Please convert the below content to rendered equations without any errors.Given:
* Logic **1 → 3 V**, Logic **0 → 0 V**
* Pull-up resistor (R_P = 2,k\Omega) to **+3 V**
* NMOS **ON resistance** (R_{ON} = 100,\Omega)
* NMOS **OFF resistance** (R_{OFF} = 100,M\Omega)
The output node (v_C) is between the **2 kΩ pull-up** and the **transistor network to ground**.
Voltage divider formula:
[
v_C = 3V \times \frac{R_{down}}{R_{up}+R_{down}}
]
where
(R_{up}=2k\Omega) and (R_{down}) is the equivalent resistance of the NMOS path to ground.
---
# 1️⃣ Circuit 1 (Single NMOS controlled by A)
### Case 1: A = 0
* Transistor OFF
* (R_{down}=100,M\Omega)
[
v_C = 3 \times \frac{100M}{100M+2000}
]
[
v_C \approx 2.99994V \approx 3V
]
Output = **1**
---
### Case 2: A = 1
* Transistor ON
* (R_{down}=100\Omega)
[
v_C = 3 \times \frac{100}{2000+100}
]
[
v_C = 3 \times \frac{100}{2100}
]
[
v_C = 0.143V
]
Output ≈ **0**
---
### Truth Table
| A | (v_C) | C |
| - | ------- | - |
| 0 | ≈3 V | 1 |
| 1 | 0.143 V | 0 |
**Logic function**
[
C = \overline{A}
]
(Inverter)
---
# 2️⃣ Circuit 2 (Two NMOS in SERIES)
Both transistors must conduct to connect the node to ground.
### Case A=0, B=0
Both OFF
[
R_{down} \approx 100M + 100M \approx 200M\Omega
]
[
v_C \approx 3V
]
Output **1**
---
### Case A=0, B=1
Top OFF → no path to ground
[
R_{down} \approx 100M\Omega
]
[
v_C \approx 3V
]
Output **1**
---
### Case A=1, B=0
Bottom OFF
[
R_{down} \approx 100M\Omega
]
[
v_C \approx 3V
]
Output **1**
---
### Case A=1, B=1
Both ON
[
R_{down}=100+100=200\Omega
]
[
v_C = 3 \times \frac{200}{2000+200}
]
[
v_C = 3 \times \frac{200}{2200}
]
[
v_C = 0.273V
]
Output ≈ **0**
---
### Truth Table
| A | B | (v_C) | C |
| - | - | ------- | - |
| 0 | 0 | ≈3 V | 1 |
| 0 | 1 | ≈3 V | 1 |
| 1 | 0 | ≈3 V | 1 |
| 1 | 1 | 0.273 V | 0 |
Logic function:
[
C = \overline{AB}
]
**NAND gate**
---
# 3️⃣ Circuit 3 (Two NMOS in PARALLEL)
Either transistor can pull the node to ground.
---
### Case A=0, B=0
Both OFF
[
R_{down}=100M||100M
]
[
R_{down}=50M\Omega
]
[
v_C = 3 \times \frac{50M}{50M+2000}
]
[
v_C \approx 2.99988V
]
Output **1**
---
### Case A=1, B=0
[
R_{down}=100||100M \approx 100\Omega
]
[
v_C = 3 \times \frac{100}{2100}
]
[
v_C = 0.143V
]
Output **0**
---
### Case A=0, B=1
Same as above
[
v_C = 0.143V
]
Output **0**
---
### Case A=1, B=1
[
R_{down}=100||100
]
[
R_{down}=50\Omega
]
[
v_C = 3 \times \frac{50}{2000+50}
]
[
v_C = 3 \times \frac{50}{2050}
]
[
v_C = 0.073V
]
Output **0**
---
### Truth Table
| A | B | (v_C) | C |
| - | - | ------- | - |
| 0 | 0 | ≈3 V | 1 |
| 0 | 1 | 0.143 V | 0 |
| 1 | 0 | 0.143 V | 0 |
| 1 | 1 | 0.073 V | 0 |
Logic function:
[
C = \overline{A + B}
]
**NOR gate**
---
# ✅ Final Results
| Circuit | Logic Function |
| ------- | -------------- |
| 1 | NOT (Inverter) |
| 2 | NAND |
| 3 | NOR |
---
✅ If you want, I can also show a **very quick shortcut method professors expect in exams (without calculating all voltages)**. It makes these problems solvable in **under 30 seconds.**
Question:
Please convert the below content to rendered equations without any errors.Given:
* Logic **1 → 3 V**, Logic **0 → 0 V**
* Pull-up resistor (R_P = 2,k\Omega) to **+3 V**
* NMOS **ON resistance** (R_{ON} = 100,\Omega)
* NMOS **OFF resistance** (R_{OFF} = 100,M\Omega)
The output node (v_C) is between the **2 kΩ pull-up** and the **transistor network to ground**.
