All ammeters of this type only measure DC. The original installation (nice meter by the way) no doubt included a series resistance, and a diode.
In the above tutorial, you can find out how to measure the internal resistance of the meter and how to use it to measure either AC or DC volts or amps.
If the internal resistance of your meter is, say 600 ohms and its full scale deflection occurs with 1 ma flowing thru it, according to Mr Ohm and his famous law, if 0.6VDC is applied, 0.6v divided by 600 ohms should result in 1ma of current thru the meter, resulting in a full scale deflection of the meter.
Further if 0db into 600 ohms (1mw) is equal to 0.776 Vrms, you should be able to work out how much voltage (with the appropriate series resistance needs to be applied for the meter to deflect to the 0db point.
Applying 5.0v dc is about 10 times too much, and could result in destroying the meter.
Also applying an AC voltage to the meter would cause it to try and deflect one way and then the other. In the case of 60 Hz, a current reversal of 60 times per second.
Please review the excellent tutorial referenced above, it will tell you everything you need to know.
Do you know anything more about this meter? When was it made, by who?
Did you determine the internal resistance of the meter was 152 ohms per the method outlined in the link that I provided?
You cannot just slap an ohm meter across it.
0db is defined the amount of voltage dropped across a 600 ohm load that would produce 1mW. If P=IE (power = Current times Voltage, then Voltage is equal to the square root of current times resistance. The square root of 0.600 (1mw * 600 ohms) is 0.775 V
As I said before, the original meter installation contained additional components besides the meter itself.
The meter with the network (that is missing) provided an impedance of 600 ohms
I really would like to know, if anyone knows exactly how 'they' configured this device in the 'olden days' to make a linear measurement of the db level.
Also, I have looked on the internet and cannot find an example of this meter.
I suspect it was probably from some early telco equipment.
Did you determine the internal resistance of the meter was 152 ohms per the method outlined in the link that I provided?
You cannot just slap an ohm meter across it.
0db is defined the amount of voltage dropped across a 600 ohm load that would produce 1mW. If P=IE (power = Current times Voltage, then Voltage is equal to the square root of current times resistance. The square root of 0.600 (1mw * 600 ohms) is 0.775 V
As I said before, the original meter installation contained additional components besides the meter itself.
The meter with the network (that is missing) provided an impedance of 600 ohms
I really would like to know, if anyone knows exactly how 'they' configured this device in the 'olden days' to make a linear measurement of the db level.
Also, I have looked on the internet and cannot find an example of this meter.
I suspect it was probably from some early telco equipment.
Please read the link, I do not have any idea what the internal resistance of the meter actually is, you have to do that by following the instructions in the link.
Hint: it will involve a variable resistor, a voltmeter and a voltage source. 5v should be ok. Also, please 'sneak up' the application of the voltage and start with the highest resistance. If you use a 5volt voltage, when you measure 2.5V across the resistor, take the resistor out of the circuit and measure it's resistance, that will be the resistance of your meter.
Please read the link, I do not have any idea what the internal resistance of the meter actually is, you have to do that by following the instructions in the link.
Hint: it will involve a variable resistor, a voltmeter and a voltage source. 5v should be ok. Also, please 'sneak up' the application of the voltage and start with the highest resistance. If you use a 5volt voltage, when you measure 2.5V across the resistor, take the resistor out of the circuit and measure it's resistance, that will be the resistance of your meter.
Please read the link, I do not have any idea what the internal resistance of the meter actually is, you have to do that by following the instructions in the link.
Hint: it will involve a variable resistor, a voltmeter and a voltage source. 5v should be ok. Also, please 'sneak up' the application of the voltage and start with the highest resistance. If you use a 5volt voltage, when you measure 2.5V across the resistor, take the resistor out of the circuit and measure it's resistance, that will be the resistance of your meter.
Thanks,in the link they were setting up an ameter so they put a resistor in parallel. But a voltmeter will need the resistor in series? Sorry for all the questions. Just trying to get this sorted. Thanks again
Thanks,in the link they were setting up an ameter so they put a resistor in parallel. But a voltmeter will need the resistor in series? Sorry for all the questions. Just trying to get this sorted. Thanks again
OK you are going to use the meter to measure random voltages between 0 and 5v to give the appearance of random power fluctuations in the manor which is located on the heath.
You need to know what the internal resistance of the meter is in order to do this without overloading the meter and to know what its display will look like.
In a series circuit of two resistors of equal value R1 and R2, the voltage measured between R2 and 0 volts will be 1/2 of the voltage applied between R1 and R2.
If R2 is not known, and R1 is a variable resistor, adjusting R1 until the voltage across the unknown resistor is 1/2 of the total voltage applied, R1 is equal to R2, and "Bob's your uncle", R2 goes from being unknown to known.
You observed when you applied about 0.5 volts across the meter the needle deflected to its maximum. This would imply that the resistance of the meter is around 500 ohms if it is a 1ma meter movement. You need to find out what it actually is.
