# What is function of this complicated circuit?

a) Explain principle of working of circuit below. Draw waveforms of voltage at points A, B, C, D and Q, as well as capacitors voltage waveforms.

b) What is minimum time constant of charging of C3 (ie value of resistance R6) so that voltage waveform of point Q are rectangular pulses

It is known: R1, R2, R3 (R2=R3), R4, R5, R6, C1=C2=C3, 5R1>>R2 and R1>>R3. Threshold voltage equals VDD/2.

Any suggestion? I did nothing because I have no idea how to start :( I'm beginner in Digital electronics. In analog electronics and in circuit theory, there are well known methods for solving circuits, but what to do here? I'm really stuck :(

EDIT: This is not homework question. I don't have final result. This was on my exam. I didn't solve it on exam and I'm trying to figure out what is solution to this problem. Every suggestion is welcome, I want to pass exam next time :)

• I love the smell of burning homework in the morning. What have you done to solve this, and where are you stuck?
– user65586
Apr 3, 2016 at 13:27
• Here is a hint: digital circuits are simply overdriven analogue circuits. Look up unbuffered CMOS gates. Apr 3, 2016 at 13:51

Divide and conquer

What does each part do?

• This is not an answer.
– pipe
Apr 3, 2016 at 14:32
• @pipe The general rule of this forum is NOT to answer homework questions but to give guidance. Apr 3, 2016 at 14:34
• @pipe, sometimes the best way to teach is for the teacher to ask questions of the student. This idea is not new. Apr 3, 2016 at 14:53
• No, the voltage at point B is completely determined by the gate that is driving it. The load connection will not change it. Apr 3, 2016 at 14:58
• @Igor (2) yes, but look up differentiator (3) yes (4) yes but look at the values of the ladder resistors for the comparators. I've divided the parts into functional blocks. The output from the first block is the input to the second and so on. So if you know what is output at B (for example) you know what you must input to the next stage. Apr 3, 2016 at 15:01

You seem to have most of it, so here are some additional hints regarding the first part.

Clearly, points A and B are driven by digital gates, so each one can only be in one of two states: connected to Vdd ("high") or connected to Vss ("low"). Furthermore, point B is simply going to be point A inverted.

Therefore, you only have two states to consider: How does the circuit behave when point A is high, and how does it behave when point A is low? What happens when it makes a transition from one state to the other?