Transformers Ideal Transformers
In general, a transformer is a multi-port ac device that converts
voltages, currents and impedances from one value to another.
This device only performs this transformation for time varying
signals. Here, we will consider the transformer circuit shown below:

The time varying current I_{p}( t) in the “primary” circuit produces a magnetic flux density B_{p}( t )in and around coil p. Similarly, the “secondary” coil produces B _{s}t . The total magnetic flux
density is the sum B( t) = B_{p}( t) + B_{s}( t)
We will assume that the transformer core has a very large
relative permeability ?_{r} . Consequently, B(t ) will almost exclusively be contained within the core. This B(t ) forms closed loops within the core. (We can think of this as a “selfshielded” core.) . The magnetic flux ? m is defined simply as the integral of B(t )over a cross section of the core:

With B(t ) contained exclusively within the core, ?_{m}( t) will be the same throughout the transformer (though it will vary with time).
The magnetic flux will be proportional to the number of coil
turns, the geometry of the coil and the current in the coil:
?_{m}_{j}(t) = N_{j} A _{j}I_{j}( t) [Wb/turn]
A_{1} is the inductance constant [H/turn2] of the core and j = p, s.
This A_{1} is provided by the manufacturer of the cores that you use for your transformers (and inductors). Table D.2 in your text lists A_{1} for various cores used in the NorCal 40A.

Note that A_{1 can be a very strong function of frequency
Induced Voltage
As we know, a time varying magnetic field through a coil of
wire produces a voltage between the ends of the coil. This
miraculous phenomenon was discovered by Michael Faraday
and is mathematically stated in Faraday’s law as
where N is the number of (identical) turns of the coil.
This emf is a “net push” around a circuit that causes electrons to
move. Voltage and emf are closely related concepts. We can
determine the induced voltage V (t ) using the following
equivalent circuit:
Applying this equivalent circuit to each side of the transformer
shown on the first page gives:
The polarity of the lumped emf source is set by the direction of
the current: a voltage source has current entering the negative
terminal. The sign of the emf source is due to the direction of ds
(by the RHR) and the assumed direction for ?m (and hence B ).
From these circuits and applying (3), the sinusoidal steady state
voltage at the primary and secondary are both of the form
V = j? N?m
where V and ?m are now phasors. Specifically, the primary and secondary voltages are
Vp = j? Np ?m
Vs = j? Ns?m
Dividing these two equations gives
where n is called the turns ratio.
Interestingly, we see here that the “output” voltage Vs can be
different in amplitude than the “input” voltage Vp
Note that if Ns > Np , the secondary voltage is larger in
amplitude than the primary voltage. Very interesting.
• If Ns > Np , called a step-up transformer,
• If Ns < Np , called a step-down transformer
Primary and Secondary Currents
Next we will consider the electrical currents in the primary and
secondary of the transformer. From (1), the magnetic flux is the
sum of the two magnetic fluxes from each coil
?m =?mp +?ms
Using (2) and noting that ?ms will be negative since the direction of the current is assumed OUT of the secondary, then ?m = Np Al Ip ? Np Al Is Solving for I p we find that
The magnetic flux is not a circuit quantity. To derive an
equivalent circuit for the transformer we need to express m ? in
terms of electrical circuit quantities.
To accomplish this, we use (5) in the first term of (11) yielding
where p L is the inductance of the primary coil. Substituting this
result back into (11) gives
This last result is extremely illuminating. We see that the current
in the primary is the sum of two parts: (1) magnetization current
and (2) transformer current.
(1) Magnetization Current. The first term in (13) does not
involve the secondary in any way. In other words, this is
the current the transformer would draw regardless of the
turns ratio of the transformer.
(2) Transformer Current. The second term in (13) directly
depends on the secondary because of the s N term. This
component of the primary current is a transformed
secondary current, in a manner similar to the voltage in (7),
though inversely.
Ideal Transformer
If the magnetization current Vp/( j?Lp) in (13) is very small in magnitude relative to the transformer current ( Ns| Np) Is then such a device is called an ideal transformer. The equations for an ideal transformer are from (7) and (13):
The circuit symbol for an ideal transformer is
Discussion
1. We can surmise from (15) that for an ideal step-up
transformer Is < I p . Therefore, while from (14) the voltage
increases by Ns / Np , the current has decreased by Np / Ns .
In the NorCal 40A, the transformer T1 is used to step up the
current from the Driver Amplifier to the Power Amplifier. For
T1, 14 Np = and 4 Ns = so that ( ) Is = (Np/ Ns) I =7/2 ? Ip . Because of this current behavior, the power input to the primary equals the po wer output from the secondary: Pp (t ) =Vp (t ) Is (t )
Therefore, the input power Pp (t ) equals the output power
Ps (t ), as would be expected.
2. With an impedance Zs connected to the secondary, then
Substituting for Vs and Is in this equation using (14) and (15)
In other words, the effective input impedance Zp,eff at the
primary terminals (the ratio Vp/Ip) is
The ideal transformer “transforms” the load impedance from
the secondary to the primary. (Remember that this is only true
for sinusoidal steady state signals.)
3. For maximum power transfer, we design a circuit so that the
load is matched to the output resistance. We can use
transformers as matching networks.
For example, in the NorCal 40A, T3 is used to transform the
output impedance from the RF Mixer (3 k?) to match the
input impedance of the IF Filter (200 ?). Using (18)
which is very close to the desired 200 ?.
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