Ohm’s Law for Ham Radio: Volts, Amps, Watts and Wire Loss

You do not need an engineering degree to run a good amateur radio station, but you do need a handful of simple relationships. The most useful of them is Ohm’s law, together with the power formulas that go with it. They tell you why a thin power cable makes your transceiver misbehave on transmit, how many volts are present on a dummy load or a feedline, why a 100 watt signal produces 70 volts on a 50 ohm cable and how to translate watts into decibels. This guide explains the formulas in plain language, then applies them to situations you will meet in the shack, with worked examples you can repeat with your own numbers.

The Three Quantities

Every circuit calculation in this guide uses three basic quantities.

• Voltage (V) is the electrical pressure that pushes charge around a circuit. It is measured in volts. A car battery or a station power supply delivers about 12 to 14 volts DC.
• Current (I) is the flow of charge, measured in amperes (amps). A transceiver on transmit may draw tens of amps.
• Resistance (R) opposes the flow of current and is measured in ohms. A dummy load is a 50 ohm resistance, and a half-wave dipole presents a resistance of about 73 ohms at its centre when it is resonant in free space.

The relation between them is the law named after Georg Ohm, which is explained on the Ohm’s law reference page. For a simple resistance, the current through it is proportional to the voltage across it. In formula form, V equals I times R. Rearranged, I equals V divided by R, and R equals V divided by I. If you know any two of the three quantities, you can find the third.

Power: Watts From Volts and Amps

Power is the rate at which energy is used or delivered, and it is measured in watts. The electric power relationship for direct current is P equals V times I. Combine it with Ohm’s law and you get two more forms that are very handy: P equals I squared times R, and P equals V squared divided by R. Each form is useful in a different situation.

The units are consistent: volts, amps, ohms and watts. If you work with milliamps or kilohms, convert to the basic units first, otherwise the answers will be wrong by factors of a thousand.

Example 1: How Much Does the Radio Draw?

A 100 watt HF transceiver does not draw 100 watts from the supply. It draws more, because the transmitter amplifier is not perfectly efficient and the rest of the power turns into heat. Many radios specify a current of around 20 amps at 13.8 volts on transmit, but always check the manual of your own model. Using the power formula, 13.8 volts times 20 amps is 276 watts of DC input. If 100 watts leave as RF, roughly 176 watts are dissipated as heat. That is why a radio needs a well-ventilated position, and why a power supply must be able to deliver the full current continuously, not just on paper.

It also shows the reason for choosing a supply with some margin. Our guide to choosing a ham radio power supply covers how to match a supply to the current of the radio and what to look for in terms of regulation and noise. A radio that appears to work but resets or loses power on voice peaks often has a supply or a cable that cannot deliver the peak current.

Example 2: Voltage Drop in the Power Cable

Copper wire has resistance, and the thinner and longer the wire, the greater the resistance. When 20 amps flows through it, the voltage at the radio is lower than at the supply, and the lost power becomes heat in the wire. Typical resistances for solid copper at room temperature are about 1.59 milliohms per foot for 12 AWG wire, about 1.0 milliohm per foot for 10 AWG, and about 0.63 milliohms per foot for 8 AWG. Remember that the current has to go out and come back, so the length that counts is twice the distance between the supply and the radio.

Take a 10 foot run, or 20 feet of wire in total, carrying 20 amps.

• 12 AWG: 20 feet at 1.588 milliohms per foot is about 0.032 ohms. The voltage drop is 20 amps times 0.032 ohms, or about 0.64 volts. The wire dissipates 20 squared times 0.032, about 12.7 watts.
• 10 AWG: the resistance is about 0.020 ohms, so the drop is about 0.40 volts and the heat is about 8 watts.
• 8 AWG: the resistance is about 0.0126 ohms, so the drop is about 0.25 volts and the heat is about 5 watts.

A drop of 0.64 volts looks small, but when the supply voltage is 13.8 volts, it takes the radio down to about 13.2 volts, and many transceivers lose output power or distort as the voltage falls. The cable is not the only resistance in the circuit. Fuses, connectors and crimps add their own. This is why good practice is to use a heavy cable, short runs and well-made connections, such as the standard connectors described in our guide to Anderson Powerpole wiring. The fuse should protect the wire from overheating, so size it for the cable, not for the radio. Always follow the manufacturer’s instructions for the radio’s own power lead.

Example 3: Voltage and Current on a Dummy Load

A transmitter delivering 100 watts into a 50 ohm load creates an RF voltage that is much higher than any battery in the shack. Use V equals the square root of P times R: the square root of 100 times 50 is the square root of 5000, which is about 70.7 volts RMS. The current is I equals the square root of P divided by R, the square root of 2, or about 1.41 amps. At 1500 watts into 50 ohms, the voltage is about 274 volts RMS and the current is about 5.5 amps. The voltage on the feedline is no longer a trivial number, and that matters for connector spacing, insulation and safety.

