# MOSFET Inverter Oscillator

Discussion in 'Homework Help' started by iGBT1200V, Feb 14, 2015.

1. ### iGBT1200V Thread Starter New Member

Oct 1, 2014
1
0

Hi all,

I found this circuit without explanation. I am able to have square wave at the output. I think I understand how this circuit works, but I know that I don´t have all the details.

When 5V gets applied at Drain om M3 (D3), M2 will turn on. This will result in current flowing into D2, and the output will be low, as it is connected to ground.

At the same time, current starts to run through R1 and R3 charging up the capacitor on pin 1. After some time, the voltage at capacitor pin 1 will reach Vth of M3. This will result in M3 gets turned on, current starts to flow through D3, causing M2 to turn off, since its gate is now connected to ground.

However, if someone can help me find an expression of the frequency the circuit can generate, and explain why R3 should be around 10 times R1 - I have been told that it will ease the calculations, cause then I only have to worry about R3*C - I will be grateful

thank you.

2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
You can build it, and check if you see waveforms like these:

Obviously, the trip points are a function of the threshold voltage of the specific NFET you use.

For the circuit above, the period (1/f) is about 2.3R3*C1, especially as R3 gets higher than 100KΩ. I would use between 100KΩ and 1megΩ for R3, and then set the desired frequency by picking C1. This also inproves the appearance of the squarewave at D2.

Last edited: Feb 14, 2015
3. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
If you just look at a circuit containing R1, R3, and M3, you have an inverter that feeds back into itself. The RC time constant that governs the resulting oscillation is R3 and the gate capacitance of M3. The gate capacitance tends to be very small and reasonably unpredictable, so we add another stage that let's us use an external capacitor as the dominant capacitance in the RC time constant.