the internal resistance of the PT, right? Lower resistance means more current... and a higher voltage reading with respect to Ground. (V=IR, here... an increase in I gets you an increase in V if R is held constant.)
If using a multimeter to take a Voltage reading with respect to ground using the circuit from Figure 4.1, it's expected, then, that the 3.3k resistor will "feel" most of that 5V as light shines on the PT. Here's the first of two videos that shows this in action:
But what happens if we change the circuit so that it matches Figure 4.2? There, the 3.3k resistor is connected directly to the 5V. The PT's Base, registering bright light or darkness, is going to control the resistance "felt" between the two points being measured. From page 13, "An increase in the base current causes a reduction in the effective internal resistance of the transistor. This is why current flowing through the transistor increases." Read that as "as the light to the Base increases, resistance inside the transistor decreases."
Once again referencing page 13, "... when you have two resistances in series, they divide the voltage drop between them, depending on their resistance relative to each other." Read that as "with the 3.3k resistor and the PT in series, they will divide the 5V in proportion to their respective resistance values."
Formula from Page 13: Vm = Vcc * (R2 / (R1 + R2)) where Vcc = 5V, and R1 = PT, R2 = 3.3k
Stay with me... As R1 decreases (approaches 0) you're basically multiplying Vcc (5V) by 1 (R2/R2) -- the 3.3k resistor gets almost all of the 5V voltage as the PT registers brighter and brighter light. Sorry for the math.
BTW, the equation holds true for the Figure 4.1 circuit, just flip it around -- R1 = 3.3k, R2 = PT. As R2 drops in bright light and approaches 0, Vm drops closer and closer to 0 as well. It won't reach 0, but as you saw in the first video, brighter light causes a decrease in voltage at Vm. You're basically taking a number that's decreasing -- R2 -- and dividing it by R1 that stays constant, resulting in a smaller and smaller number multipled by Vcc. See the video below:
Take away from this? Voltage division is an interesting animal! Perform these two tests yourself and see it in action. I think you'll find that the concept of voltage division will begin to clear up. (And you'll be able to add the PT component to your arsenal of cool components to find new and interesting uses for!)
My Experiment 4 Part 2 post involves the 555 Timer chip... will try to tackle that before Friday.
This little circuit is perfect for helping understand voltage division. In the first test, the phototransistor's Collector is connected straight to 5V (regulated). The 3.3k resistor connects the PT's Emitter to Ground. We know that light lowers the internal resistance of the PT, right? Lower resistance means more current... and a higher voltage reading with respect to Ground. (V=IR, here... an increase in I gets you an increase in V if R is held constant.)
If using a multimeter to take a Voltage reading with respect to ground using the circuit from Figure 4.1, it's expected, then, that the 3.3k resistor will "feel" most of that 5V as light shines on the PT. Here's the first of two videos that shows this in action:
But what happens if we change the circuit so that it matches Figure 4.2? There, the 3.3k resistor is connected directly to the 5V. The PT's Base, registering bright light or darkness, is going to control the resistance "felt" between the two points being measured. From page 13, "An increase in the base current causes a reduction in the effective internal resistance of the transistor. This is why current flowing through the transistor increases." Read that as "as the light to the Base increases, resistance inside the transistor decreases."
Once again referencing page 13, "... when you have two resistances in series, they divide the voltage drop between them, depending on their resistance relative to each other." Read that as "with the 3.3k resistor and the PT in series, they will divide the 5V in proportion to their respective resistance values."
Formula from Page 13: Vm = Vcc * (R2 / (R1 + R2)) where Vcc = 5V, and R1 = PT, R2 = 3.3k
Stay with me... As R1 decreases (approaches 0) you're basically multiplying Vcc (5V) by 1 (R2/R2) -- the 3.3k resistor gets almost all of the 5V voltage as the PT registers brighter and brighter light. Sorry for the math.
BTW, the equation holds true for the Figure 4.1 circuit, just flip it around -- R1 = 3.3k, R2 = PT. As R2 drops in bright light and approaches 0, Vm drops closer and closer to 0 as well. It won't reach 0, but as you saw in the first video, brighter light causes a decrease in voltage at Vm. You're basically taking a number that's decreasing -- R2 -- and dividing it by R1 that stays constant, resulting in a smaller and smaller number multipled by Vcc. See the video below:
Take away from this? Voltage division is an interesting animal! Perform these two tests yourself and see it in action. I think you'll find that the concept of voltage division will begin to clear up. (And you'll be able to add the PT component to your arsenal of cool components to find new and interesting uses for!)
My Experiment 4 Part 2 post involves the 555 Timer chip... will try to tackle that before Friday.
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