p33_065.pdf

65. The ampliﬁer is connected across the primary windings of a transformer and the resistor R is connected

across the secondary windings. If I s is the rms current in the secondary coil then the average power

delivered to R is P avg = I s R . Using I s =( N p /N s ) I p ,weobtain

P avg = I p N p

N s

Next, we ﬁnd the current in the primary circuit. This is eﬀectively a circuit consisting of a generator

and two resistors in series. One resistance is that of the ampliﬁer ( r ), and the other is the equivalent

resistance R eq of the secondary circuit. Therefore,

I p = E rms

r + R eq = E rms

r +( N p /N s ) 2 R

where Eq. 33-82 is used for R eq .Consequently,

P avg = E

2 ( N p /N s ) 2 R

[ r +( N p /N s ) 2 R ] 2

Now, we wish to ﬁnd the value of N p /N s such that P avg is a maximum. For brevity, let x =( N p /N s ) 2 .

Then

2 Rx

( r + xR ) 2

P avg = E

and the derivative with respect to x is

dP avg

= E

xR )

( r + xR ) 3

−

Recalling x =( N p /N s ) 2 , we conclude that the maximum power is achieved for N p /N s = √ x = 10. The

diagram below is a schematic of a transformer with a ten to one turns ratio. An actual transformer

would have many more turns in both the primary and secondary coils.

to ampliﬁer

to load resistance R

2 R ( r

This is zero for x = r/R = (1000Ω) / (10Ω) = 100. We note that for small x , P avg increases linearly with

x , and for large x it decreases in proportion to 1 /x .Thus x = r/R is indeed a maximum, not a minimum.