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could one of you do me a small favor?
Some knucklehead in misc.transport.trucking is showing off what a real ass
he is in the way of knowledge. That should take you to the proper thread. If not, look for the post from "mad dog" posted at 9:35 am. I kid you not, these are his words: quote Keep the P=IxE formula in mind when choosing a mobile amplifier. Power = Current x Voltage or Current = Power / Voltage P=Power or Wattage I =Current draw E=Energy or Voltage Example: Tommy truck driver wants to install a 1200 watt spurious emissions generator on his Kenworth with a 12VDC electrical system. We just learned that Current draw is calculated by dividing the Voltage into the Wattage so if the 1200 watt cataract producer has a Input Voltage of 12 Volts then the Current draw will be............................... One Hundred Amps (100). Yes folks you read that right, that is more current draw than most alternators can produce.So please be realistic when adding accessories to your already marginal electrical system or you may find yourself playing the "Prince of Darkness" on a very wet and rainy night. -- 714 Sandpile, The Mad Dog wavin' good bye /quote He is wrongfully assuming that RF power output is equal to saying 12volts*100amps. Assuming the rated output power is 1200 watts. As I have dealt with radios for years, I know for a fact that this is only an idiot's explanation. He's big bad mister trucker trying to show off how smart he is. I have tried to explailn it to him and he will not believe anything other than what he has written. So if one of you could come in to the thread and explain how damn wrong he is, I would appreciate it. thanks for the time. |
Richard wrote:
He is wrongfully assuming that RF power output is equal to saying 12volts*100amps. Assuming the rated output power is 1200 watts. As I have dealt with radios for years, I know for a fact that this is only an idiot's explanation. He's big bad mister trucker trying to show off how smart he is. I have tried to explailn it to him and he will not believe anything other than what he has written. ... Actually, he's probably low. That would be true only if the efficiency of the device is 100% and most of them are probably no where near that efficient. So explain how he's wrong, other than that? |
Richard wrote:
He is wrongfully assuming that RF power output is equal to saying 12volts*100amps. Assuming the rated output power is 1200 watts. As I have dealt with radios for years, I know for a fact that this is only an idiot's explanation. He's big bad mister trucker trying to show off how smart he is. I have tried to explailn it to him and he will not believe anything other than what he has written. ... Actually, he's probably low. That would be true only if the efficiency of the device is 100% and most of them are probably no where near that efficient. So explain how he's wrong, other than that? |
In New Youirk -- We Say "Forget A Bout It." Why talk to a concrete wall.
If you try to correct all the false stuff on the radio groups or anywhere else -- you would have to hire a full time team of a dozen people (at least). Caveat Lector -- Let the reader beware and leave it go at that. Regarding the power calculation -- see URL: http://www.ameritron.com/ameritron/p...rodid=ALS-500M This is a 400 watt CW (500W PEP) mobile amplifier and it "requires 14 VDC at 80 amperes peak current for PA transistors and separate line for 14 - 16 VDC at 4 amperes for control and bias circuits.That ought to help. Tis a matter of efficiency You do the math Elmer E Ing --------------------------------------------------- "Richard" anom@anom wrote in message ... Some knucklehead in misc.transport.trucking is showing off what a real ass he is in the way of knowledge. That should take you to the proper thread. If not, look for the post from "mad dog" posted at 9:35 am. I kid you not, these are his words: quote Keep the P=IxE formula in mind when choosing a mobile amplifier. Power = Current x Voltage or Current = Power / Voltage P=Power or Wattage I =Current draw E=Energy or Voltage Example: Tommy truck driver wants to install a 1200 watt spurious emissions generator on his Kenworth with a 12VDC electrical system. We just learned that Current draw is calculated by dividing the Voltage into the Wattage so if the 1200 watt cataract producer has a Input Voltage of 12 Volts then the Current draw will be............................... One Hundred Amps (100). Yes folks you read that right, that is more current draw than most alternators can produce.So please be realistic when adding accessories to your already marginal electrical system or you may find yourself playing the "Prince of Darkness" on a very wet and rainy night. -- 714 Sandpile, The Mad Dog wavin' good bye /quote He is wrongfully assuming that RF power output is equal to saying 12volts*100amps. Assuming the rated output power is 1200 watts. As I have dealt with radios for years, I know for a fact that this is only an idiot's explanation. He's big bad mister trucker trying to show off how smart he is. I have tried to explailn it to him and he will not believe anything other than what he has written. So if one of you could come in to the thread and explain how damn wrong he is, I would appreciate it. thanks for the time. |
In New Youirk -- We Say "Forget A Bout It." Why talk to a concrete wall.
