Question

5. (5pts) The figure is a CMOS clock circuit using an inverter. Because R1 resistor is ten times larger than R2, we can neglect any current through R2. Then, you can approximate this circuit as a familiar simple RC transient circuit. Mathematically, this approximation leads to a simple differential equation according to Kirchhoff's junction rule: cdiz-n.W-Y. Since Z is constant, this equation R1 IM W .001uF R2 100k UIC 4069 UID 4069 becomes dwwr The solution of this equation is X Another important feature is that the circuit makes a transition at W conditions immediately after each transition. 12Vcc. You need to find initial a) At the time t-0, let's start with the following initial conditions: W(r) Then, the constant C Vec. The +3/2Vco solution is W) Ve As the charges accumulate and +1/2Vco W decreases. the switching occurs at W ½ Vec Let's call this transition time tr, Find tr -1/2Vco by solving the following eq Voe 12-+Express 0 tr b) Y also transits from GND to +Vec and Z from +Vcc to GND. That means that mathematically at the time t-tri ,the initial conditions are: Z-0, 1/2Vcc Y Vec, and W-·½ Vcc because W-Z- You can find that the solution W() is given by wo-VAs W(t) increases with time, there gets another transition when W 1/2Vcc at t-tr2. Find (tr2- tri) by solving the r answer in terms of R2, C and r2-trl This time Z transits from GND to +Vec and Y in the opposite way. This means that W(-2+1 /2Vec) becomes +32 Vcc The solution is W (t)-+ 31 accumulate and W decreases, the switching occurs when W c) e-ra 4 As the charges Vccat ttr3 Express tr3-t2 d) Based on the above analysis in (b), the wave pattern of W can be summarized as follows. From this plot, determine the period (T) and oscillation frequency (). e) With R2-100kΩ and C-0001 μF, calculate f.

Answer 1

Solution:

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