VIPSolutions1.4 A silicon diode has Cjp = 15 fF. It is biased by a 43-kQ) resistor connected between the cathode of the diode and the input signal, as shown in Fig. P1.4. Initially, the input is 3 V, and then at time 0 it changes to 0 V. Estimate the time it takes for the output voltage to change from 3 V to 0.9 V (i.e., the At_sqq, time). Repeat for an input voltage change from 0 V to 3 V and an output voltage change from 0 V to 2.1 V. Com- pare your answers to those obtained using a SPICE simulation.
Question:
1.4 A silicon diode has Cjp = 15 fF. It is biased by a 43-kQ) resistor connected between the cathode of the
diode and the input signal, as shown in Fig. P1.4. Initially, the input is 3 V, and then at time 0 it changes to
0 V. Estimate the time it takes for the output voltage to change from 3 V to 0.9 V (i.e., the At_sqq, time).
Repeat for an input voltage change from 0 V to 3 V and an output voltage change from 0 V to 2.1 V. Com-
pare your answers to those obtained using a SPICE simulation.
This is AI generated Response.
Asked by: Adapaka Jagadeesh
Created at: 2026-02-14 09:37:55
Asked by: Adapaka Jagadeesh
Created at: 2026-02-14 09:37:55
Final Answer:
Rendered
Raw Text
Did you like this answer?
# Solution to Problem 1.4
## Given Data
- **Junction capacitance, \( C_{jo} = 15 \) fF = \( 15 \times 10^{-15} \) F**
- **Resistor, \( R = 43 \) kΩ = \( 43 \times 10^3 \) Ω**
- **Initial input: 3 V, then changes to V at \( t = \)**
## Circuit Description
The circuit is an RC discharge/charge circuit with the diode's junction capacitance and the resistor.
## (a) Output Voltage Change from 3 V to .9 V
This is a classic RC discharge problem. The voltage across the capacitor as it discharges:
\[
V(t) = V_ e^{-t/RC}
\]
Where:
- \( V_ = 3 \) V (initial voltage)
- \( V(t) = .9 \) V (final voltage of interest)
- \( R = 43 \times 10^3 \) Ω
- \( C = 15 \times 10^{-15} \) F
We need to find \( t \) when \( V(t) = .9 \) V:
\[
.9 = 3 e^{-t/RC}
\]
\[
\frac{.9}{3} = e^{-t/RC}
\]
\[
.3 = e^{-t/RC}
\]
\[
\ln(.3) = -\frac{t}{RC}
\]
\[
t = -RC \ln(.3)
\]
Calculate \( RC \):
\[
RC = 43 \times 10^3 \times 15 \times 10^{-15} = 645 \times 10^{-12} = 6.45 \times 10^{-10} \text{ s}
\]
Now:
\[
t = -6.45 \times 10^{-10} \ln(.3)
\]
Calculate \(\ln(.3) = -1.204\):
\[
t = -6.45 \times 10^{-10} \times (-1.204)
\]
\[
t = 7.76 \times 10^{-10} \text{ s} = .776 \text{ ns}
\]
---
## (b) Input Voltage Change from V to 3 V (Output: V to 2.1 V)
For charging from V towards 3 V:
\[
V(t) = V_f (1 - e^{-t/RC})
\]
Where \( V_f = 3 \) V.
We want \( V(t) = 2.1 \) V:
\[
2.1 = 3(1 - e^{-t/RC})
\]
\[
\frac{2.1}{3} = 1 - e^{-t/RC}
\]
\[
.7 = 1 - e^{-t/RC}
\]
\[
e^{-t/RC} = .3
\]
So the time is the same as before:
\[
t = -RC \ln(.3) = .776 \text{ ns}
\]
---
## **Summary Table**
| Case | Voltage Change | Time (ns) |
|------------------------|-------------------------|-------------|
| Discharge | 3 V → .9 V | .776 |
| Charge | V → 2.1 V | .776 |
---
## **SPICE Simulation**
You should simulate the RC circuit in SPICE with the given values and compare the results. The analytical and SPICE results should closely match, as the circuit is idealized.
---
### **Key Equations**
\[
V(t) = V_ e^{-t/RC} \quad \text{(Discharge)}
\]
\[
V(t) = V_f (1 - e^{-t/RC}) \quad \text{(Charge)}
\]
\[
RC = 6.45 \times 10^{-10} \text{ s}
\]
\[
t = -RC \ln(.3) \approx .776 \text{ ns}
\]
---
Let me know if you need a SPICE netlist or simulation guide!
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :