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I understand (and correct me if I'm wrong) that transmission power follows Ohm's law ($I = \frac{V}{R}$), that is that if I have a 50 Ω antenna that I know is transmitting at 1 W, then the current is:

$$ V = IR\\ P = VI \\ P = (IR)R = I^2R\\ \frac{P}{R} = I^2\\ I = \sqrt{\frac{P}{R}}=\sqrt{\frac{1}{50}}=0.141\\ I = 0.141\,\mathrm{A} $$

Then the voltage is

$$ V=IR=0.141\,\mathrm{A} \cdot 50\,\Omega=7.07\,\mathrm{A}\\ V=7.07\,\mathrm{A} $$

Is that voltage RMS, peak to peak, or something else?

If it's RMS, then peak to peak voltage is:

$$ V_{\text{PEAK}}=V_{\text{RMS}}\cdot\sqrt{2}= 7.07\cdot\sqrt{2}=7.07\cdot1.414=10\\ V_{\text{PEAK}}=10\,\mathrm{V}\\ V_{PP} = V_{PEAK}\cdot2=10\cdot2=20\,\mathrm{V} $$

Am I correct in saying that I need a 20V peak to peak sinewave signal to generate 1W RF output on the 50Ω antenna?

Is the maximum current I need to design for in my final amplifier 141 mA, or do I need to use $V_{\text{PP}}$ to calculate the maximum current? ($I=\frac{V}{R}=\frac{20\,\mathrm{V}}{50\,\Omega}=0.400\,\mathrm{A}$)

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The AC voltage and current you get from P=IE or Ohm's law in AC circuits are RMS values. I'll explain why.

If you have a constant voltage into a resistor, the power is $P=E^2/R$. That is, power is proportional to the square of voltage. The same is true for a constant current through a resistor: $P=I^2R$. Either of these equations can be derived by by substituting some arrangement of Ohm's law into $P=IE$.

Now say you have AC into a resistor. If you want to calculate the power, you can measure the voltage and current at each instant, multiply them together to get power, then average all these instants together to get the mean power.

What if you measure just the voltage? Assuming a resistive load, you can still calculate power from voltage measured at some instant by $P=E^2/R$. However, if you simply take the arithmetic mean of the voltages at each instant, then try to calculate power with that average, you miss an important fact: doubling the voltage quadruples the power. RMS is simply a way of averaging voltage that doesn't miss that.

We can make a simplified example. Let's just consider three instants at which we measured voltage and current into a 50 ohm load:

voltage  current  power
1V       0.02A    0.02W
2V       0.04A    0.08W
3V       0.06A    0.18W

At each of these instants, $P=IE$ and $E/I = 50 \Omega$. Let's calculate some different kinds of averages:

$$ \begin{align} P_{mean} &= \frac{0.02 + 0.08 + 0.18}{3} &\approx 0.0933\mathrm{W} \\ V_{mean} &= \frac{1+2+3}{3} &= 2\mathrm{V} \\ V_{RMS} &= \sqrt{\frac{1^2 + 2^2 + 3^2}{3}} &\approx 2.16\mathrm{V} \\ I_{mean} &= \frac{0.02 + 0.04 + 0.06}{3} &= 0.04\mathrm{A} \\ I_{RMS} &= \sqrt{\frac{0.02^2 + 0.04^2 + 0.06^2}{3}} &\approx 0.0432\mathrm{A} \\ \end{align}$$

Now then, for which of these does $P=IE$ hold?

$$ \begin{align} V_{mean} I_{mean} &= 2\mathrm{V} \cdot 0.04\mathrm{A} &= 0.08\mathrm{W} &\text{ wrong!} \\ V_{RMS} I_{RMS} &= 2.16\mathrm{V} \cdot 0.0432\mathrm{A} &= 0.0933\mathrm{W} &\text{ correct!} \end{align} $$

You are right that if you want to know the peak current your amplifier will need to supply, you need to convert this RMS value to something else. However, you got it backwards: $V_p = \sqrt{2} V_{RMS}$. So for the example above, your amplifier would be required to supply $\sqrt{2} \cdot 2.16\mathrm{V} = 3.05\mathrm{V}$ peak. Being AC, it would have to supply up to +3.05V and also down to -3.05V. Thus, the peak to peak amplitude is 6.10V.

Same math applies to current.

Note that this math of converting between peak and RMS values with $\sqrt{2}$ only applies to sine waves, which our example wasn't, so there was some wrongness there. However in radio, our transmissions are close enough to sine waves that assuming they are is usually a good enough approximation.

Note also that being able to RMS averages in equations like $P=IE$ or $P=I^2R$ only works for resistive loads. It works for reactive loads if what you care about is apparent power and not real power, and it doesn't work at all for non-linear loads.

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  • $\begingroup$ Ok, so to summarize, 1) Yes, the voltage is RMS, and 2) Yes, you need to convert to peak to peak (and I messed up the math, fixed), and 3) No, you need to use Vpp and Ipp to calculate maximum current in the amplifier. Is this correct? $\endgroup$ – Adam Davis Jan 28 '14 at 19:31
  • $\begingroup$ @AdamDavisKD8OAS pretty much...whether you need peak or peak-to-peak is going to depend somewhat on your particular amplifier topology. I think it most instances you will care about peak. Take some current for example, peak 100mA means also -100mA will be necessary, for 200mA peak-to-peak. But peak current is usually limited by a component's ability to dissipate heat, and peak power is reached at 100mA or -100mA -- it never sees 200mA. Likely, the negative current will probably be flowing through a different component than the positive current; this is certainly true of class AB amps. $\endgroup$ – Phil Frost - W8II Jan 28 '14 at 19:36

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