If you put a 1000 ohm variable resistor in series with the meter and apply 5.0v across the whole thing (MAKE SURE THE VARIABLE RESISTOR IS SET TO 1000 ohms) and decrease the variable resistor until you measure 2.5 V across it, the resistance of the variable resistor will be equal to the internal resistance of the meter.
Lets ASSUME (the actual value that you discover may be different) that the value you measure on the variable resistor 500 ohms, how then, do you cause the meter to fully deflect when 5.0 volts are applied?
Place a 4.2K in series with 300 ohms resistor (4.5 k is not a standard value, but 4.2k and 300 ohms are). 5 Volts /(4500 + 500 ohms) would as Mr Ohm has proved many times would allow 1ma of current to flow thru the meter.
Anyway, the point is to place a resistor in series with the meter movement that will allow a current sufficient to deflect the meter to its maximum without pegging it. All other voltages between 0v and 5v should cause the meter to deflect in a linear manner.
If you want to measure voltages corresponding to the numbers printed on the dial, a much more sophisticated network would have to be attached. I would suggest googling "logarithmic amplifiers"
Again, do you have any information on the history of this device? I would really like to know any details. Your picture of it was not clear enough to read all of its markings, could you note them for me, and or point me to any links?
OK you are going to use the meter to measure random voltages between 0 and 5v to give the appearance of random power fluctuations in the manor which is located on the heath.
You need to know what the internal resistance of the meter is in order to do this without overloading the meter and to know what its display will look like.
In a series circuit of two resistors of equal value R1 and R2, the voltage measured between R2 and 0 volts will be 1/2 of the voltage applied between R1 and R2.
If R2 is not known, and R1 is a variable resistor, adjusting R1 until the voltage across the unknown resistor is 1/2 of the total voltage applied, R1 is equal to R2, and "Bob's your uncle", R2 goes from being unknown to known.
You observed when you applied about 0.5 volts across the meter the needle deflected to its maximum. This would imply that the resistance of the meter is around 500 ohms if it is a 1ma meter movement. You need to find out what it actually is.
If you put a 1000 ohm variable resistor in series with the meter and apply 5.0v across the whole thing (MAKE SURE THE VARIABLE RESISTOR IS SET TO 1000 ohms) and decrease the variable resistor until you measure 2.5 V across it, the resistance of the variable resistor will be equal to the internal resistance of the meter.
Lets ASSUME (the actual value that you discover may be different) that the value you measure on the variable resistor 500 ohms, how then, do you cause the meter to fully deflect when 5.0 volts are applied?
Place a 4.2K in series with 300 ohms resistor (4.5 k is not a standard value, but 4.2k and 300 ohms are). 5 Volts /(4500 + 500 ohms) would as Mr Ohm has proved many times would allow 1ma of current to flow thru the meter.
Anyway, the point is to place a resistor in series with the meter movement that will allow a current sufficient to deflect the meter to its maximum without pegging it. All other voltages between 0v and 5v should cause the meter to deflect in a linear manner.
If you want to measure voltages corresponding to the numbers printed on the dial, a much more sophisticated network would have to be attached. I would suggest googling "logarithmic amplifiers"
Again, do you have any information on the history of this device? I would really like to know any details. Your picture of it was not clear enough to read all of its markings, could you note them for me, and or point me to any links?
OK you are going to use the meter to measure random voltages between 0 and 5v to give the appearance of random power fluctuations in the manor which is located on the heath.
You need to know what the internal resistance of the meter is in order to do this without overloading the meter and to know what its display will look like.
In a series circuit of two resistors of equal value R1 and R2, the voltage measured between R2 and 0 volts will be 1/2 of the voltage applied between R1 and R2.
If R2 is not known, and R1 is a variable resistor, adjusting R1 until the voltage across the unknown resistor is 1/2 of the total voltage applied, R1 is equal to R2, and "Bob's your uncle", R2 goes from being unknown to known.
You observed when you applied about 0.5 volts across the meter the needle deflected to its maximum. This would imply that the resistance of the meter is around 500 ohms if it is a 1ma meter movement. You need to find out what it actually is.
If you put a 1000 ohm variable resistor in series with the meter and apply 5.0v across the whole thing (MAKE SURE THE VARIABLE RESISTOR IS SET TO 1000 ohms) and decrease the variable resistor until you measure 2.5 V across it, the resistance of the variable resistor will be equal to the internal resistance of the meter.
Lets ASSUME (the actual value that you discover may be different) that the value you measure on the variable resistor 500 ohms, how then, do you cause the meter to fully deflect when 5.0 volts are applied?
Place a 4.2K in series with 300 ohms resistor (4.5 k is not a standard value, but 4.2k and 300 ohms are). 5 Volts /(4500 + 500 ohms) would as Mr Ohm has proved many times would allow 1ma of current to flow thru the meter.
Anyway, the point is to place a resistor in series with the meter movement that will allow a current sufficient to deflect the meter to its maximum without pegging it. All other voltages between 0v and 5v should cause the meter to deflect in a linear manner.
