Basic Circuits Concepts

Superposition theorem

Statement: In a linear, bilateral network with more than one independent source, the current (or voltage) in any element equals the algebraic sum of the currents (or voltages) produced by each source acting alone, with all other independent sources replaced by their internal resistances.

• An ideal voltage source is replaced by a short circuit.
• An ideal current source is replaced by an open circuit.
• Dependent sources are left in the circuit.
• It applies to currents and voltages, not to power (power is not linear).

Example: A 10 V source in series with 2 Ω feeds node X; a 2 A current source also feeds node X; a 3 Ω resistor connects X to ground. Find the current in 3 Ω.

10 V alone (2 A open): I1 = 10/(2+3)      = 2 A
2 A alone (10 V short): I2 = 2 × 2/(2+3)  = 0.8 A
Total: I = I1 + I2 = 2 + 0.8              = 2.8 A

Maximum power to RL

Remove RL and find the Thevenin equivalent across node B and ground.

Thevenin resistance (20 V source shorted, 4 A source opened):

5 Ω || 10 Ω = (5×10)/(5+10) = 3.333 Ω
Rth = 10 + 3.333            = 13.333 Ω

Thevenin voltage (RL open): the 4 A source current has no path except through the 10 Ω (A–B) into node A. KCL at node A (currents leaving A):

(VA − 20)/5 + VA/10 − 4 = 0
0.2VA − 4 + 0.1VA − 4   = 0
0.3VA                   = 8
VA                      = 26.667 V
Vth = VB = VA + 4 × 10  = 26.667 + 40 = 66.667 V

(The 4 A flows from B to A through the 10 Ω, so B is 40 V above A.)

Maximum power transfer: power is maximum when RL = Rth = 13.333 Ω.

Pmax = Vth² / (4 Rth)
     = (66.667)² / (4 × 13.333)
     = 4444.4 / 53.333
     = 83.33 W

Answer: RL = 13.33 Ω, Pmax = 83.33 W (load current = 66.667/26.667 = 2.5 A).

Thevenin's theorem: any linear two-terminal network can be replaced by a voltage source Vth in series with a resistance Rth, where Vth is the open-circuit voltage at the terminals and Rth is the resistance seen from the terminals with independent sources replaced by their internal resistances.

Take the bottom line as reference. Node P is fixed at 12 V by the battery (ideal), so the 4 Ω across the battery does not affect the rest of the circuit.

 P   2Ω    Q
 o--/\/\---o----+-----+
 |         |    |     |
12V  4Ω   6Ω    RL    (open for Vth)
 |         |    |
 |       +10V   |
 o---------o----+

Step 1: Thevenin voltage (RL removed)

Current flows from P (12 V) through 2 Ω and 6 Ω into the + terminal of the 10 V battery:

I   = (12 − 10)/(2 + 6) = 2/8 = 0.25 A
Vth = VQ = 12 − I × 2  = 12 − 0.5 = 11.5 V
check: VQ = 10 + 0.25 × 6 = 11.5 V

Step 2: Thevenin resistance

Short both batteries. The 4 Ω is shorted by the 12 V source, so from Q we see 2 Ω in parallel with 6 Ω:

Rth = (2 × 6)/(2 + 6) = 1.5 Ω

Step 3: Load current (RL = 2 kΩ)

IL = Vth/(Rth + RL)
   = 11.5/(1.5 + 2000)
   = 5.746 × 10⁻³ A

Answer: IL ≈ 5.75 mA (power in the 2 kΩ load ≈ 66 mW).

Step 4: Maximum power transfer

Maximum power is delivered when RL = Rth:

RL  = Rth = 1.5 Ω
Pmax = Vth²/(4 Rth) = (11.5)²/(4 × 1.5)
     = 132.25/6 = 22.04 W

Answer: RL = 1.5 Ω for maximum power; Pmax = 22.04 W.

The equivalent circuit is 11.5 V in series with 1.5 Ω feeding RL. With RL = 2 kΩ the load is badly mismatched, so it takes only about 0.3% of the maximum possible power.

An amplifier or two-port network is described by the ratio of an output quantity to an input quantity. The name depends on which quantities are used.

Trans-conductance (gm)

The ratio of output current to input voltage:

gm = I_out / V_in

• Unit: siemens (S), or mA/V.
• "Trans" means the two quantities are at different ports.
• Example: for a MOSFET or JFET, gm = ΔI_D / ΔV_GS at constant V_DS. A device with gm = 2 mA/V gives a 2 mA drain current change for a 1 V gate voltage change.
• An amplifier described this way is a trans-conductance amplifier (voltage in, current out).

Trans-impedance (Rm or Zm)

The ratio of output voltage to input current:

Rm = V_out / I_in

• Unit: ohm (Ω), or V/mA.
• Example: a photodiode amplifier converts a small diode current into an output voltage; with Rm = 100 kΩ, 10 µA input gives 1 V output.
• An amplifier described this way is a trans-impedance (trans-resistance) amplifier.

Gain

The ratio of an output quantity to the same kind of input quantity, so it has no unit:

Gain is often expressed in decibels: Av(dB) = 20 log₁₀ Av, Ap(dB) = 10 log₁₀ Ap. Example: Av = 100 means 40 dB.