# Generating three voltages with a MCU

I'm going to try and generate black and white NTSC video, using a micro-controller. According to this:

https://instruct1.cit.cornell.edu/courses/ee476/video/

It is possible to do it using two pins of Atmel Mega644. I looked at the datasheet and noted that the IO is 1.8 V. For NTSC you just need three voltages sent at the correct time, 0V, 0.3V, and 1V.

What I don't understand is how does this circuit generate the 0, .3 and 1V. When I do the calculations it doesn't really add up.

I simulated this using TI TINA simulator, 2 batteries and switches inplace of the IO ports, and trying all four combinations of the IO, I can get 0, .3V but not 1V.

Can someone using math explain how you can get 0 V, .3 V, and 1 V using the circuit shown in the link above. I also, I guess what I don't understand is the uC IO equivalent circuit for simulating in a SPICE program, is it just a battery set to the IO voltage? Or maybe an OP AMP with the output set to the IO voltage can someone please explain this.

Any help is appreciated.

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The ATmega644 works fine with 5 V (which is also close enough to the output voltage of the microcontroller which itself is rated at minimal 4.2 V on page 316 of the datasheet) and at that voltage using a $330 \mbox{ } \Omega$ and $75 \mbox{ } \Omega$ voltage divider, you get 0.926 V which is close enough to 1 V for most purposes.
Using the $1 \mbox{ } k\Omega$ and $75 \mbox{ } \Omega$ voltage divider, you get around 0.349 V which is for most purposes close enough to 0.3 V. And when both pins are off, you get 0 V.
@user968102 Then the $330 \mbox{ }\Omega$ and $1 \mbox{ } k \Omega$ resistors are in parallel and they have the $75 \mbox{ } \Omega$ resistor connected in series with them. The resulting voltage is around 1.16 V at the top of the $75 \mbox{ } \Omega$ resistor . –  AndrejaKo Apr 7 '12 at 18:36