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I need to know how to calculate the voltage drop across a load connected in series to the secondary coil of a transformer.

I know the resistance (Rs) and inductance (Ls) of the secondary coil (and hence impedance (Zs) for any given frequency), I know the voltage generated by the coil (Vs), and I know the impedance of the wires (Zw) and the load (Zl).

My thought is that I can calculate the current in the secondary circuit using:

$$ V_s = V_c $$

The voltage that drops across the whole circuit is the same as the voltage generated by the secondary coil, and;

$$ V_c = I_c Z_c $$

Where Zc is the combined impedance of the circuit:

$$ Zc = (R_s + R_w + R_l) + j(X_s + X_w + X_l) $$

$$ X = X_L - X_C $$

$$ X_L = 2 \pi fL $$

$$ X_C = \frac{1}{2 \pi fC} $$

Once I know the current in the secondary circuit I can calculate the voltage drop across each component using V = I x Z.

I'm not confident with this at all, could some one advise?

Many Thanks.

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2 Answers 2

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The impedance seen at the output of an ideal transformer is not the impedance of the secondary side inductor.

A good practical example of this is an Ethernet transformer. It might have a typical inductance of like \$350~\mathrm{\mu H}\$, operate at more than \$100~\mathrm{MHz}\$, and drive a \$100~\Omega\$ transmission line.

If we just used the impedance of a \$350~\mathrm{\mu H}\$ inductor at \$100~\mathrm{MHz}\$ we would have like \$220~\mathrm{k\Omega}\$ of output impedance and would be completely unable to drive any usable signal onto wires which look like \$100~\Omega\$ at those frequencies.

Actually, the inductance appears in parallel with the source or load, respectively. As long as the inductive reactance is large enough not to care, we can ignore it, as an ideal transformer.

For an ideal transformer the impedance seen at the output side is,

$$Z_o = Z_s N^2$$

Where:
\$Z_s\$ is the output impedance of whatever is driving the transformer.
\$N\$ is the transformer turns ratio.

schematic

simulate this circuit – Schematic created using CircuitLab

Consider the schematic above. There is a sinusoidal voltage source with an output resistance \$R_s\$ of \$2~\Omega\$. In this example \$N=10\$ so \$R_s\$ looks like \$200~\Omega\$ at the output side and the voltage across the load resistor \$R_L\$ is:

$$V_\textrm{L} = V_\textrm{s} \frac{R_\textrm{L}}{R_\textrm{s} + R_\textrm{L}} = (5~\textrm{V}) (10) \frac{(100~\Omega)}{(2~\Omega) (10)^2 + (100~\Omega)} = 16.66~\textrm{V}$$

Taking into account the turns ratio, we use the secondary-reflected equivalent value for Rs here.

If you want to make the models a little more accurate, there is some output impedance related to the "leakage inductance". But leakage inductance on a good transformer is a very small percentage of the actual coil inductance.

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How to calculate the voltage drop across a load connected in series to the secondary coil of a transformer

OK. Let's proceed.

The voltage that drops across the whole circuit is the same as the voltage generated by the secondary coil

Yes.

$$ V_c = I_c Z_c $$

Where Zc is the combined impedance of the circuit:

Good.

Once I know the current in the secondary circuit I can calculate the voltage drop across each component using V = I x Z.

The way a transformer is "traditionally" used, the voltage across the primary is given, and the voltage across the secondary is (to a first approximation) the voltage across the primary times (or divided by, depending upon your point of view) the turns ratio. Thus, in a traditional configuration, if you know the primary voltage, and you know the turns ratio, you have an approximation of the the secondary voltage. In many cases, a very good approximation. From that approximation of the secondary voltage and knowledge of the load, you calculate the secondary current. What you are trying to do is somewhat the reverse of what is usually done.

There is another way to use a transformer which is to drive the primary with a given current rather than a given voltage. This is often done when measuring AC current, and a transformer designed for this use, are simply used in this manner, is called a current transformer. With a current transformer, we make the approximation that the current through the secondary is related to the current in the primary by the turns ratio. Then, knowing the current in the secondary and the load, you can proceed to calculate the voltages in your secondary circuit. You would use this method only if you are using a transformer that is driven by a given current rather than a given voltage. If you are using the transformer "traditionally", use the first method described above.

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  • \$\begingroup\$ Thank you for your answer. To be a bit more specific, was I correct in including the impedance of secondary coil in total impedance of the circuit? This implies that some of the voltage generated by the coil is dropped across the coil, which implies only a proportion of the voltage generated by the coil is dropped across the cable and load? \$\endgroup\$
    – Ben Booth
    Oct 5, 2021 at 9:05
  • \$\begingroup\$ I'm trying to understand the effect of a change to the secondary impedance on the voltage drop across the load, when the secondary voltage is kept the same; by adjusting the primary coil design or primary signal as required. So the voltage of the secondary coil is the same, so is the impedance of the wires and load, the only change to the secondary circuit is the impedance of the secondary coil, we can ignore the primary coil because it can be whatever we need it to be to get the required secondary voltage. \$\endgroup\$
    – Ben Booth
    Oct 5, 2021 at 13:24

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