Question

Notify Me Bookmark

The circuit in Figure 1 is referred to as a

The circuit in Figure 1 is referred to as a

Image text
The circuit in Figure 1 is referred to as a

Detailed Answer

1 0

Please enter your email here. You will get email when this question is answered by our tutor.

Answer
Doubtrix Logo

This Problem has been solved!

You'll get a detailed, step-by-step and expert verified solution.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Login Sign Up
  • Student Reviews:
  • (1)

Similar/Related Questions

Non-inverting Amplifier The circuit labeled "a" shown in Figure 2 is a non-inverting amplifier. The relationship between the input voltage Vin and the output voltage Vout is Vot = Vin⋅(1 + R2 R1). The gain of this circuit is (1 + R2 R1), which is an ideal circuit for sensors that have high source impedance and thus would be affected by the current draw of the data acquisition system. If R2 = 0 and R1 is removed, then the gain of this non-inverting amplifier is unity. Such a circuit is usually referred to as a follower or a buffer, which can be used to isolate input voltage and output signals. It can serve as protection between the input and output sides. One major advantage of a buffer is that it can boost the current of the input signal to avoid loading effect or distortion for input signals.
Solution
0 0

The circuit in Figure 1 is a level shift circuit. It achieves a DC level shift between the input and the output. The value of this shift is determined by the current I0. Assume Xd = 0, y = 0.4, 2|φf| = 0.6 V, VT0 = 0.43 V, kn = 115 μA/V2 and λ = 0. a. Suppose we want the nominal level shift between Vi and Vo to be 0.6 V in the circuit in Figure 1-a. Neglecting the body effect, calculate the width of M2 to provide this level shift (Hint: first relate Vi to Vo in terms of Io). b. Now assume that an ideal current source replaces M2 (Figure 1-b). The NMOS transistor M1 experiences a shift in VT due to the body effect. Find VT as a function of V0 for V0 ranging from 0 to 2.5 V with 0.5 V intervals. Plot VT Vs. Vo c. Plot Vo vs. Vi as Vo varies from 0 to 2.5 V with 0.5 V intervals. Plot two curves: one neglecting the body effect and one accounting for it. How does the body effect influence the operation of the level converter? d. At Vo (with body effect) = 2.5 V , find Vo (ideal) and thus determine the maximum error introduced by the body effect. (a) (b) Figure 1 - NMOS Level Shifter
Solution
0 0

Charge sharing is an important problem in dynamic circuits. Switching exposes a fixed amount of charge to a different capacitive environment. This results in node voltage change. After pre-charge stage, the output node voltage of the dynamic logic circuit in Figure 1 will be equal to VDD. During the evaluate phase, assume A = 1 and B=0. Output node will not connect to ground but the total charge on the output node (Q = C.V) will be shared by parasitic capacitance (C1) and the load capacitance (CL). So that the output voltage will drop and be equal to M1 source voltage. a. The circuit is Figure 1 is pre-charged with VDD = 3V. During evaluation stage, M1 turns on and M2 is off (A=1 and B=0). CL = 250fF, C1 = 125fF and Vt0,n=0.55V. Calculate the final voltage after charge sharing. Is M1 still conducting at this voltage? b. Repeat question a for C1 = 40fF. Calculate the final voltage after charge sharing. Is M1 still conducting at this voltage? If not, what is the value of the output voltage at this scenario? c. With A = 1 and B = 0, the output of the circuit in Fig 1 should be logic high. However, you have seen that logic high does not mean VDD for this example. Output voltage depends on the parasitic capacitance at the source node of M2. For a dynamic 2 input NAND gate, what is the maximum value that the Figure 1 output voltage can reach when A= 1 and B = 0?

Solution
0 0