How long should antenna wire be to receive amateur band?

[dietermoreno]

How long should antenna wire be to receive amateur band?

I have a Grundig radio receiver that is hybrid super heterodyne and phase locked loop with matrix keypad. I recieved it from my grandparents who used it to listen to German radio stations and now they no longer use it because it has been replaced by streaming media on the internet.

It can receive LW, MW, and SW bands in AM. It can receive VHF band in FM. It can not receive VHF band in AM. It can receive side bands with a switch to select from AM to USB or LSB. The local oscillator (BFO) can also be played with. The band width can also be played with. The manual says that amatures use the side bands.

The receiver is in my second floor of my house which is a finished attic. There is a hallway in the finished attic that runs from the music room to my bedroom door. I would run the antenna wire on the ceiling along the hallway from the dipole of the telescopic antenna of the receiver to my bedroom door. This distance would be about 12 metres.

Would 12 metres allow me to receive up to 48 metres since an effective antenna must be at least one quarter of the wavelength?

 

[davenn]

A 12m wire will let you receive anything in the HF bands, just not overly efficiently

What you really ned to ask yourself is .... is there one particular band I would prefer to do the most listening on ?

if there is one in particular then do a 1/2 wave dipole for that band for best efficiency but it will still happily receive the other bands with just a little less efficiency

The main HF amateur bands are from 80m ( 3.5MHz) to 10m (30MHz) yes there is a 160m band but I wouldnt worry about it.
The other way would be to do a dipole for say the 20m band, ie. roughly middle of the range and you will get reasonable response from 80m to 30m

Dave

 

[dietermoreno]

What is a 20 m half wave dipole in noob language?

Do you mean have a 10 m long wire and connect it to a dipole made out of scrunched up aluminum foil?

 

[duke37]

A 20m halfwave dipole will be a wire or aluminium tubing 10m long.

This should be split in the middle and connected to some feeder cable. 50 ohm or 75 ohm co-ax is often used and will be adequate for reception. Much more accurate construction and matching is necessary for transmission.

The dipole should be kept away from noise sources (switch mode power supplies etc) and the ends should be insulated, plastic cord will do this.

I do not know what you would use scrunched up aluminium for other than a barbie.

If your receiver is sensitive then a simple random wire will do, the signal level will be lower but the noise will also be lower so the signal/noise ratio wil be not much worse.

 

[dietermoreno]

Okay well after reading on some forums of the dangers of connecting an antenna wire to ground on an outlet plate, not to mention that then your antenna would pick up all the noise of all of your appliance, I decided that is both dangerous and will cause too much interference to connect the antenna wire ground to the ground on an outlet plate.

Unfortunately the pipes in my house are PVC.

On the bright side, my radio receiver is right next to a door that opens to an outdoor stair case which leads up to the attic from the side yard. So I could run the antenna wire down the outdoor stair case and bury it in the ground several inches to establish ground and the run down the stair case also establishes the necessary antenna length.

 

[(*steve*)]

The antenna wire IS NOT connected to ground.

 

[davenn]

connecting the antenna to the ground will just ground any signal that the antenna picks up, why on earth would you want to do that ?
it would defeat the whole purpose of having an external antenna

the external antenna goes to the input circuit of the receiver. If it doesnt have an external antenna input connection, then a few turns of the wire around the ferrite rod antenna inside the radio will improve reception


Dave

 

[dietermoreno]

That's weird that you said that grounding an antenna will attenuate the signal, because I found dozens of search results on forums saying that grounding an antenna is the key to improving reception of said antenna.


Well here in this image of regenerative receiver skematic it shows that the antenna is grounded:

In the image it looks like the antenna is split and has one end go to ground and the other go into the air.

So is that what I want to do?

I could have two wires running down the outdoor flight of stairs. One wire is burried several inches in the ground, while the other wire hangs in the air. Would this work?

Yes my radio has a jack to connect an aerial, but first I want to start with an absolutely homemade antenna madefrom just some wire at the hardware store. So I will wind it around the telescopic antenna.

So it would be like this:

telescopic antenna

^
^
^
signal
^
nut and bolt ---->---->----->
^ V
^ return
^ V
open air V
^ V
^ V
^ V
RF wave propogation signal V
V
V
V
ground



You can see my diagram correctly if you quote me.

The direction of the arrows in my diagram is reversed for transmission.

My radio receiver can not transmit, it can only receive.



Well so maybe I will understand antenna principles better if I look at simpler circuit.