If you try to correct all the false stuff on the radio groups or anywhere else -- you would have to hire a full time team of a dozen people (at least). Caveat Lector -- Let the reader beware and leave it go at that. Regarding the power calculation -- see URL: http://www.ameritron.com/ameritron/p...rodid=ALS-500M This is a 400 watt CW (500W PEP) mobile amplifier and it "requires 14 VDC at 80 amperes peak current for PA transistors and separate line for 14 - 16 VDC at 4 amperes for control and bias circuits.That ought to help. Tis a matter of efficiency You do the math Elmer E Ing --------------------------------------------------- "Richard" anom@anom wrote in message ... Some knucklehead in misc.transport.trucking is showing off what a real ass he is in the way of knowledge. That should take you to the proper thread. If not, look for the post from "mad dog" posted at 9:35 am. I kid you not, these are his words: quote Keep the P=IxE formula in mind when choosing a mobile amplifier. Power = Current x Voltage or Current = Power / Voltage P=Power or Wattage I =Current draw E=Energy or Voltage Example: Tommy truck driver wants to install a 1200 watt spurious emissions generator on his Kenworth with a 12VDC electrical system. We just learned that Current draw is calculated by dividing the Voltage into the Wattage so if the 1200 watt cataract producer has a Input Voltage of 12 Volts then the Current draw will be............................... One Hundred Amps (100). Yes folks you read that right, that is more current draw than most alternators can produce.So please be realistic when adding accessories to your already marginal electrical system or you may find yourself playing the "Prince of Darkness" on a very wet and rainy night. -- 714 Sandpile, The Mad Dog wavin' good bye /quote He is wrongfully assuming that RF power output is equal to saying 12volts*100amps. Assuming the rated output power is 1200 watts. As I have dealt with radios for years, I know for a fact that this is only an idiot's explanation. He's big bad mister trucker trying to show off how smart he is. I have tried to explailn it to him and he will not believe anything other than what he has written. So if one of you could come in to the thread and explain how damn wrong he is, I would appreciate it. thanks for the time. |
"Z.Z." wrote in message ... Richard wrote: He is wrongfully assuming that RF power output is equal to saying 12volts*100amps. Assuming the rated output power is 1200 watts. As I have dealt with radios for years, I know for a fact that this is only an idiot's explanation. He's big bad mister trucker trying to show off how smart he is. I have tried to explailn it to him and he will not believe anything other than what he has written. ... Actually, he's probably low. That would be true only if the efficiency of the device is 100% and most of them are probably no where near that efficient. So explain how he's wrong, other than that? He's mixing RF power with line power. As I tried to explain it to him, if you were to put a voltmeter at the RF connector and keyed down, you would not see 12 volts at 100 amps as he claims. In order to do that, you'd need an antenna cable an inch thick for the center wire. Plus the fact, that his theory suggests the antenna is receiving that much juice, would ultimately fry the antenna every time you keyed down. Standard RF coax such as RG8 does not handle more than a few amps at the most simply for the fact of it's size. Plus the fact that internal wiring of the amplifier would also have to be capable of handling the extreme amperage. I don't think circuit boards can handle it. Let alone any transistors, resistors, or capacitors. Do they even make a 1200 watt resistor? I also tried to compare his theory with that of an inverter. The rated output power is by no means any where near what the rated input power is. Obviously, to many people just do not understand the bare basics of electronics and radios. |
