Radio
Moderators: kgudger, bfinio, MadelineB, Moderators
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dafender
- Posts: 3
- Joined: Sat Jan 03, 2009 7:09 pm
- Occupation: Teacher
- Project Question: Why 2000 ohm headphones must be used in a crystal reveiver
- Project Due Date: Jan 30
- Project Status: I am finished with my experiment and analyzing the data
Radio
Please explain why a 2000 ohm impediance earphone is needed in a crystal set instead of 8 ohm head sets
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Craig_Bridge
- Former Expert
- Posts: 1297
- Joined: Mon Oct 16, 2006 11:47 am
Re: Radio
You did not give us much to go on in terms of particulars. Can you point us to a circuit you are trying to build?
Theory of power transfer efficiency:
If the source and load impedances match, optimal power will be transfered between the source and the load. In your case, the headset is the load and the source would be some crystal radio circuit.
From my experience with 8 ohm speakers and headphones:
8 ohm headsets are typically designed to be replacements for 8 ohm speakers which typically take 1/8 watt or more of power to be heard from several feet away. In order not to be overly loud sitting right next to your ears, 8 ohm headsets are typically not very efficient at converting electrical power into sound vibrations. In this case, the electrical source is ususally some aplifier designed to drive 8 ohm speakers/headphones.
High impedance headphones (2,000 ohms and above) are typically designed to be efficient at converting small amounts of electrical power into sound to be heard at very small distances.
From this, I can only assume that the crystal radio circuit that you are looking at calls for 2,000 ohm headsets because it does not have an audio amplifier capable of driving 8 ohm speakers and it was designed to be used with high impedance high efficiency headphones without amplification.
If you really want to drive 8 ohm speakers or headsets from a circuit that has an low power high impedance output, you could certainly buy or design and build an audio amplifier that has a high input impedance and an 8 ohm output impedance that has a volume control but it will require some source of power (battery or one of these small plug in charger units).
Theory of power transfer efficiency:
If the source and load impedances match, optimal power will be transfered between the source and the load. In your case, the headset is the load and the source would be some crystal radio circuit.
From my experience with 8 ohm speakers and headphones:
8 ohm headsets are typically designed to be replacements for 8 ohm speakers which typically take 1/8 watt or more of power to be heard from several feet away. In order not to be overly loud sitting right next to your ears, 8 ohm headsets are typically not very efficient at converting electrical power into sound vibrations. In this case, the electrical source is ususally some aplifier designed to drive 8 ohm speakers/headphones.
High impedance headphones (2,000 ohms and above) are typically designed to be efficient at converting small amounts of electrical power into sound to be heard at very small distances.
From this, I can only assume that the crystal radio circuit that you are looking at calls for 2,000 ohm headsets because it does not have an audio amplifier capable of driving 8 ohm speakers and it was designed to be used with high impedance high efficiency headphones without amplification.
If you really want to drive 8 ohm speakers or headsets from a circuit that has an low power high impedance output, you could certainly buy or design and build an audio amplifier that has a high input impedance and an 8 ohm output impedance that has a volume control but it will require some source of power (battery or one of these small plug in charger units).
-Craig
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philip
- Posts: 1
- Joined: Wed Jan 14, 2009 7:11 pm
- Occupation: student
- Project Question: Make your own low power am radio transmitter
- Project Due Date: January 17, 2009
- Project Status: I am conducting my experiment
Re: Radio
I am building a make your own low power AM radio transmitter I found on Science buddies for my science project and I have followed the instructions on wiring up the breadboard as per the specs. but when I power up the circuit nothing is working, can someone help me resolve why this is not working. I have follwed to the tee the exact diagram but I am at a loss. I think i hane not put jumpers on, and in the correct place.
Can someone guide me through of where I might be going wrong. My project is due on friday of this week and this thing is driving me up a wall.
Thank You for your help a
Can someone guide me through of where I might be going wrong. My project is due on friday of this week and this thing is driving me up a wall.
Thank You for your help a
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Craig_Bridge
- Former Expert
- Posts: 1297
- Joined: Mon Oct 16, 2006 11:47 am
Re: Radio
Assuming this is the project you are referring to: https://www.sciencebuddies.org/science- ... p024.shtml
We maybe able to help; however, you would have a better change at getting it done if you knew of an electronic technician or electrical engineer in your neighborhood that could help with trouble shooting via this forum just isn't going to happen quickly.
Do you have a multi-meter? If not, try and find somebody who has one that you can have help you.
Have you measured the voltage between pin 14 and pin 7 on the crystal oscillator package? You should get a reading of about +6 VDC if you put the + lead on pin 14 and the - lead on pin 7. If you don't get something close, you need to disconnect one of the battery leads and measure continuity with the ohms scale to verify the connections between the end of the wire that connects to the + battery lead through the matching transformer to pin 14 of the oscillator. It should read close to 0 ohms indicating you have good connections. If not, test along the way and see where the connections are bad. Measure the continuity of the wire from the - battery connection to pin 7. This one should measure 0 ohms.
What city do you live in and what frequency oscillator did you purchase?
If you are in the listening area of AM radio stations near 1000 KHz (for the 1 MHz) or 1230 KHz (for the 1.2288 MHz), you will have trouble with this project. Knowing your genera area will allow us to look up where the nearest radio stations with assigned frequencies near your oscillator frequence and determine find out what their power levels are and what their transmission patterns and operating hours are.
Do you have a working AM radio that will tune to 1000 KHz or 1230 KHz with a non-digital dial? If not, you really need a non-digital tuned AM radio for this project.