So here is the simplest radio transmitter circuit -- the spark gap transmitter:

So with the guitar spark gap transmitter in my previous post, the ground wire burried outside in the ground should be connected to the ground wire inside my guitar, and the antenna wire should be connected to an instrument cable connected to my guitar output jack.

 

[(*steve*)]

You are grounding the ground connection, not the antenna connection.

If you had even the smallest understanding you would note that there is no low impedance connection between the two at the operating frequency of the antenna (even if there is a DC connection)

 

[dietermoreno]

You are grounding the ground connection, not the antenna connection.

If you had even the smallest understanding you would note that there is no low impedance connection between the two at the operating frequency of the antenna (even if there is a DC connection)

Oh! Okay!

So I had it right with my guitar spark gap transmitter analogy but wrong with connecting it to my radio receiver.

I see that of course the antenna should not be grounded because then the AC signal transmitted or received will just flow to ground and no signal will be transmitted or received since AC always "wants" to go to ground.

I have no clue what the people in those dozens of forums posts were thinking when they were talking about grounding the antenna to improve signal quality.

What should be grounded is the ground of the transmitter/receiver, NOT the RF signal output of the transmitter/ RF signal input of the receiver!

Since my radio receiver is already plugged into a wall outlet, it is already grounded, so there is no need to ground it.

Maybe intentionally grounding is only needed in battery powered radios since they are not already grounded by the wall outlet connected to the ground and that is what those forum posts were talking about.

Now that I think about it, all of the YouTube videos where I saw people sticking a wire into the ground of a wall outlet to use as the ground wire for a radio receiver, those videos were all of crystal receivers, which are battery powered.

 

[davenn]

Maybe intentionally grounding is only needed in battery powered radios since they are not already grounded by the wall outlet connected to the ground and that is what those forum posts were talking about.

NO, the active part of an antenna is NEVER directly grounded.

Dave

 

[dietermoreno]

Exactly. In those YouTube videos of crystal receivers which use a wire stuck in the ground of the wall outlet, the ground in the wall outlet is used to ground one end of the inductor to create a difference in potential that can be sent on a wire as an AC RF signal before it is demodulated into an audio signal by the crystal.

The antenna is never connected to the ground, because then that would ground any RF signal received resulting in no RF signals reaching the radio receiver.

I'm not sure what the resistor does in this schematic.

Perhaps the resistor is designed to increase the load on the circuit so that more voltage will flow through the circuit creating a greater amplitude audio signal (louder audio signal). Maybe that is why it says to use high impedence headphones instead of today's hi-fi low impedence headphones.

Unfortunately, I think that more load on the circuit would reduce the current, so I think adding a battery into the circuit would offset the current losses of the resistor while maintaining the load which will increase the amplitude of the audio signal.

I wonder if my guitar would have RF gain (voltage) increased if I buy a cheap small instrument cable and saw away at it exposing the shield and wire and connect a 100K resistor to the wire and then complete the circuit with another sawed open cheap small instrument cable connected to a male-to-male 1/4 inch instrument cable adapter and then I connect the instrument cable to my guitar amp and my guitar amp demodulates the RF which the RF gain (voltage) has been increased by the resistor and the current loss has been offset by I make it an active guitar with a 9 volt battery where the 100K resistor is connected.

 

[(*steve*)]

...so that more voltage will flow through the circuit...

Voltage doesn't flow.

I think adding a battery into the circuit would offset the current losses of the resistor while maintaining the load which will increase the amplitude of the audio signal.

No.

I wonder if my guitar would have RF gain (voltage) increased if I buy a cheap small instrument cable and saw away at it exposing the shield and wire and connect a 100K resistor to the wire and then complete the circuit with another sawed open cheap small instrument cable connected to a male-to-male 1/4 inch instrument cable adapter and then I connect the instrument cable to my guitar amp and my guitar amp demodulates the RF which the RF gain (voltage) has been increased by the resistor and the current loss has been offset by I make it an active guitar with a 9 volt battery where the 100K resistor is connected.

WOW, Just WOW.

No!

 

[davenn]

I'm not sure what the resistor does in this schematic.

The crystal earpiece doesn't pass DC the way a magnetic earphone does. You need a DC return for the diode to work properly - hence connecting a resistor across the earphone(s). You might have to experiment with the value.
That 100k is a lot higher than what many circuits use, 47 to 82 k seems more common.
If you are using really high impedance magnetic headphones/earphones, then the resistor isnt necessary

You CANNOT use low impedance hear(ear) phones, they wont work

Dave

 

[dietermoreno]

Well so what if I was using modern hi fi headphones and my crystal guitar experiment wasn't working, except it only works amplified by my guitar amp with the Audacity software to remove all of the crap because it pics up so much crap, like it picks up more crap than signal, I think that's called SNR.

