SME
Circuit Analysis II · Problem type
Turns ratio for impedance matching
Use this for “An amplifier with 800 Ω output resistance drives an 8 Ω speaker. Find the turns ratio for maximum power.”
Diagram
Vₛ
=
Vₚ
×
Nₛ/Nₚ
=
10
×
50
500
Vₛ = 1 V
The alternating current in the primary makes a changing field in the iron, which induces the voltage in the secondary.
This is an example. Type your own values to start a new problem.
a
=
Rₛ (Ω)
RL (Ω)
Nₚ
=
a
×
Nₛ
E
Source voltage
example
Vₚ
Primary voltage
calculated
P
Load power
calculated
Formulas
a
=
√(
Rₛ
RL
)
10
=
√(
800
8
)
Nₚ
=
a
×
Nₛ
500 = 10 × 50
Vₚ
=
E
2
10
=
20
2
P
=
E
2
4 × Rₛ
0.125
=
20
2
4 × 800
Assumptions
•
Ideal transformer: no losses, and all the flux in the core links both coils, so each turn carries the same voltage.
•
Matched: the load seen through the transformer, a² × RL, equals Rₛ, so the load gets the most power.
•
Matched, the primary takes half the source voltage; the other half drops across Rₛ.
Step-by-step
WE KNOW
Rₛ = 800 Ω, RL = 8 Ω, Nₛ = 50, E = 20 V
FIND
turns ratio (a), primary turns (Nₚ), primary voltage (Vₚ), load power (P)
Step 1 · Find turns ratio (a)
Use a = √(Rₛ ÷ RL)
The
load
is
seen
as
a
2
×
RL;
set
that
equal
to
Rₛ
and
take
the
square
root.
a
=
√(
800
8
)
a
=
√
100
a
=
10
Step 2 · Find primary turns (Nₚ)
Use Nₚ = a × Nₛ
The
turns
ratio
is
Nₚ
Nₛ
,
so
multiply
by
the
secondary
turns.
Nₚ = 10 × 50
Nₚ = 500
Step 3 · Find primary voltage (Vₚ)
Use Vₚ = E ÷ 2
Rₛ and the matched primary, also Rₛ, split the source voltage equally.
Vₚ
=
20
2
Vₚ = 10 V
Step 4 · Find load power (P)
Use P = E² ÷ (4 × Rₛ)
The
current
is
E
2Rₛ
and
the
matched
primary
is
Rₛ,
so
the
power
is
that
current
squared
times
Rₛ.
P
=
20
2
4 × 800
P
=
20
2
3,200
P
=
400
3,200
P
=
0.125
W
Check
Put the numbers back in. Both sides should match.
10
=
√(
800
8
)
✓
500 = 10 × 50 ✓
10
=
20
2
✓
0.125
=
20
2
4 × 800
✓
Topic check
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