"Z.Z." wrote in message ... Richard wrote: He is wrongfully assuming that RF power output is equal to saying 12volts*100amps. Assuming the rated output power is 1200 watts. As I have dealt with radios for years, I know for a fact that this is only an idiot's explanation. He's big bad mister trucker trying to show off how smart he is. I have tried to explailn it to him and he will not believe anything other than what he has written. ... Actually, he's probably low. That would be true only if the efficiency of the device is 100% and most of them are probably no where near that efficient. So explain how he's wrong, other than that? He's mixing RF power with line power. As I tried to explain it to him, if you were to put a voltmeter at the RF connector and keyed down, you would not see 12 volts at 100 amps as he claims. In order to do that, you'd need an antenna cable an inch thick for the center wire. Plus the fact, that his theory suggests the antenna is receiving that much juice, would ultimately fry the antenna every time you keyed down. Standard RF coax such as RG8 does not handle more than a few amps at the most simply for the fact of it's size. Plus the fact that internal wiring of the amplifier would also have to be capable of handling the extreme amperage. I don't think circuit boards can handle it. Let alone any transistors, resistors, or capacitors. Do they even make a 1200 watt resistor? I also tried to compare his theory with that of an inverter. The rated output power is by no means any where near what the rated input power is. Obviously, to many people just do not understand the bare basics of electronics and radios. |
Yon knucklehead is assuming 100% efficiency or doesn't know the meaning of
the term. So ask Yon knucklehead how one builds an amplifier that yields 1200 Watts of DC Power into 1200 Watts of RF power ?? Lots of folks would like to know how to do that. Power out divided by power in times 100 = % efficiency "Richard" anom@anom wrote in message ... Some knucklehead in misc.transport.trucking is showing off what a real ass he is in the way of knowledge. That should take you to the proper thread. If not, look for the post from "mad dog" posted at 9:35 am. I kid you not, these are his words: quote Keep the P=IxE formula in mind when choosing a mobile amplifier. Power = Current x Voltage or Current = Power / Voltage P=Power or Wattage I =Current draw E=Energy or Voltage Example: Tommy truck driver wants to install a 1200 watt spurious emissions generator on his Kenworth with a 12VDC electrical system. We just learned that Current draw is calculated by dividing the Voltage into the Wattage so if the 1200 watt cataract producer has a Input Voltage of 12 Volts then the Current draw will be............................... One Hundred Amps (100). Yes folks you read that right, that is more current draw than most alternators can produce.So please be realistic when adding accessories to your already marginal electrical system or you may find yourself playing the "Prince of Darkness" on a very wet and rainy night. -- 714 Sandpile, The Mad Dog wavin' good bye /quote He is wrongfully assuming that RF power output is equal to saying 12volts*100amps. Assuming the rated output power is 1200 watts. As I have dealt with radios for years, I know for a fact that this is only an idiot's explanation. He's big bad mister trucker trying to show off how smart he is. I have tried to explailn it to him and he will not believe anything other than what he has written. So if one of you could come in to the thread and explain how damn wrong he is, I would appreciate it. thanks for the time. |
Yon knucklehead is assuming 100% efficiency or doesn't know the meaning of
the term. So ask Yon knucklehead how one builds an amplifier that yields 1200 Watts of DC Power into 1200 Watts of RF power ?? Lots of folks would like to know how to do that. Power out divided by power in times 100 = % efficiency "Richard" anom@anom wrote in message ... Some knucklehead in misc.transport.trucking is showing off what a real ass he is in the way of knowledge. That should take you to the proper thread. If not, look for the post from "mad dog" posted at 9:35 am. I kid you not, these are his words: quote Keep the P=IxE formula in mind when choosing a mobile amplifier. Power = Current x Voltage or Current = Power / Voltage P=Power or Wattage I =Current draw E=Energy or Voltage Example: Tommy