Resistance of power source line and primary circuit

7. RESISTANCE IN PRIMARY AND SECONDARY CIRCUITS

by Sam Belkin, MSEE

Suppose that the portable benchtop machine is operating from a 120 V single phase AC line. The welding rate is 4 welds 9 cycles each, and the duty cycle D.C. = 1 %. Assume that the resistance of the welded parts is 1 mOhm (10^-3 Ohm), the turn ratio of transformer n = 22.2, the resistance of primary winding R1 = 0.4 Ohm, and the resistance of secondary winding plus resistance of the electrodes and connections R2 = 0.8 mOhm (0.8 x 10^-3 Ohm). Also the welding machine is connected to a power line with a 6 ft long AWG 16 cord, and the leakage inductance Ls1 is negligible. Latter means that the impedance of Ls1 at 60 Hz much less than the primary circuit resistance (Rps + R1). First let us determine the resistance Rps which for this case is equal to the resistance of the two 6 ft long AWG 16 wires. From the wire chart we found that the AWG 16 wire has a resistance of 3.67 x 10^-3 Ohm/ft. Therefore for the two 6 ft wires Rps = 3.67 x 10^-3 x 6 x 2 = 44 x 10^-3 Ohm. Now we know all values for the equivalent circuit ( see Figure 2 in "How to determine parameters of transformer") and can easily calculate the parameters we can expect from the given machine.

The current flow in the primary circuit at the welding time:

Ipmax = Vp / [Rps + R1 + (R2 x n²) + (Rw x n²)] = 120 / [44 x 10^-3 + 0.4 + (0.8 x 10^-3 x 22.2²) + (10^-3 x 22.2²)] = 90.15 A.

Average current flow in the primary circuit:

Ip = Ipmax x DC = 90.15 x 0.01 = 9.0 A.

Voltage drop across the power cord:

Vpc = Ipmax x Rps = 90.15 x 44 x 10^-3 = 4 V.

Power loss into power cord:

Ppc = Ip² x Rps = 9.0² x 44 x 10^-3 = 3.6 W.

Note that the average current is significantly less than the maximum value. This average value may be used for the determination of the power cord wires' area. But for the voltage drop we are using the primary current value at the welding time (Ipmax).

Effective voltage applied to the primary winding:

Vpe = Vp x Rw x n² / [Rps + R1 + (R2 x n²) + (Rw x n²)] = Ipmax x Rw x n² = 90.15 x 10^-3 x 22.2² = 44.43 V.

Note that only 44.43 V from 120 V or just about 37 % are used for the welding. Another 67 % of the AC line voltage are lost and are practically converted to heat. These numbers show how good is the efficiency of the welding machine and we will talk about it later.

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Resistance of the secondary circuit

Now it is time to analyze the secondary circuit. The open circuit output voltage can be easily found from the transformer ratio formula:

Vo = Vp / n = 120 / 22.2 = 5.4 V.

Effective secondary voltage applied to the welded parts:

Vse = Vpe / n = 44.43 / 22.2 = 2 V.

Welding current for given machine and welded parts:

Iw = Vse / Rw = 2 / 10^-3 = 2000 A.

Voltage drop across the secondary circuit resistance R2:

V_R2 = Iw x R2 = 2000 x 0.8 x 10^-3 = 1.6 V.

From the 5.4 V of open circuit voltage only 2 V are used for welding. 1.6 V were lost in the secondary circuit. Lets see where are the other 1.8 V. Definitely the primary circuit is responsible for that. The voltage drop across the primary circuit resistance Rps + R1 may be found as

V_R1ps = Ipmax x (Rps + R1) = 90.15 x [(44 x 10^-3) + 0.4] = 40 V.

This 40 V drop will give

V_R1ps/ n = 40 / 22.2 = 1.8 V

drop in the secondary circuit. This number shows that the primary circuit delivers more losses than the secondary, and that apparently a designer has enough room there for improvements. A good way to evaluate a design is to measure the efficiency of a transformer.

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