Question

Question 1 (4 Marks) 1) In the circuit shown, find / and V, using the ideal diode model. 240 2) Consider the given circuit a. Design R for an output voltage of 1.5 V. Assume identical diodes with land Is 104 A. b. Explain why the circuit could be used for supply noise regulation

Answer 1

1) We Replaced The Diodes with Short Circuits.

Drop across the Ideal Diode Model- Voltage Diode 2 Zero (0) Ви хемония Diodec as short circuits we get the circuit 24h 777 -31 - 1 * 1000 1 2 3mA V=VA-020-001

2. a.

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If we Start Writing KVL Equations then We can get that -

$\small 6 V - I \times R = V_0 \hspace{0.25cm}........ (1)$
Now I is the same current that flowing through the Diode.

$\small I = Diode \hspace{0.10cm}Current = I_d$

Now we can say The Potential Drop Across the two Diodes is 1.5 V. As the Diodes are similar then We can assume that Drop Across the

So the Diode Current Id is given by -

$\small I_d = I_s \times (e^{\frac{V_D}{0.026}}-1) \hspace{0.25cm}........ (2)$

Given VD  = 1.5 V then Putting in the Equation 2 we get

$\small I_d = 10^{-14} \times (e^{\frac{1.5/2}{0.026}}-1) \hspace{0.25cm}........ (2) \newline = I_d \approx 10^{-14} \times (e^{\frac{0.75}{0.026}}) \newline \approx 0.033 A$

Putting back the Value of Id in Equation 1 we get -

$\small 6-1.5 = I \times R \newline R = \frac{4.5}{I} R = \frac{4.5}{0.033} \approx 137 \Omega$

b. If there is Noise in the input side, then Current Variations can be seen. But the Potential Drop across the Two Diodes (V0=1.5 V) would be constant. The Variation in Current Changes Can't do much in the Output V0 .