truck driver wants to install a 1200 watt spurious emissions generator on his Kenworth with a 12VDC electrical system. We just learned that Current draw is calculated by dividing the Voltage into the Wattage so if the 1200 watt cataract producer has a Input Voltage of 12 Volts then the Current draw will be............................... One Hundred Amps (100). Yes folks you read that right, that is more current draw than most alternators can produce.So please be realistic when adding accessories to your already marginal electrical system or you may find yourself playing the "Prince of Darkness" on a very wet and rainy night. -- 714 Sandpile, The Mad Dog wavin' good bye /quote He is wrongfully assuming that RF power output is equal to saying 12volts*100amps. Assuming the rated output power is 1200 watts. As I have dealt with radios for years, I know for a fact that this is only an idiot's explanation. He's big bad mister trucker trying to show off how smart he is. I have tried to explailn it to him and he will not believe anything other than what he has written. So if one of you could come in to the thread and explain how damn wrong he is, I would appreciate it. thanks for the time. |
Oh boy -- lets assume an antenna is totally resistive -- 50 ohms or better
still a dummy load of 50 ohms that exhibits no inductance or capacitance. Now since Power in watts = I squared R where I is the current and R is a pure 50 ohm resistance transpose and solve for I = square root of P over R I get about 4.9 amperes RF current now E=I times R so 4.9 times 50 = 244 volts RF volts That's what you should see at the antenna. Try that on old knucklehead. With any inductive reactance or capacitive reactance --- different ball game. Gurus check my math please Elmer E Ing "Richard" anom@anom wrote in message ... "Z.Z." wrote in message ... Richard wrote: He is wrongfully assuming that RF power output is equal to saying 12volts*100amps. Assuming the rated output power is 1200 watts. As I have dealt with radios for years, I know for a fact that this is only an idiot's explanation. He's big bad mister trucker trying to show off how smart he is. I have tried to explailn it to him and he will not believe anything other than what he has written. ... Actually, he's probably low. That would be true only if the efficiency of the device is 100% and most of them are probably no where near that efficient. So explain how he's wrong, other than that? He's mixing RF power with line power. As I tried to explain it to him, if you were to put a voltmeter at the RF connector and keyed down, you would not see 12 volts at 100 amps as he claims. In order to do that, you'd need an antenna cable an inch thick for the center wire. Plus the fact, that his theory suggests the antenna is receiving that much juice, would ultimately fry the antenna every time you keyed down. Standard RF coax such as RG8 does not handle more than a few amps at the most simply for the fact of it's size. Plus the fact that internal wiring of the amplifier would also have to be capable of handling the extreme amperage. I don't think circuit boards can handle it. Let alone any transistors, resistors, or capacitors. Do they even make a 1200 watt resistor? I also tried to compare his theory with that of an inverter. The rated output power is by no means any where near what the rated input power is. Obviously, to many people just do not understand the bare basics of electronics and radios. |
Oh boy -- lets assume an antenna is totally resistive -- 50 ohms or better
still a dummy load of 50 ohms that exhibits no inductance or capacitance. Now since Power in watts = I squared R where I is the current and R is a pure 50 ohm resistance transpose and solve for I = square root of P over R I get about 4.9 amperes RF current now E=I times R so 4.9 times 50 = 244 volts RF volts That's what you should see at the antenna. Try that on old knucklehead. With any inductive reactance or capacitive reactance --- different ball game. Gurus check my math please Elmer E Ing "Richard" anom@anom wrote in message ... "Z.Z." wrote in message ... Richard wrote: He is wrongfully assuming that RF power output is equal to saying 