ABCD parameters: what they tell us about a transmission line

In this guide we treat a whole transmission line as one simple box with four numbers, and we use diagrams and a worked example to see what each number means.

Want to play with it? Our interactive ABCD parameters lab lets us change a line and watch A, B, C and D update. It works well next to this article.

Why treat a line as a box?

A transmission line can be hundreds of kilometres long. Along every metre, the wire has resistance and inductance. Between the wires, and between each wire and the ground, there is capacitance. So the voltage and current change a little at every point along the line.

We could describe all of this with equations for every metre. But a grid engineer usually needs much less. We need to know what happens at the two ends. How much voltage must the power plant side send? And what voltage and current arrive at the city?

So here is the question for this article. Can we replace hundreds of kilometres of wire with just four numbers? The answer is yes, because the line is linear. This means that if we double the current, the voltage drops also double. The animation at the top of the page shows the idea. We hide the line inside a box and look only at its two ends.

The two-port model

A two-port network is any circuit with two pairs of terminals. Each pair is a port. For a line, port 1 is the sending end, the power plant side. Port 2 is the receiving end, for example the busbar of a city substation. The figure below shows the box and the four quantities at its ports.

At the sending end we have the voltage V_s and the current I_s. At the receiving end we have V_r and I_r. Note the direction of the currents. I_s flows into the box. I_r flows out of the box, toward the load.

Because the line is linear, the sending-end quantities are simple sums of the receiving-end quantities:

The same two equations in matrix form are:

These are phasor equations. A phasor is a complex number that holds both the size and the phase angle of an AC quantity. So A, B, C and D are complex numbers too. For a three-phase line we use per-phase values: the line-to-neutral voltage and the line current.

What A, B, C and D mean

Each parameter has a clear physical meaning. We can find each one with a simple test at the receiving end. In the first test the receiving end is open, so no current leaves it. In the second test it is short-circuited, so its voltage is zero. The figure below shows both tests.

So now we can read the first equation in plain words. The sending voltage is the no-load part, A·V_r, plus a load part, B·I_r. The load part is the voltage drop that the load current causes in the series impedance. It is Ohm's law, V = I·Z, applied to the line.

There are two quick checks that every correct ABCD matrix of a line must pass. The first holds for any passive, reciprocal network. Passive means it has no sources inside. Reciprocal means it behaves the same in both directions, which is true for lines, cables and normal transformers.

The second check holds for a symmetrical network, one that looks the same from both ends. A uniform line is symmetrical. So for a line:

Together, these two checks mean that a line has only three independent parameters, not four. So if a calculation gives AD − BC far from 1 + j0, there is an arithmetic error somewhere. But careful: a chain that includes a transformer at one end is not symmetrical. For that chain, A is not equal to D, and that is correct.

Short, medium and long lines

To find A, B, C and D we first need a model of the line. We start from the line constants per kilometre. The series impedance per kilometre is z = r + jx, where r is the resistance and x the inductive reactance. The shunt admittance per kilometre is y = jb, where b is the capacitive susceptance. For a line of length l, the totals are:

How we use these totals depends on the length of the line. The figure below shows the three common models side by side.

Short line (under about 80 km). The charging current is small, so we ignore the capacitance. The line is just a series impedance. The sending and receiving currents are equal:

Medium line (about 80 to 250 km). We use the nominal-π model. We put the whole series impedance Z in the middle, and half of the shunt admittance, Y/2, at each end. The circuit looks like the Greek letter π. Then:

Long line (over about 250 km). Now the lumped model is not accurate enough. The resistance, inductance and capacitance are spread evenly along the whole line. We solve the line equations exactly and get:

Here γ is the propagation constant, which tells us how the voltage changes per kilometre. Z_c is the characteristic impedance, also called the surge impedance. For overhead lines it is about 400 Ω. Both come from the per-kilometre constants:

All three models pass both checks: A = D, and AD − BC = 1. The long-line formulas also contain the others. For a short enough line, cosh γl is close to 1 + YZ/2, and the exact model gives almost the same numbers as the nominal π. We will see this in the worked example.

Cascading: multiply the matrices

Real power reaches a city through several parts in a row. For example, a generator step-up transformer T1 raises 13.8 kV to transmission voltage. Then a line L carries the power. Then a substation transformer T2 lowers the voltage again. Each part is a two-port with its own ABCD matrix. We call such a chain a cascade.

This is where ABCD parameters are most useful. The output port of one block is the input port of the next. So the matrix of the whole chain is simply the product of the matrices, taken in the same order as the power flows. The figure below shows the idea.

