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Andrew Kesterson
2026-06-04 18:53:30 -04:00
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05-photoresistors/README.md
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@@ -44,7 +44,7 @@ Okay, hear me out. Consider this circuit, wired up on a breadboard:
+5V --> 10k Resistor --> Wire --> GND
```
On a 5v current, a 10k resistor should give us an output of about 2.5v, right? Right.
Our intuitio here says that, on a 5v current, a 10k resistor should give us an output of about 2.5v, right? Right, let's go with that.
* Put your voltmeter red probe on the +5V rail side of the resistor, and the black probe on GND. You will see 5v.
* Put your voltmeter red probe on the downstream side of the resistor, away from the +5V rail. Put your black probe on GND. What do you think you will see? 2.5v? No. You will see 0v.
@@ -70,7 +70,7 @@ A voltage divider does exactly what it says: it creates a point in the circuit a
Now take your voltmeter and put the black probe on ground. Put the red probe on the wire between the two 10k resistors. What do you read? 2.5 volts!
Because the 2 10k resistors are wired in series, there is a potential difference between the upstream (supply) leg of R1 to the upstream (supply) of R2. The total potential difference measured across the entire circuit remains constant - tapping the supply side of R1 and the ground side of R2 would show a total 5v difference. Since R1 is no longer the only resistance in the circuit, current continues to flow past R1 towards the potential difference of R2 downstream. This allows us to place a tap (the bare wire) in between R1 and R2, where we can detect current passing by. And by placing another 10k resistor in as R2, we can detect the potential difference between the tap (at the junction between R1 and R2) and the GND. And because R2 is a 10k resistor, we get a reading of 2.5v.
Because the 2 10k resistors are wired in series, there is a potential difference between the upstream (supply) leg of R1 to the upstream (supply) of R2. The total potential difference measured across the entire circuit remains constant - tapping the supply side of R1 and the ground side of R2 would show a total 5v difference. Since R1 is no longer physically connected directly to ground and instead leads to another component, current continues to flow past R1 towards R2 downstream. This allows us to place a tap (the bare wire) in between R1 and R2, where we can detect voltage (being the potential difference between this point and ground). And by placing another 10k resistor in as R2, we can detect the potential difference between the tap (at the junction between R1 and R2) and the GND. And because R2 is a 10k resistor, we get a reading of 2.5v.
$$ Vout = Vin * \frac{R2}{R1 + R2} $$