12volts*100amps. Assuming the rated output power is 1200 watts. As I have dealt with radios for years, I know for a fact that this is only an idiot's explanation. He's big bad mister trucker trying to show off how smart he is. I have tried to explailn it to him and he will not believe anything other than what he has written. ... Actually, he's probably low. That would be true only if the efficiency of the device is 100% and most of them are probably no where near that efficient. So explain how he's wrong, other than that? He's mixing RF power with line power. As I tried to explain it to him, if you were to put a voltmeter at the RF connector and keyed down, you would not see 12 volts at 100 amps as he claims. In order to do that, you'd need an antenna cable an inch thick for the center wire. Plus the fact, that his theory suggests the antenna is receiving that much juice, would ultimately fry the antenna every time you keyed down. Standard RF coax such as RG8 does not handle more than a few amps at the most simply for the fact of it's size. Plus the fact that internal wiring of the amplifier would also have to be capable of handling the extreme amperage. I don't think circuit boards can handle it. Let alone any transistors, resistors, or capacitors. Do they even make a 1200 watt resistor? I also tried to compare his theory with that of an inverter. The rated output power is by no means any where near what the rated input power is. Obviously, to many people just do not understand the bare basics of electronics and radios. |
"Elmer E Ing" wrote in message news:VM_Ua.12050$ff.5596@fed1read01... Oh boy -- lets assume an antenna is totally resistive -- 50 ohms or better still a dummy load of 50 ohms that exhibits no inductance or capacitance. Now since Power in watts = I squared R where I is the current and R is a pure 50 ohm resistance transpose and solve for I = square root of P over R I get about 4.9 amperes RF current now E=I times R so 4.9 times 50 = 244 volts RF volts That's what you should see at the antenna. Try that on old knucklehead. With any inductive reactance or capacitive reactance --- different ball game. Gurus check my math please RF power is not electrical power. As audio power is not electrical power. Square root of P or 1200 watts in our case = 34.641 34.641/50 = 0.6928 amps. E=IIR (0.6928*0.6928)*50 = 23.99 volts. Therfor, the wire can easily handle the power. Use of a calculator helps. Your theory was correct. The decimal point was not. |
"Elmer E Ing" wrote in message news:VM_Ua.12050$ff.5596@fed1read01... Oh boy -- lets assume an antenna is totally resistive -- 50 ohms or better still a dummy load of 50 ohms that exhibits no inductance or capacitance. Now since Power in watts = I squared R where I is the current and R is a pure 50 ohm resistance transpose and solve for I = square root of P over R I get about 4.9 amperes RF current now E=I times R so 4.9 times 50 = 244 volts RF volts That's what you should see at the antenna. Try that on old knucklehead. With any inductive reactance or capacitive reactance --- different ball game. Gurus check my math please RF power is not electrical power. As audio power is not electrical power. Square root of P or 1200 watts in our case = 34.641 34.641/50 = 0.6928 amps. E=IIR (0.6928*0.6928)*50 = 23.99 volts. Therfor, the wire can easily handle the power. Use of a calculator helps. Your theory was correct. The decimal point was not. |
Richard wrote:
Oh boy -- lets assume an antenna is totally resistive -- 50 ohms or better still a dummy load of 50 ohms that exhibits no inductance or capacitance. Now since Power in watts = I squared R where I is the current and R is a pure 50 ohm resistance transpose and solve for I = square root of P over R I get about 4.9 amperes RF current now E=I times R so 4.9 times 50 = 244 volts RF volts That's what you should see at the antenna. Try that on old knucklehead. With any inductive reactance or capacitive reactance --- different ball game. Gurus check my math please RF power is not electrical power. As audio power is not electrical power. ????????????? What kind of power is RF? Magical? At which frequency does the electric current stop following the