Voltage divider formula:
[
v_C = 3V \times \frac{R_{down}}{R_{up}+R_{down}}
]
where
(R_{up}=2k\Omega) and (R_{down}) is the equivalent resistance of the NMOS path to ground.
---
# 1️⃣ Circuit 1 (Single NMOS controlled by A)
### Case 1: A = 0
* Transistor OFF
* (R_{down}=100,M\Omega)
[
v_C = 3 \times \frac{100M}{100M+2000}
]
[
v_C \approx 2.99994V \approx 3V
]
Output = **1**
---
### Case 2: A = 1
* Transistor ON
* (R_{down}=100\Omega)
[
v_C = 3 \times \frac{100}{2000+100}
]
[
v_C = 3 \times \frac{100}{2100}
]
[
v_C = 0.143V
]
Output ≈ **0**
---
### Truth Table
| A | (v_C) | C |
| - | ------- | - |
| 0 | ≈3 V | 1 |
| 1 | 0.143 V | 0 |
**Logic function**
[
C = \overline{A}
]
(Inverter)
---
# 2️⃣ Circuit 2 (Two NMOS in SERIES)
Both transistors must conduct to connect the node to ground.
### Case A=0, B=0
Both OFF
[
R_{down} \approx 100M + 100M \approx 200M\Omega
]
[
v_C \approx 3V
]
Output **1**
---
### Case A=0, B=1
Top OFF → no path to ground
[
R_{down} \approx 100M\Omega
]
[
v_C \approx 3V
]
Output **1**
---
### Case A=1, B=0
Bottom OFF
[
R_{down} \approx 100M\Omega
]
[
v_C \approx 3V
]
Output **1**
---
### Case A=1, B=1
Both ON
[
R_{down}=100+100=200\Omega
]
[
v_C = 3 \times \frac{200}{2000+200}
]
[
v_C = 3 \times \frac{200}{2200}
]
[
v_C = 0.273V
]
Output ≈ **0**
---
### Truth Table
| A | B | (v_C) | C |
| - | - | ------- | - |
| 0 | 0 | ≈3 V | 1 |
| 0 | 1 | ≈3 V | 1 |
| 1 | 0 | ≈3 V | 1 |
| 1 | 1 | 0.273 V | 0 |
Logic function:
[
C = \overline{AB}
]
**NAND gate**
---
# 3️⃣ Circuit 3 (Two NMOS in PARALLEL)
Either transistor can pull the node to ground.