For two blocks, matrix 1 followed by matrix 2, the product gives:

Engineers use this idea in many places. A series capacitor bank in the middle of a line is one more block in the chain. So is a shunt reactor at the receiving end. Power-flow programs use related network models, such as the Y bus, to describe the whole grid. And protection and planning engineers use ABCD parameters to find how a line behaves at no load and full load.

Worked example: a 150 km, 220 kV line

Now we put everything together. A three-phase, 50 Hz overhead line is 150 km long and runs at 220 kV. These are typical constants for a 220 kV line with one conductor per phase:

We want the sending-end voltage and the voltage regulation. A length of 150 km is a medium line, so we use the nominal-π model. The four cards below show the method.

Step 1: totals.

Step 2: the ABCD parameters. First we find the product YZ, because it appears in A and C.

|A| = 0.9874, a little less than 1. This small difference matters later. Now we check the result:

Step 3: the sending voltage. We take the receiving voltage as the reference, at 0°. Per phase it is:

The load current follows from the three-phase power, P = √3 × V × I × power factor. A lagging power factor of 0.95 means the current lags the voltage by cos⁻¹ 0.95 ≈ 18.19°.

Now we work out the two parts of V_s = A·V_r + B·I_r:

The figure below draws this sum as phasors. Note how the voltage drop B·I_r points almost straight up. This is because B is mostly reactance.

As a line-to-line value, the sending end must be at:

The sending current is I_s = C·V_r + D·I_r ≈ 261.2 A at −6.9°. This is smaller than I_r, because the line's own charging current supplies part of the reactive power of the load. The sending-end power is about 102.6 MW, so about 2.6 MW is lost in the line, and the efficiency is about 97.5%.

Step 4: the regulation. Now we remove the load and keep V_s the same. With I_r = 0, the first equation gives V_s = A·V_r. So the no-load receiving voltage is:

How good is the nominal π? The exact long-line formulas give A = 0.9874 + j0.0025 and B = 11.90 + j59.76 Ω. With these, V_s = 233.0 kV and the regulation is 7.3%. So for 150 km the nominal π is very close. A short-line model, which ignores C, would give V_s = 235.8 kV. That is 2.7 kV too high, because it misses the charging current.

Voltage regulation and the Ferranti effect

Voltage regulation tells us how much the receiving voltage changes between no load and full load, while the sending voltage stays the same:

Here V_r,NL is the receiving voltage at no load and V_r,FL at full load. The ABCD parameters give V_r,NL directly:

For a line, |A| is less than 1. So at no load, the receiving end is at a higher voltage than the sending end. This is the Ferranti effect. It happens because the charging current of the line capacitance flows through the line inductance. The voltage across that inductance adds to the voltage, instead of reducing it.

The rise grows quickly with length. For a lossless line, A = cos βl, where β is the phase constant (the imaginary part of γ). For moderate lengths, this gives a handy rule, with x and b per kilometre:

For our line, 0.40 × 2.8 × 10⁻⁶ × 150² / 2 ≈ 1.3%. That matches 1/|A| = 1/0.9874. Because the rise grows with the square of the length, a line twice as long has about four times the rise. Move the slider below to see this.

At 150 km the rise is small. At 300 km it is about 5%. Above about 430 km the no-load voltage would go past 245 kV. That is the highest voltage that equipment for a 220 kV system is designed for. So long lines need shunt reactors, inductors connected from line to ground. They absorb the charging current and hold the voltage down at light load.

Key takeaways

• ABCD parameters model a whole line as a two-port: V_s = A·V_r + B·I_r and I_s = C·V_r + D·I_r, using per-phase phasors.
• A and D have no units, B is in ohms and C is in siemens. A and C come from the open-circuit test, B and D from the short-circuit test.
• For a passive line AD − BC = 1, and for a symmetrical line A = D. Use these to check every answer.
• Short line: A = D = 1, B = Z, C = 0. Nominal π: A = D = 1 + YZ/2, B = Z, C = Y(1 + YZ/4). Long line: A = cosh γl, B = Z_c sinh γl, C = sinh γl / Z_c.
• For blocks in a row, multiply their matrices in the same order as the power flows.
• At no load V_r = V_s/|A|. Because |A| < 1, long lines show the Ferranti effect and need shunt reactors.

This is the last step of our power system basics path. To practise, try the interactive lab. Then see how whole networks are described with the Z bus and Y bus.