laws of physics? Square root of P or 1200 watts in our case = 34.641 34.641/50 = 0.6928 amps. E=IIR (0.6928*0.6928)*50 = 23.99 volts. Therfor, the wire can easily handle the power. Use of a calculator helps. Your theory was correct. The decimal point was not. Are you serious? And you wanted *us* to get on that newsgroup and make asses of ourselves like you just did? Better check your math and your understanding of the Ohm's law, the Power equation and the efficiency of the AB class amplifiers. U= I*R, P= U*I = P= I^2*R or P= U^2/R = I= SQR(P/R) and *not* I= SQR(P)/R like you mistakenly claim Elmer's calculation is 100% correct. The trucker is right. You are wrong. 73 .... WA7AA -- Anti-spam measu look me up on qrz.com if you need to reply directly |
Richard wrote:
Oh boy -- lets assume an antenna is totally resistive -- 50 ohms or better still a dummy load of 50 ohms that exhibits no inductance or capacitance. Now since Power in watts = I squared R where I is the current and R is a pure 50 ohm resistance transpose and solve for I = square root of P over R I get about 4.9 amperes RF current now E=I times R so 4.9 times 50 = 244 volts RF volts That's what you should see at the antenna. Try that on old knucklehead. With any inductive reactance or capacitive reactance --- different ball game. Gurus check my math please RF power is not electrical power. As audio power is not electrical power. ????????????? What kind of power is RF? Magical? At which frequency does the electric current stop following the laws of physics? Square root of P or 1200 watts in our case = 34.641 34.641/50 = 0.6928 amps. E=IIR (0.6928*0.6928)*50 = 23.99 volts. Therfor, the wire can easily handle the power. Use of a calculator helps. Your theory was correct. The decimal point was not. Are you serious? And you wanted *us* to get on that newsgroup and make asses of ourselves like you just did? Better check your math and your understanding of the Ohm's law, the Power equation and the efficiency of the AB class amplifiers. U= I*R, P= U*I = P= I^2*R or P= U^2/R = I= SQR(P/R) and *not* I= SQR(P)/R like you mistakenly claim Elmer's calculation is 100% correct. The trucker is right. You are wrong. 73 .... WA7AA -- Anti-spam measu look me up on qrz.com if you need to reply directly |
I is NOT the the square root of power. It is the square root of (power
divided by R) You first have to divide the power by the resistance THEN take the square root. And E=I x R not I x I x R Back to Ohms Law --see URL: http://www.angelfire.com/pa/baconbacon/page2.html and AC RMS power is the same as DC power. Provided the circuit has no inductance and capacitance and RMS values are used -- see OHMS LAW URL above and URL: http://www1.jaycar.com.au/images_uploaded/ohmpower.pdf About now I think you are putting me on, so the old Elmer is exit stage left. Elmer E Ing ------------------------------------------------ "Richard" anom@anom wrote in message ... "Elmer E Ing" wrote in message news:VM_Ua.12050$ff.5596@fed1read01... Oh boy -- lets assume an antenna is totally resistive -- 50 ohms or better still a dummy load of 50 ohms that exhibits no inductance or capacitance. Now since Power in watts = I squared R where I is the current and R is a pure 50 ohm resistance transpose and solve for I = square root of P over R I get about 4.9 amperes RF current now E=I times R so 4.9 times 50 = 244 volts RF volts That's what you should see at the antenna. Try that on old knucklehead. With any inductive reactance or capacitive reactance --- different ball game. Gurus check my math please RF power is not electrical power. As audio power is not electrical power. Square root of P or 1200 watts in our case = 34.641 34.641/50 = 0.6928 amps. E=IIR (0.6928*0.6928)*50 = 23.99 volts. Therfor, the wire can easily handle the power. Use of a calculator helps. Your theory was correct. The decimal point was not. |
I is NOT the the square root of power. It is the square root of (power
