People tell me I shall not use my CB-radio station without an antenna connected or it will result in damage of the device. Is this true and why is this?
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6\$\begingroup\$ Indeed, it can. When testing, you should operate the transmitter into a resistive dummy load. For the legal CB power limit this will not be an expensive item and can be air cooled - but in other cases it could be a large resistor submerged in a container of oil. \$\endgroup\$– Chris StrattonOct 23, 2017 at 12:05
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\$\begingroup\$ Used to work for a company that built radios and base station transmitters, and in the lab we always had to use large attenuators and dummy loads. It would damage the transmitters - saw it happen. \$\endgroup\$– JPhi1618Oct 23, 2017 at 18:04
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\$\begingroup\$ With no antenna, where do you think the power goes when you transmit? \$\endgroup\$– David SchwartzOct 24, 2017 at 3:18
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3\$\begingroup\$ @DavidSchwartz My idea was: No antenna, no current which gets drawn from the transmitter. But apparently current flows regardless of antenna connected. \$\endgroup\$– arminbOct 24, 2017 at 7:25
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\$\begingroup\$ Back in the old days when we hams built our own, I used a light bulb for a dummy load. The brightness of the light was a confirmation that my rig was putting out power. But a CB radio cannot be tuned to match a load, so that is probably not an option for you. \$\endgroup\$– richard1941Oct 27, 2017 at 0:09
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3 Answers
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You could potentially break some radio transmitters by operating them without the antenna connected.
Several things combine to make this possible.
First, it's difficult to make power at RF frequencies, so the power devices are often fairly fragile, and run near their limiting conditions.
Secondly, radio signals passing down a transmission line behave in a way most non-engineers don't expect. They get reflected from an open circuit, that is, from a connector without the antenna connected.
The antenna provides a load, absorbing the energy travelling down the transmission line from the power device to the antenna. If that energy is reflected, then for a well matched transmitter, it has the potential to double the voltage seen at the transmitting device. Depending on the length of the line, it might alternatively double the current, which usually isn't as bad. So a transistor that's already close to its limits could be pushed beyond them.
If the transmitter's output device is poorly matched, then there's the potential for much higher voltage magnification.
Usually, the transmitter will be designed so that the power device is sufficiently within its ratings that it won't fail into any load, as blowing up into no load is not very nice behaviour. But sometimes, especially for low cost and high power amplifiers, like your CB kicker for instance, it won't be, and needs that load to prevent reflections.
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2\$\begingroup\$ Good answer - would benefit from description of antenna matching and VSWR. \$\endgroup\$– FlorisOct 23, 2017 at 12:37
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1\$\begingroup\$ It is possible to divert any reflected power into a dummy load, eg. with a circulator. Such precision devices are not cheap, and would not be found in a consumer CB:( \$\endgroup\$ Oct 23, 2017 at 21:11
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7\$\begingroup\$ I've deliberately not delved into transmission line theory. If you're an RF engineer, you don't need this answer. I figure that at the level of the OP, I've said just enough that's true, and doesn't have to be unlearned if you want to take it to the next level. \$\endgroup\$– Neil_UKOct 24, 2017 at 10:10
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Just to complement the excellent answer of Neil_UK and stress the fact that at RF frequencies voltages and currents don't really behave as those nice entities you know from KCL and KVL.
You must drop Kirchhoff's laws and get your hands dirty with transmission lines theory, where the same concepts of voltage and current become a lot weirder!
In other words, voltages and currents behave more like EM waves that are generated at the transmitter and travel along the cable until they reach the antenna, which "allow them to radiate into space".
Disconnecting the antenna is like putting a brick wall (well, more like a mirror in reality) in front of the wave. The power has to go somewhere, and the open end of the cable cannot make it radiate efficiently, so it bounces back almost entirely, until it reaches the final stage of the transmitter, which is not supposed to sink that power (it should operate as a source).
Hence, as Neil pointed out, if there are no protection devices to dump the reflected wave somewhere, all the reflected power will be dissipated inside the amp, which is not usually designed to do that. POOF! (magic smoke escapes!)
In a sense it's like a rock concert: the band on stage is not deafened to death because the huge loudspeakers are directed toward the people gathered in front of them in a big open space. Put a big wall just before the stage and the band will be blown away by the reflected soundwaves!
answered Oct 23, 2017 at 18:52
LorenzoDonati4Ukraine-OnStrikeLorenzoDonati4Ukraine-OnStrike
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8\$\begingroup\$ I really like the rock concert analogy and the mental image it gives. \$\endgroup\$ Oct 23, 2017 at 23:42
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2\$\begingroup\$ ...and if the transmitter is well-designed but not able to sink all the power, it will at least have protective circuits that cut back power when the reflection is detected, to hopefully limit the damage (and in the best of cases, avert damage entirely). Still not a good thing to have happen to the equipment, but better than dumping all that power (whether in the form of voltage or current) from the wrong side into the RF power amplifier stage! \$\endgroup\$– userOct 24, 2017 at 11:32
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2\$\begingroup\$ @JeffBowman Peace and love, man! :-) .... (or also peace and matched load) :-D \$\endgroup\$ Oct 24, 2017 at 18:39
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What destroys finals is avalanche breakdown, often due to parasitic BJT latching, not reflected power. When the output power amplifier detects the high reflected voltage, it enables the foldback circuit, which reduces output power somewhat proportionally, protecting the finals. Yes, transmitting without an antenna can destroy your finals, but due to the high voltage needed to maintain the high current flowing through the pullup inductor, since it can't flow through the non-existent load. The excessive voltage then causes the FET to go into avalanche.