If there isn't an AM station at the same frequency in your area and there is power to the oscillator, you should be able to tune your radio through the oscillator frequency and hear a squeal that will start off as a high pitch and get lower and disappear and then come back and increase in pitch until you can't hear it as you VERY VERY slowly tune the radio through the oscillator frequency. If you get this behavior, then you should be able to tune to between the two low pitch points and drive the transformer with an audio source and have a working project.
We maybe able to help; however, you would have a better change at getting it done if you knew of an electronic technician or electrical engineer in your neighborhood that could help with trouble shooting via this forum just isn't going to happen quickly.
Do you have a multi-meter? If not, try and find somebody who has one that you can have help you.
Have you measured the voltage between pin 14 and pin 7 on the crystal oscillator package? You should get a reading of about +6 VDC if you put the + lead on pin 14 and the - lead on pin 7. If you don't get something close, you need to disconnect one of the battery leads and measure continuity with the ohms scale to verify the connections between the end of the wire that connects to the + battery lead through the matching transformer to pin 14 of the oscillator. It should read close to 0 ohms indicating you have good connections. If not, test along the way and see where the connections are bad. Measure the continuity of the wire from the - battery connection to pin 7. This one should measure 0 ohms.
What city do you live in and what frequency oscillator did you purchase?
If you are in the listening area of AM radio stations near 1000 KHz (for the 1 MHz) or 1230 KHz (for the 1.2288 MHz), you will have trouble with this project. Knowing your genera area will allow us to look up where the nearest radio stations with assigned frequencies near your oscillator frequence and determine find out what their power levels are and what their transmission patterns and operating hours are.
Do you have a working AM radio that will tune to 1000 KHz or 1230 KHz with a non-digital dial? If not, you really need a non-digital tuned AM radio for this project.
If there isn't an AM station at the same frequency in your area and there is power to the oscillator, you should be able to tune your radio through the oscillator frequency and hear a squeal that will start off as a high pitch and get lower and disappear and then come back and increase in pitch until you can't hear it as you VERY VERY slowly tune the radio through the oscillator frequency. If you get this behavior, then you should be able to tune to between the two low pitch points and drive the transformer with an audio source and have a working project.
-Craig
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dafender
- Posts: 3
- Joined: Sat Jan 03, 2009 7:09 pm
- Occupation: Teacher
- Project Question: Why 2000 ohm headphones must be used in a crystal reveiver
- Project Due Date: Jan 30
- Project Status: I am finished with my experiment and analyzing the data
Re: Radio
What I really wanted to know is the physics behind the use of 2000 ohm ear piece instead of 8 ohm head piece when building a crystal set. No audio amp is intended for use. Can you give me the math or electricial formulas for comparing the 8 ohm vs the 2000 ohm head sets so I can grasp why a higher resistance is needed to hear the audio. Thanks Dave
Craig_Bridge wrote:You did not give us much to go on in terms of particulars. Can you point us to a circuit you are trying to build?
Theory of power transfer efficiency:
If the source and load impedances match, optimal power will be transfered between the source and the load. In your case, the headset is the load and the source would be some crystal radio circuit.
From my experience with 8 ohm speakers and headphones:
8 ohm headsets are typically designed to be replacements for 8 ohm speakers which typically take 1/8 watt or more of power to be heard from several feet away. In order not to be overly loud sitting right next to your ears, 8 ohm headsets are typically not very efficient at converting electrical power into sound vibrations. In this case, the electrical source is ususally some aplifier designed to drive 8 ohm speakers/headphones.
High impedance headphones (2,000 ohms and above) are typically designed to be efficient at converting small amounts of electrical power into sound to be heard at very small distances.
From this, I can only assume that the crystal radio circuit that you are looking at calls for 2,000 ohm headsets because it does not have an audio amplifier capable of driving 8 ohm speakers and it was designed to be used with high impedance high efficiency headphones without amplification.
If you really want to drive 8 ohm speakers or headsets from a circuit that has an low power high impedance output, you could certainly buy or design and build an audio amplifier that has a high input impedance and an 8 ohm output impedance that has a volume control but it will require some source of power (battery or one of these small plug in charger units).
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kgudger
- Moderator
- Posts: 421
- Joined: Wed Sep 17, 2008 9:20 pm
- Occupation: electronic engineer
- Project Question: n/a
- Project Due Date: n/a
- Project Status: Not applicable
Re: Radio
Hi:
Essentially the power you get from a crystal tuner is extremely small. To be able to hear this, you need an earphone with the ability to make sounds from very little power. An 8 ohm earphone, like from an MP3 player, would not qualify, as it's used to getting a lot of power from the player.
The main equation that counts here is just ohm's law - the output of the tuner stage, which is an inductor, tends to look like a current source (over a small range of operation), so the larger resistance will give you a larger voltage (force) to drive the earphone. This is a gross simplification. The other issue here, as was mentioned, is matching the impedance of the driving circuit with the output circuit (the tuner with the earphone). This is already demonstrated in the oscillator circuit, where you use a transformer to match the output of the music player with the input of the oscillator. In that case, without the transformer, you would drag the player's output down so far it would be unusable. HTH.
Keith
Essentially the power you get from a crystal tuner is extremely small. To be able to hear this, you need an earphone with the ability to make sounds from very little power. An 8 ohm earphone, like from an MP3 player, would not qualify, as it's used to getting a lot of power from the player.
The main equation that counts here is just ohm's law - the output of the tuner stage, which is an inductor, tends to look like a current source (over a small range of operation), so the larger resistance will give you a larger voltage (force) to drive the earphone. This is a gross simplification. The other issue here, as was mentioned, is matching the impedance of the driving circuit with the output circuit (the tuner with the earphone). This is already demonstrated in the oscillator circuit, where you use a transformer to match the output of the music player with the input of the oscillator. In that case, without the transformer, you would drag the player's output down so far it would be unusable. HTH.
Keith