7.7 Regulation, Turns Ratio, Zener Bias & Conductance

7.7 Regulation, Turns Ratio, Zener Bias & Conductance

Quick Answer: % regulation = (V_nl − V_fl) / V_fl × 100. Turns ratio = √(Z_source / Z_load), so 500:10 → 7.1 : 1. Zener limiting resistor R = (V_in − V_z) / I_midpoint. Conductance G = I / V, unit the siemens (S), formerly the mho. Series impedance magnitude = √(R² + X²).

Sub-topic 3-B-018 (Calculations) is the miscellany of the electrical-math topic — five unrelated formulas plus one impedance-magnitude repeat. Each is a single step, so the whole sub-topic is winnable in a few minutes of drill.

Conductance and the siemens

What is the conductance (G) of a circuit if 6 amperes of current flows when 12 volts DC is applied? 0.50 siemens (mhos).

Conductance is simply the reciprocal of resistance:

G = 1 / R = I / V

Here G = 6 A ÷ 12 V = 0.5 S. Equivalently, R = 12/6 = 2 Ω, and 1/2 = 0.5 S. Either route gets there.

Why two names? The unit was originally the mho — "ohm" spelled backwards, a piece of nineteenth-century engineering humour — and was formally renamed the siemens in 1971 after Werner von Siemens. Both names appear in the pool's own answer text, so recognise either. The mho symbol is an upside-down omega and still appears on older meters and schematics.

When is conductance the more natural quantity? For parallel networks. Resistances in parallel require the reciprocal-of-sum-of-reciprocals dance; conductances in parallel simply add. Three 0.5 S branches in parallel total 1.5 S, which is 0.667 Ω — far quicker than the resistance route. Transistor data sheets use the same idea when they specify transconductance in siemens (section 9.2).

Percentage voltage regulation

% regulation = (V_no-load − V_full-load) / V_full-load × 100

The denominator is the full-load voltage. That is the detail candidates get wrong, and the pool tests the formula in both directions:

Forward: no load 12 V, full load 10 V → (12 − 10)/10 × 100 = 20%

Reverse: full load 200 V, regulation 25% → V_nl = 200 × 1.25 = 250 V

Interpret the number properly: lower regulation percentage is better. A perfect supply holds its output constant as load current rises, giving 0%. A 20% figure means the output sags a fifth below its unloaded value when fully loaded — poor for a regulated bench supply, unremarkable for an unregulated transformer-and-rectifier design. Section 15.2 covers the circuits that improve it.

Impedance-matching transformer turns ratio

What turns ratio does a transformer need in order to match a source impedance of 500 ohms to a load of 10 ohms? 7.1 to 1.

The key relationship is that a transformer transforms impedance as the square of the turns ratio:

Z_primary / Z_secondary = (N_p / N_s)², so N_p / N_s = √(Z_p / Z_s)

Here: 500 / 10 = 50, and √50 = 7.07 ≈ 7.1 : 1.

The single most common error is answering 50 : 1 — the impedance ratio, not the turns ratio. Take the square root. A quick sanity check: a 4:1 turns ratio gives a 16:1 impedance ratio, not 4:1.

This connects to the maximum power transfer theorem from section 6.7. You match impedances because maximum power reaches the load when load impedance equals source impedance, and a transformer is the classic way to make an unequal pair look equal.

Zener current-limiting resistor

A 1-watt, 10-volt Zener diode with I_min = 5 mA, I_max = 95 mA and Z = 8 Ω is used as part of a voltage regulator in a 20-V supply. Approximately what size current-limiting resistor sets its bias to the midpoint of its operating range? 200 ohms.

Work it in three steps:

1. Midpoint current: (5 mA + 95 mA) / 2 = 50 mA
2. Voltage across the resistor: the supply minus the Zener voltage = 20 − 10 = 10 V
3. Ohm's law: R = 10 V / 0.05 A = 200 Ω

The 8-Ω Zener impedance and the 1-watt rating are deliberately supplied and deliberately unused in this calculation — they are there to check whether you know which parameters the question actually needs. (They matter elsewhere: 1 W ÷ 10 V = 100 mA absolute maximum, consistent with the stated 95 mA I_max.)

Series impedance magnitude

What is the magnitude of the impedance of a series AC circuit having 6 ohms resistance, 17 ohms inductive reactance, and zero capacitive reactance? 18 ohms.

|Z| = √(6² + 17²) = √(36 + 289) = √325 = 18.03 ≈ 18 Ω. This is the same Pythagorean step as section 7.6; note the magnitude is only slightly above the larger of the two terms, which is the usual pattern when one term dominates.

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