Circuit : Andy Collinson
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How far away can a transmitter be heard? The question depends on many factors, transmitter power, antenna, antenna gain, height, efficiency, SWR and probagation conditions. There are also losses in the antenna, antenna feedline etc. I'll start with this example circuit:
The general equation for calculating field strength is below:
| E = | √ 30 Pt | |
| d |
Where d is distance in meters, E is the field strength in V/m and Pt is the total power from the transmitter. By finding your radio receivers sensitivity,(usually in the manual) then the equation can be transposed for distance:
| d = | √ 30 Pt | |
| E |
The next step is to calculate transmitter power. The 2 stage circuit above works from a 9 volt battery, its output frequency was measured to be 107.2MHz. The final common emitter stage of this circuit, develops power in the tank circuit, which is transferred to the antenna, in this case a 30cm telescopic whip. Most of the power is developed in the coil, there are three ways to calculate this:
P = VI cos θP = I2Z
| P = | V2 | |
| Z |
At resonance the voltage and current in the tank circuit are in phase. To estimate power the impedance of the tank circuit and either the voltage across it or series current are required. The problem in measuring thigh frequency RF voltages or currents is that most meters do not give accurate results at high frequencies. To estimate the AC collector current first calculate the DC collector current. The two values are slightly different, but as this is only an approximation, the error is not significant. To find the collector current, measure the DC voltage across the emitter resistor and use ohm's law. (Do not measure the collector volatge as the impedance in the tank circuit is a maximum and will be shunted by the meters impedance. In my circuit, VE was measured at 2.99V across the 470 ohm emitter resistor. As IE = IC then collector current is:
This value will be substituted for the ac collector current. The impedance of the tank circuit is now found. At resonance the impedance is given by the following equation, where R is the DC resistance of the coil in the tank circuit. R was measured at 0.1 ohm. :
| Z = | L | = | 0.15 x 10-6 | = 100k |
| CR | 15 x 10-12 x 0.1 |
Output power will be worked out as I2Z but first the equivalent output circuit for the transmitter must now be drawn to calculate overall impedance. The tank circuit ( impedance 100K) is in parallel with the output impedance hoe of the transistor. hoe varies with collector current, but at 6mA is about 15k, see this output characteristic plot. Looking at the schematic there is a 3.3pf capacitor across base and emitter in series with a 470 ohm resistor, one end of the resistor is grounded. The power rails are decoupled with a 22n capacitor which is considered a short circuit at RF. The equivalent output circuit now looks as below:
The 3.3p capacitor has an impedance of around 450 ohms at 107.2 MHz. The overall impedance is therefore:-
40k // 100k // (450+470) = 891.3 ohms
Having now found the impedance, the approximate power in the tank circuit can be calculated:- P = I2Z = 6.36x10-3 x 891.3 = 0.036 Watt
Estimated Distance
Having now calculated transmitter power, Pt an ideal range of the transmitter can be worked out. If the radio receiver has a known sensitivity, e.g.
50uV/meter the distance the signal could be received is :
| d= | √30 Pt | = | √ 30 x 0.036 | = 20600 Metres |
| E | 50 x 10-6 |
This answer is an ideal answer. It assumes that all power in the tank circuit is transferred to the antenna without loss, the impedance of the tank circuit is matched to the antenna, there is no standing wave ration and no propagation effects. This is never the case as losses always occur in a real circuit due to dielectric loss, imperfect impedance matching, and other factors.
Realistic DistanceIn a commercial transmitter, the quality of components, and efficiency of the antenna are much better than anything built by a home constructor. When making a transmitter the hobbyist will use small lengths of wire or telescopic rods. These are never very efficient, and you may find that only about 1 or 2% of the energy in the tank circuit gets coupled to the transmitter. This has an impact on the distance that the signal may be received, and may be a small fraction of the calculated distance.
Without properly calibrated equipment is is difficult to estimate the losses. If the antenna is only 1% efficient then in our case, the transmitted distance would be 1% of the calculated value or just 206 metres. If the receiver used has a greater sensitivity than 50uV/M then the signal may be heard over a greater distance.