A Beginner's Guide to Circuits by Oyvind Nydal Dahl

Author:Oyvind Nydal Dahl
Language: eng
Format: epub
Publisher: No Starch Press
Published: 2019-10-24T16:00:00+00:00

This calculation is based on Ohm’s law, which describes the relationship among voltage, resistance, and current. You can learn more about Ohm’s law at https://www.build-electronic-circuits.com/ohms-law/.

The voltage on the upper side of the resistor is easy. It’s 9 V because it is connected to the positive terminal of the battery, but what about the lower side? Since you’re looking for the maximum current that can flow through the transistor, it only makes sense to look at the current when the transistor is turned on. When the transistor is on, the voltage on the base of the transistor is around 0.7 V.

So you have 9 V on one side and 0.7 V on the other. This means you have 8.3 V across the resistor (R1). Using Ohm’s law, you can divide 8.3 V by 100,000 Ω (= 100 kΩ) to get 0.000083 A (= 0.083 mA), so the maximum current that can flow through the LED and into the collector of the transistor is 100 times larger, or 8.3 mA.

Since R1 is limiting the current to a value that is safe for the LED, in this circuit you could actually skip the R3 resistor, whose job is also to limit the current to the LED.

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