This is a simple circuit to demonstrate the forward voltage drop of three types of diodes:
- A Silicon Diode (1N34)
- A Germanium Diode (1N4148)
- A 3mm LED (Light Emitting Diode)
Pressing switch 1 (S1) turns on the LED (LED2). The LED is protected with a 2.7k Ohm resistor (R2) and with no other switches turned on, will be the brightest it will get. Now, one at a time, press switches S2, S3 and S4 and watch how the LED dims with each. Some diodes (D1 and D2) will dim the LED a little, while the LED switch will dim the main LED a lot more than the previous two diodes. Give it a try!
On this silicon diode (D1 – 1N34), the voltage drop is 0.6V.
On the germanium diode (D2 – 1N4148), the voltage drop is 0.2V.
The LED’s voltage drop is 1.6V.
Voltage drops across resistors, potentiometers, transistors and of course diodes. With a 9 volt battery, a voltage drop of 0.6 or even 1.6V is not noticeable. However, with lower power circuits, such as one powered by a 3.7 V Lithium Ion battery, 1.6V is pretty close to half your voltage and a 0.6V drop is over 16% of your voltage used up in a diode. In those cases, it might make sense to go with the germanium diode with has a lower forward voltage drop.