How do I convert it to work with modern hi fi headphones?

Does the 100k resistor convert it to work with modern hi fi headphones?

Oh and is the reason it picks up so much crap because the tuning isn't accurate, and any radio receiver uses the phenomenon of resonance to pick up less crap and greatly amplify the resonant frequency?

So if I added a vari-cap connected to the ground wire of my guitar connected to the output jack of my guitar, then that would improve the tuning so it picks up less crap?

And why wouldn't adding a battery offset the current loss of the 100K resistor?

I have it right that the whole motivation for high impedence headphones with crystal set is amplification by the means of increasing the load to increase the voltage in accordance with Ohm's Law, correct?

 

[BobK]

Increasing the load does not increase the voltage, it decreases the current.

The 100K resistor is there because crystal headphones act like a capacitor and will just charge up to the max voltage seen. The resistor bleeds off this charge and allows them to see the audio signal.

Bob

 

[dietermoreno]

Increasing the load does not increase the voltage, it decreases the current.

The 100K resistor is there because crystal headphones act like a capacitor and will just charge up to the max voltage seen. The resistor bleeds off this charge and allows them to see the audio signal.

Bob

Oh okay. Several websites say that the resistor is needed to bleed off charge, but they don't explain why this is needed.

So the reason it is needed is that the diode is not selective in what it demodulates, it demodulates EVERYTHING, but the resistor filters out the higher current demodulated, allowing the weak audio signal current to pass.

So that is what I was doing wrong why I am picking up so much more crap than signal, because I didn't have a resistor, so I was demodulating EVERYTHING and hearing EVERYTHING, instead of only hearing the weak audio signal.

So adding a battery across the resistor would defeat the purpose of the resistor, which the purpose of the resistor is indeed to decrease current to allow only the weak audio signals to pass while higher current can not pass. Its a current divider.

 

[davenn]

Does the 100k resistor convert it to work with modern hi fi headphones?

So the reason it is needed is that the diode is not selective in what it demodulates, it demodulates EVERYTHING, but the resistor filters out the higher current demodulated, allowing the weak audio signal current to pass.

Oh okay. Several websites say that the resistor is needed to bleed off charge, but they don't explain why this is needed.

Dieter
you obviously didnt even read what I wrote about the resistor ? ... I told you why it was needed

So that is what I was doing wrong why I am picking up so much more crap than signal, because I didn't have a resistor, so I was demodulating EVERYTHING and hearing EVERYTHING, instead of only hearing the weak audio signal.

this is just total rubbish doesnt mean anything
why do we even bother to answer your questions when you just ignore the answers and still continue to make oddball comments ??
it really gets frustrating. We are all trying to help you learn, but its a difficult process when you keep going against the flow


Dave

 

[(*steve*)]

So the reason it is needed is that the diode is not selective in what it demodulates, it demodulates EVERYTHING

You're right up to here, because the detector has almost *NOTHING* to do with selectiviety.

Did you look to the left? To those inductors and the variable capacitor? Sometimes the capacitor is called a "tuning capacitor" -- did you ever wonder why?

, but the resistor filters out the higher current demodulated, allowing the weak audio signal current to pass.

And then you descend into meaningless rubbish.

Honestly, when it's been explained to you at least twice already, why do you persist in ignoring the explanation?

 

[dietermoreno]

I wasn't ignoring the explanation.

I was confirming the explanation.

I was confirming that the crystal demodulates EVERYTHING, including mains hum, lights turning on and off in your and your neighbor's house, motor hum, lightning, and even the sun (in my car when I use a cassette tape adapter to connect a digital music player to my car stereo system using an unshielded pair of wires, I noticed that when no music is playing there is a hiss heard on my car stereo and that hiss is quieter at night; of course the hiss from the sun is the least of my problems compared to the whine from my spark plugs creating a spark gap transmitter).

I was confirming that since the crystal demodulates EVERYTHING, then a resistor is needed to block off the strong current of all the crap so that only the weak audio signals are heard instead of all the crap.

So then after I made those confirmations, I said that that is the reason why my previous attempts at crystal receivers only picked up hum instead of signals with the signals only being audible after alot of cleaning up to remove the hum in Audacity software.