---
### Case A=0, B=0
Both OFF
[
R_{down}=100M||100M
]
[
R_{down}=50M\Omega
]
[
v_C = 3 \times \frac{50M}{50M+2000}
]
[
v_C \approx 2.99988V
]
Output **1**
---
### Case A=1, B=0
[
R_{down}=100||100M \approx 100\Omega
]
[
v_C = 3 \times \frac{100}{2100}
]
[
v_C = 0.143V
]
Output **0**
---
### Case A=0, B=1
Same as above
[
v_C = 0.143V
]
Output **0**
---
### Case A=1, B=1
[
R_{down}=100||100
]
[
R_{down}=50\Omega
]
[
v_C = 3 \times \frac{50}{2000+50}
]
[
v_C = 3 \times \frac{50}{2050}
]
[
v_C = 0.073V
]
Output **0**
---
### Truth Table
| A | B | (v_C) | C |
| - | - | ------- | - |
| 0 | 0 | ≈3 V | 1 |
| 0 | 1 | 0.143 V | 0 |
| 1 | 0 | 0.143 V | 0 |
| 1 | 1 | 0.073 V | 0 |
Logic function:
[
C = \overline{A + B}
]
**NOR gate**
---
# ✅ Final Results
| Circuit | Logic Function |
| ------- | -------------- |
| 1 | NOT (Inverter) |
| 2 | NAND |
| 3 | NOR |
---
✅ If you want, I can also show a **very quick shortcut method professors expect in exams (without calculating all voltages)**. It makes these problems solvable in **under 30 seconds.**
This is AI generated Response. Asked by: Salaar Created at: 2026-03-06 21:33:51
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# NMOS Pull-Down Circuits: Rendered Equations
## Given Parameters
- Logic **1**: \( 3\,\mathrm{V} \)
- Logic ****: \( \,\mathrm{V} \)
- Pull-up resistor: \( R_P = 2\,\mathrm{k}\Omega \) to \( +3\,\mathrm{V} \)
- NMOS ON resistance: \( R_{ON} = 100\,\Omega \)
- NMOS OFF resistance: \( R_{OFF} = 100\,\mathrm{M}\Omega \)
- Output node voltage (\( v_C \)) is between the pull-up and the NMOS network to ground.
**Voltage divider formula:**
\[
v_C = 3\,\mathrm{V} \times \frac{R_{down}}{R_{up} + R_{down}}
\]
where
\[
R_{up} = 2\,\mathrm{k}\Omega
\]
and \( R_{down} \) is the equivalent resistance to ground via the NMOS network.
---
## 1️⃣ Circuit 1: Single NMOS (Controlled by A)
### Case 1: \( A = \) (Transistor OFF)
\[
R_{down} = 100\,\mathrm{M}\Omega
\]
\[
v_C = 3 \times \frac{100\,\mathrm{M}}{100\,\mathrm{M} + 200}
\]
Since \( 200 \ll 100\,\mathrm{M} \),
\[
v_C \approx 2.99994\,\mathrm{V} \approx 3\,\mathrm{V}
\]
**Output:** 1
---
### Case 2: \( A = 1 \) (Transistor ON)
\[
R_{down} = 100\,\Omega
\]
\[
v_C = 3 \times \frac{100}{200 + 100}
\]
\[
v_C = 3 \times \frac{100}{210}
\]
\[
v_C = .143\,\mathrm{V}
\]
**Output:**
---
### Truth Table
| \(A\) | \(v_C\) | \(C\) |
|-------|--------------|-------|
| | \(\approx 3\) V | 1 |
| 1 | \(.143\) V | |
**Logic function:**
\[
C = \overline{A}
\]
(Inverter)
---
## 2️⃣ Circuit 2: Two NMOS in SERIES
Both must conduct to pull low.
### Case 1: \( A = , B = \)
\[
R_{down} \approx 100\,\mathrm{M} + 100\,\mathrm{M} = 200\,\mathrm{M}\Omega
\]
\[
v_C \approx 3\,\mathrm{V}
\]
**Output:** 1
---
### Case 2: \( A = , B = 1 \)
Top transistor OFF:
\[
R_{down} \approx 100\,\mathrm{M}\Omega
\]
\[
v_C \approx 3\,\mathrm{V}
\]
**Output:** 1
---
### Case 3: \( A = 1, B = \)
Bottom transistor OFF:
\[
R_{down} \approx 100\,\mathrm{M}\Omega
\]
\[
v_C \approx 3\,\mathrm{V}
\]
**Output:** 1
---
### Case 4: \( A = 1, B = 1 \)
Both ON:
\[
R_{down} = 100 + 100 = 200\,\Omega
\]
\[
v_C = 3 \times \frac{200}{200 + 200}
\]
\[
v_C = 3 \times \frac{200}{220}
\]
\[
v_C = .273\,\mathrm{V}
\]
**Output:**
---
### Truth Table
| \(A\) | \(B\) | \(v_C\) | \(C\) |
|-------|-------|--------------|-------|
| | | \(\approx 3\) V | 1 |
| | 1 | \(\approx 3\) V | 1 |
| 1 | | \(\approx 3\) V | 1 |
| 1 | 1 | \(.273\) V | |
**Logic function:**
\[
C = \overline{AB}
\]
(NAND gate)
---
## 3️⃣ Circuit 3: Two NMOS in PARALLEL
Either can pull low.
