Will this circuit work as I predict it to?

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I want to design a basic Oscillator circuit with an NPN transistor.

My predictions:

When C1 charges up to 0.7 volts, it will open up the npn transistor and allow 3 v dc to flow to the LED. C1 is then discharged and the process repeats itself.

My schematic:
Schematic of a simple transistor oscillator: 3 V DC supply through resistor R1, capacitor C1 across the base and emitter of an NPN transistor, and an LED with ground in the collector path.
 
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  • Schematic of a transistor oscillator: 3 V DC supply through resistor R1, capacitor C1 across the base and emitter of an NPN transistor, with an LED and ground on the collector side.
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No, sorry, it won't do that.

There is no reason the capacitor would discharge.

The voltage across the capacitor rises to about 0.6 volts above the LED voltage and then the transistor turns on and the LED lights up.

So, if the LED was a white one and it needed 3.5 volts to turn on, then the voltage across the capacitor would rise to about 3.5 volts + 0.6 volts or 4.1 volts.

This voltage would then stay on the capacitor and the LED would light up.
 
Schmoopsiepoo said:
I want to design a basic Oscillator circuit with an NPN transistor.
You need two transistors to make the basic multivibrator-type digital oscillator.
 
There is a phase shift oscillator which gives a sine wave out, but it requires a very high gain transistor:
Schematic of a phase shift oscillator using a BC 547 NPN transistor, with three 22n capacitors and three resistors (two 10k, one 1M), a 22k collector resistor, and a 6v battery supply, labelled PHASE SHIFT OSCILLATOR.


There is this one (below) which gives enough output to drive a speaker.

http://dl.dropbox.com/u/4222062/NPN%20PNP%20OSC.PNG

The waveform shown is across the 8 ohm resistor which represents the speaker.
 
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vk6kro said:
The waveform shown is across the 8 ohm resistor which represents the speaker.
Make the 100kΩ resistor variable to vary the frequency.