The same arithmetic applies to the dummy load itself, which must be able to dissipate the power as heat. Our dummy load guide explains the ratings, the duty cycle and the cooling. A load rated for 100 watts continuous is not necessarily suitable for a long digital transmission at that level.

You can build a 50 ohm load from ordinary non-inductive resistors. Two 100 ohm resistors in parallel make 50 ohms. Four 200 ohm resistors in parallel also make 50 ohms, and share the power equally, so each one dissipates a quarter of the total. For parallel resistors of equal value, divide the value by the number of resistors. For resistors in series, add the values.

Example 4: Why Ladder Line Has High Voltage

Because P equals V squared over R, a higher impedance means a higher voltage for the same power. A 100 watt signal on a 450 ohm balanced feedline, when it is matched, corresponds to a voltage of the square root of 100 times 450, which is about 212 volts RMS. The peak is about 1.4 times higher. When the SWR is high, the voltages at certain points on the line rise further. That is why open wire feeders need adequate spacing and why you should never touch the line during transmission. In the other direction, a low impedance load such as a very low radiation resistance antenna needs large currents, which is why connections and loading coils for short antennas must be low in resistance.

Antennas, Resistance and Efficiency

An antenna presents a feedpoint resistance made of two parts: the radiation resistance, which represents power sent into space, and the loss resistance, which represents power turned into heat in wire, coils, ground and nearby objects. Efficiency is the radiation resistance divided by the total. If an electrically short antenna has a radiation resistance of 5 ohms and a loss resistance of 15 ohms, its efficiency is 5 divided by 20, or 25 percent, and the other 75 percent of the transmitter power is wasted as heat. The formulas are the same ones you use for a cable. This is the reason that short antennas with poor ground systems perform badly even when the SWR is low.

Power Ratios and Decibels

Ratios of power are normally expressed in decibels because the numbers are more manageable and because the ear and the S-meter respond to them in a roughly logarithmic way. The decibel for a power ratio is 10 times the base 10 logarithm of the ratio of the two powers. A few values are worth remembering:

• 3 dB is a factor of about 2 in power.
• 6 dB is a factor of about 4 in power, or 2 in voltage.
• 10 dB is a factor of 10 in power.
• 20 dB is a factor of 100 in power, or 10 in voltage.

For voltage or current ratios across the same resistance the multiplier is 20 instead of 10, because power goes with the square of the voltage. If you wonder why an antenna with 3 dB of gain is described as doubling the effective power, this is the answer. Our guide to dBi and dBd shows how antenna gain figures are expressed and compared. A line loss of 3 dB means that half the power is lost, and a loss of 10 dB means that only one tenth of the power reaches the antenna.

Practical Habits

• Write down the units. Most mistakes in electronics come from mixing milliamps and amps, or watts and milliwatts.
• Estimate before measuring. A quick calculation tells you whether a measurement makes sense. If the cable should drop 0.3 volts and the meter shows 2 volts, look for a bad connection.
• Use a meter with the right range. Measure the voltage at the radio terminals while transmitting, not at the supply. The difference is the drop in the cable.
• Remember that RF is different. Ohm’s law applies to resistances. For reactive loads and transmission lines, the voltage and the current can be out of step, and the real power also depends on the phase.
• Do not trust a rating blindly. A fuse, a connector or a switch is rated for a particular current and voltage. Check the real working conditions against them.
• Battery use. For portable work, calculate the energy you need. Our article on battery and power options for portable operating uses the same arithmetic to estimate how long a battery will last.

The Bottom Line

Ohm’s law and the power formulas are the simplest and most reusable tools in amateur radio. With them you can work out how much current your radio draws, how much voltage you lose in a cable, how many volts are on a dummy load or a feedline, how efficient an antenna is and what a number of decibels really means. Keep the table of formulas where you can see it, write down the units, calculate before you measure, and use heavy cable and good connections wherever the current is high. A few minutes of arithmetic often saves hours of troubleshooting and a good deal of money.

Frequently Asked Questions

What is Ohm’s law in simple terms?

The voltage across a resistance equals the current through it multiplied by the resistance. V equals I times R, so with any two values you can calculate the third.

How do I calculate power from volts and amps?

Multiply them. Power in watts equals volts times amps, so 13.8 volts at 20 amps is 276 watts of DC input power.

How much voltage is on a 50 ohm line at 100 watts?

About 70.7 volts RMS and 1.41 amps, found with V equals the square root of P times R. At 1500 watts it is about 274 volts RMS.

Why does a thin power cable cause problems?

It has more resistance, so at high current it drops more voltage and wastes more power as heat. The radio then sees a lower voltage on transmit and may lose output or distort.

Does Ohm’s law apply to antennas?

It applies to the resistive parts, such as radiation resistance and loss resistance. Efficiency is the ratio of radiation resistance to the total feedpoint resistance, though reactive effects also matter.

How do I convert watts to decibels?

Take 10 times the base 10 logarithm of the power ratio. A factor of 2 is about 3 dB, a factor of 10 is 10 dB and a factor of 100 is 20 dB.