### Case 1: \( A = , B = \) (Both OFF)
\[
R_{down} = 100\,\mathrm{M} \parallel 100\,\mathrm{M} = 50\,\mathrm{M}\Omega
\]
\[
v_C = 3 \times \frac{50\,\mathrm{M}}{50\,\mathrm{M} + 200}
\]
\[
v_C \approx 2.99988\,\mathrm{V}
\]
**Output:** 1
---
### Case 2: \( A = 1, B = \) (One ON)
\[
R_{down} = 100\,\Omega \parallel 100\,\mathrm{M}\Omega \approx 100\,\Omega
\]
\[
v_C = 3 \times \frac{100}{210}
\]
\[
v_C = .143\,\mathrm{V}
\]
**Output:**
---
### Case 3: \( A = , B = 1 \) (Same as above)
\[
v_C = .143\,\mathrm{V}
\]
**Output:**
---
### Case 4: \( A = 1, B = 1 \) (Both ON)
\[
R_{down} = 100\,\Omega \parallel 100\,\Omega = 50\,\Omega
\]
\[
v_C = 3 \times \frac{50}{200 + 50}
\]
\[
v_C = 3 \times \frac{50}{205}
\]
\[
v_C = .073\,\mathrm{V}
\]
**Output:**
---
### Truth Table
| \(A\) | \(B\) | \(v_C\) | \(C\) |
|-------|-------|--------------|-------|
| | | \(\approx 3\) V | 1 |
| | 1 | \(.143\) V | |
| 1 | | \(.143\) V | |
| 1 | 1 | \(.073\) V | |
**Logic function:**
\[
C = \overline{A + B}
\]
(NOR gate)
---
## ✅ Summary Table
| Circuit | Logic Function |
|---------|---------------|
| 1 | NOT |
| 2 | NAND |
| 3 | NOR |
Question:
Please give me the answer same as it is without any changes and errors
This is AI generated Response. Asked by: Salaar Created at: 2026-03-06 21:35:16
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# NMOS Pull-Down Circuits: Rendered Equations
## Given Parameters
- Logic **1**: \( 3\,\mathrm{V} \)
- Logic **0**: \( 0\,\mathrm{V} \)
- Pull-up resistor: \( R_P = 2\,\mathrm{k}\Omega \) to \( +3\,\mathrm{V} \)
- NMOS ON resistance: \( R_{ON} = 100\,\Omega \)
- NMOS OFF resistance: \( R_{OFF} = 100\,\mathrm{M}\Omega \)
- Output node voltage (\( v_C \)) is between the pull-up and the NMOS network to ground.
**Voltage divider formula:**
\[
v_C = 3\,\mathrm{V} \times \frac{R_{down}}{R_{up} + R_{down}}
\]
where
\[
R_{up} = 2\,\mathrm{k}\Omega
\]
and \( R_{down} \) is the equivalent resistance to ground via the NMOS network.
---
## 1️⃣ Circuit 1: Single NMOS (Controlled by A)
### Case 1: \( A = 0 \) (Transistor OFF)
\[
R_{down} = 100\,\mathrm{M}\Omega
\]
\[
v_C = 3 \times \frac{100\,\mathrm{M}}{100\,\mathrm{M} + 2000}
\]
Since \( 2000 \ll 100\,\mathrm{M} \),
\[
v_C \approx 2.99994\,\mathrm{V} \approx 3\,\mathrm{V}
\]
**Output:** 1
---
### Case 2: \( A = 1 \) (Transistor ON)
\[
R_{down} = 100\,\Omega
\]
\[
v_C = 3 \times \frac{100}{2000 + 100}
\]
\[
v_C = 3 \times \frac{100}{2100}
\]
\[
v_C \approx 0.143\,\mathrm{V}
\]
**Output:** 0
---
### Truth Table
| \(A\) | \(v_C\) | \(C\) |
|-------|--------------|-------|
| 0 | \(\approx 3\) V | 1 |
| 1 | \(0.143\) V | 0 |
**Logic function:**
\[
C = \overline{A}
\]
(Inverter)
---
## 2️⃣ Circuit 2: Two NMOS in SERIES
Both must conduct to pull low.