divided by R) You first have to divide the power by the resistance THEN take the square root. And E=I x R not I x I x R Back to Ohms Law --see URL: http://www.angelfire.com/pa/baconbacon/page2.html and AC RMS power is the same as DC power. Provided the circuit has no inductance and capacitance and RMS values are used -- see OHMS LAW URL above and URL: http://www1.jaycar.com.au/images_uploaded/ohmpower.pdf About now I think you are putting me on, so the old Elmer is exit stage left. Elmer E Ing ------------------------------------------------ "Richard" anom@anom wrote in message ... "Elmer E Ing" wrote in message news:VM_Ua.12050$ff.5596@fed1read01... Oh boy -- lets assume an antenna is totally resistive -- 50 ohms or better still a dummy load of 50 ohms that exhibits no inductance or capacitance. Now since Power in watts = I squared R where I is the current and R is a pure 50 ohm resistance transpose and solve for I = square root of P over R I get about 4.9 amperes RF current now E=I times R so 4.9 times 50 = 244 volts RF volts That's what you should see at the antenna. Try that on old knucklehead. With any inductive reactance or capacitive reactance --- different ball game. Gurus check my math please RF power is not electrical power. As audio power is not electrical power. Square root of P or 1200 watts in our case = 34.641 34.641/50 = 0.6928 amps. E=IIR (0.6928*0.6928)*50 = 23.99 volts. Therfor, the wire can easily handle the power. Use of a calculator helps. Your theory was correct. The decimal point was not. |
Your assuming the input and the ouput power are the same, not so. For most
amps the effiiciency is about 50 to 60% of the input. For solid state figure 50%. So in order to figure the current needed to run the amp you need to assume a 50% rule. he was only partly right. So you take his answer and divide it by 0.5 and you get 200Amps. That's right folks. 73 Dan N0FQN "Richard" anom@anom wrote in message ... Some knucklehead in misc.transport.trucking is showing off what a real ass he is in the way of knowledge. That should take you to the proper thread. If not, look for the post from "mad dog" posted at 9:35 am. I kid you not, these are his words: quote Keep the P=IxE formula in mind when choosing a mobile amplifier. Power = Current x Voltage or Current = Power / Voltage P=Power or Wattage I =Current draw E=Energy or Voltage Example: Tommy truck driver wants to install a 1200 watt spurious emissions generator on his Kenworth with a 12VDC electrical system. We just learned that Current draw is calculated by dividing the Voltage into the Wattage so if the 1200 watt cataract producer has a Input Voltage of 12 Volts then the Current draw will be............................... One Hundred Amps (100). Yes folks you read that right, that is more current draw than most alternators can produce.So please be realistic when adding accessories to your already marginal electrical system or you may find yourself playing the "Prince of Darkness" on a very wet and rainy night. -- 714 Sandpile, The Mad Dog wavin' good bye /quote He is wrongfully assuming that RF power output is equal to saying 12volts*100amps. Assuming the rated output power is 1200 watts. As I have dealt with radios for years, I know for a fact that this is only an idiot's explanation. He's big bad mister trucker trying to show off how smart he is. I have tried to explailn it to him and he will not believe anything other than what he has written. So if one of you could come in to the thread and explain how damn wrong he is, I would appreciate it. thanks for the time. |
Your assuming the input and the ouput power are the same, not so. For most
amps the effiiciency is about 50 to 60% of the input. For solid state figure 50%. So in order to figure the current needed to run the amp you need to assume a 50% rule. he was only partly right. So you take his answer and divide it by 0.5 and you get 200Amps. That's right folks. 73 Dan N0FQN "Richard" anom@anom wrote in message ... Some knucklehead in misc.transport.trucking is showing off what a real ass he is in the way of knowledge. That should take you to the proper thread. If not, look for the post from "mad dog" posted at 9:35 am. I kid you not, these are his words: quote Keep the P=IxE formula in mind when choosing a mobile amplifier. Power = Current x Voltage or Current = Power / Voltage P=Power or Wattage I =Current draw E=Energy or Voltage Example: Tommy truck driver wants to install a 1200 watt spurious emissions generator on his Kenworth with a 12VDC electrical system. We just learned that Current draw is calculated by dividing the Voltage into the Wattage so if the 1200 watt cataract producer has a Input Voltage of 12 Volts then the Current draw will be............................... One Hundred Amps (100). Yes folks you read that right, that is more current draw than most alternators can produce.So please be realistic when adding accessories to your already marginal electrical system or you may find yourself playing the "Prince of Darkness" on a very wet and rainy night. -- 714 Sandpile, The Mad Dog wavin' good bye /quote He is wrongfully assuming that RF power output is equal to saying 12volts*100amps. Assuming the rated output power is 1200 watts. As I have dealt with radios for years, I know for a fact that this is only an idiot's explanation. He's big bad mister trucker trying to show off how smart he is. I have tried to explailn it to him and he will not believe anything other than what he has written. So if one of you could come in to the thread and explain how damn wrong he is, I would appreciate it. thanks for the time. |
On Sun, 27 Jul 2003 22:16:15 -0500, "Richard" anom@anom wrote:
"Elmer E Ing" wrote in message news:VM_Ua.12050$ff.5596@fed1read01... Oh boy -- lets assume an antenna is totally resistive -- 50 ohms or better still a dummy load of 50 ohms that exhibits no inductance or capacitance. Now since Power in watts = I squared R where I is the current and R is a pure 50 ohm resistance transpose and solve for I = square root of P over R I get about 4.9 amperes RF current now E=I times R so 4.9 times 50 = 244 volts RF volts That's what you should see at the antenna. Try that on old knucklehead. With any inductive reactance or capacitive reactance --- different ball game. Gurus check my math please RF power is not electrical power. As audio power is not electrical power. Square root of P or 1200 watts in our case = 34.641 34.641/50 = 0.6928 amps. E=IIR (0.6928*0.6928)*50 = 23.99 volts. Therfor, the wire can easily handle the power. Use of a calculator helps. Your theory was correct. The decimal point was not. Elmer's calculations were correct. Power(P) = voltage (E) x current (I), and voltage (E) = current (I) x resistance (R) so... P = I * I * R and I = square root( P / R ) = sqr( 1200 / 50 ) = sqr( 14 ) ~= 4.898979 Amps and E = I * R = 4.9 * 50 ~= 245 volts RG-58C/U (Belden 8262) is rated at 1,400 volts RMS. The 20 AWG center conductor of RG-58C/U (Belden 8262) is good for about 6 amps RMS. Ref: http://ecom.belden.com/static/ZZBLDN...TA.HTM?P0=8262 73 de Leigh W3NLB |
On Sun, 27 Jul 2003 22:16:15 -0500, "Richard" anom@anom wrote:
"Elmer E Ing" wrote in message news:VM_Ua.12050$ff.5596@fed1read01... Oh boy -- lets assume an antenna is totally resistive -- 50 ohms or better still a dummy load of 50 ohms that exhibits no inductance or capacitance. Now since Power in watts = I squared R where I is the current and R is a pure 50 ohm resistance transpose and solve for I = square root of P over R I get about 4.9 amperes RF current now E=I times R so 4.9 times 50 = 244 volts RF volts That's what you should see at the antenna. Try that on old knucklehead. With any inductive reactance or capacitive reactance --- different ball game. Gurus check my math please RF power is not electrical power. As audio power is not electrical power. Square root of P or 1200 watts in our case = 34.641 34.641/50 = 0.6928 amps. E=IIR (0.6928*0.6928)*50 = 23.99 volts. Therfor, the wire can easily handle the power. Use of a calculator helps. Your theory was correct. The decimal point was not. Elmer's calculations were correct. Power(P) = voltage (E) x current (I), and voltage (E) = current (I) x resistance (R) so... P = I * I * R and I = square root( P / R ) = sqr( 1200 / 50 ) = sqr( 14 ) ~= 4.898979 Amps and E = I * R = 4.9 * 50 ~= 245 volts RG-58C/U (Belden 8262) is rated at 1,400 volts RMS. The 20 AWG center conductor of RG-58C/U (Belden 8262) is good for about 6 amps RMS. Ref: http://ecom.belden.com/static/ZZBLDN...TA.HTM?P0=8262 73 de Leigh W3NLB |
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