### Case 1: \( A = 0, B = 0 \)
Both OFF:
\[
R_{down} \approx 100\,\mathrm{M} + 100\,\mathrm{M} = 200\,\mathrm{M}\Omega
\]
\[
v_C \approx 3\,\mathrm{V}
\]
**Output:** 1
---
### Case 2: \( A = 0, B = 1 \)
Top transistor OFF:
\[
R_{down} \approx 100\,\mathrm{M}\Omega
\]
\[
v_C \approx 3\,\mathrm{V}
\]
**Output:** 1
---
### Case 3: \( A = 1, B = 0 \)
Bottom transistor OFF:
\[
R_{down} \approx 100\,\mathrm{M}\Omega
\]
\[
v_C \approx 3\,\mathrm{V}
\]
**Output:** 1
---
### Case 4: \( A = 1, B = 1 \)
Both ON:
\[
R_{down} = 100 + 100 = 200\,\Omega
\]
\[
v_C = 3 \times \frac{200}{2000 + 200}
\]
\[
v_C = 3 \times \frac{200}{2200}
\]
\[
v_C \approx 0.273\,\mathrm{V}
\]
**Output:** 0
---
### Truth Table
| \(A\) | \(B\) | \(v_C\) | \(C\) |
|-------|-------|--------------|-------|
| 0 | 0 | \(\approx 3\) V | 1 |
| 0 | 1 | \(\approx 3\) V | 1 |
| 1 | 0 | \(\approx 3\) V | 1 |
| 1 | 1 | \(0.273\) V | 0 |
**Logic function:**
\[
C = \overline{AB}
\]
(NAND gate)
---
## 3️⃣ Circuit 3: Two NMOS in PARALLEL
Either can pull low.
### Case 1: \( A = 0, B = 0 \) (Both OFF)
\[
R_{down} = 100\,\mathrm{M} \parallel 100\,\mathrm{M} = 50\,\mathrm{M}\Omega
\]
\[
v_C = 3 \times \frac{50\,\mathrm{M}}{50\,\mathrm{M} + 2000}
\]
\[
v_C \approx 2.99988\,\mathrm{V}
\]
**Output:** 1
---
### Case 2: \( A = 1, B = 0 \)
\[
R_{down} = 100\,\Omega \parallel 100\,\mathrm{M}\Omega \approx 100\,\Omega
\]
\[
v_C = 3 \times \frac{100}{2100}
\]
\[
v_C \approx 0.143\,\mathrm{V}
\]
**Output:** 0
---
### Case 3: \( A = 0, B = 1 \) (Same as above)
\[
v_C \approx 0.143\,\mathrm{V}
\]
**Output:** 0
---
### Case 4: \( A = 1, B = 1 \) (Both ON)
\[
R_{down} = 100\,\Omega \parallel 100\,\Omega = 50\,\Omega
\]
\[
v_C = 3 \times \frac{50}{2000 + 50}
\]
\[
v_C = 3 \times \frac{50}{2050}
\]
\[
v_C \approx 0.073\,\mathrm{V}
\]
**Output:** 0
---
### Truth Table
| \(A\) | \(B\) | \(v_C\) | \(C\) |
|-------|-------|--------------|-------|
| 0 | 0 | \(\approx 3\) V | 1 |
| 0 | 1 | \(0.143\) V | 0 |
| 1 | 0 | \(0.143\) V | 0 |
| 1 | 1 | \(0.073\) V | 0 |
**Logic function:**
\[
C = \overline{A + B}
\]
(NOR gate)
---
## ✅ Summary Table
| Circuit | Logic Function |
|---------|---------------|
| 1 | NOT |
| 2 | NAND |
| 3 | NOR |
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