If you want to measure voltages corresponding to the numbers printed on the dial, a much more sophisticated network would have to be attached. I would suggest googling "logarithmic amplifiers"
Again, do you have any information on the history of this device? I would really like to know any details. Your picture of it was not clear enough to read all of its markings, could you note them for me, and or point me to any links?
OK you are going to use the meter to measure random voltages between 0 and 5v to give the appearance of random power fluctuations in the manor which is located on the heath.
You need to know what the internal resistance of the meter is in order to do this without overloading the meter and to know what its display will look like.
In a series circuit of two resistors of equal value R1 and R2, the voltage measured between R2 and 0 volts will be 1/2 of the voltage applied between R1 and R2.
If R2 is not known, and R1 is a variable resistor, adjusting R1 until the voltage across the unknown resistor is 1/2 of the total voltage applied, R1 is equal to R2, and "Bob's your uncle", R2 goes from being unknown to known.
You observed when you applied about 0.5 volts across the meter the needle deflected to its maximum. This would imply that the resistance of the meter is around 500 ohms if it is a 1ma meter movement. You need to find out what it actually is.
If you put a 1000 ohm variable resistor in series with the meter and apply 5.0v across the whole thing (MAKE SURE THE VARIABLE RESISTOR IS SET TO 1000 ohms) and decrease the variable resistor until you measure 2.5 V across it, the resistance of the variable resistor will be equal to the internal resistance of the meter.
Lets ASSUME (the actual value that you discover may be different) that the value you measure on the variable resistor 500 ohms, how then, do you cause the meter to fully deflect when 5.0 volts are applied?
Place a 4.2K in series with 300 ohms resistor (4.5 k is not a standard value, but 4.2k and 300 ohms are). 5 Volts /(4500 + 500 ohms) would as Mr Ohm has proved many times would allow 1ma of current to flow thru the meter.
Anyway, the point is to place a resistor in series with the meter movement that will allow a current sufficient to deflect the meter to its maximum without pegging it. All other voltages between 0v and 5v should cause the meter to deflect in a linear manner.
If you want to measure voltages corresponding to the numbers printed on the dial, a much more sophisticated network would have to be attached. I would suggest googling "logarithmic amplifiers"
Again, do you have any information on the history of this device? I would really like to know any details. Your picture of it was not clear enough to read all of its markings, could you note them for me, and or point me to any links?
OK you are going to use the meter to measure random voltages between 0 and 5v to give the appearance of random power fluctuations in the manor which is located on the heath.
You need to know what the internal resistance of the meter is in order to do this without overloading the meter and to know what its display will look like.
In a series circuit of two resistors of equal value R1 and R2, the voltage measured between R2 and 0 volts will be 1/2 of the voltage applied between R1 and R2.
If R2 is not known, and R1 is a variable resistor, adjusting R1 until the voltage across the unknown resistor is 1/2 of the total voltage applied, R1 is equal to R2, and "Bob's your uncle", R2 goes from being unknown to known.
You observed when you applied about 0.5 volts across the meter the needle deflected to its maximum. This would imply that the resistance of the meter is around 500 ohms if it is a 1ma meter movement. You need to find out what it actually is.
If you put a 1000 ohm variable resistor in series with the meter and apply 5.0v across the whole thing (MAKE SURE THE VARIABLE RESISTOR IS SET TO 1000 ohms) and decrease the variable resistor until you measure 2.5 V across it, the resistance of the variable resistor will be equal to the internal resistance of the meter.
Lets ASSUME (the actual value that you discover may be different) that the value you measure on the variable resistor 500 ohms, how then, do you cause the meter to fully deflect when 5.0 volts are applied?
Place a 4.2K in series with 300 ohms resistor (4.5 k is not a standard value, but 4.2k and 300 ohms are). 5 Volts /(4500 + 500 ohms) would as Mr Ohm has proved many times would allow 1ma of current to flow thru the meter.
Anyway, the point is to place a resistor in series with the meter movement that will allow a current sufficient to deflect the meter to its maximum without pegging it. All other voltages between 0v and 5v should cause the meter to deflect in a linear manner.
If you want to measure voltages corresponding to the numbers printed on the dial, a much more sophisticated network would have to be attached. I would suggest googling "logarithmic amplifiers"
Again, do you have any information on the history of this device? I would really like to know any details. Your picture of it was not clear enough to read all of its markings, could you note them for me, and or point me to any links?
I set up the circuit with a POT and did som measuring. @5v I got the meter to full deflection (5v) the POT gave me a reading of 3.3k ohms. I also read voltage across it and it read 4.15v. My current throught the circuit was 1.24 ma. The meter had a voltage of .187v across it. So if I take the .187v and divide it by .00124 that should give me my resistance, correct. 150 ohms?
OK you are going to use the meter to measure random voltages between 0 and 5v to give the appearance of random power fluctuations in the manor which is located on the heath.
You need to know what the internal resistance of the meter is in order to do this without overloading the meter and to know what its display will look like.
In a series circuit of two resistors of equal value R1 and R2, the voltage measured between R2 and 0 volts will be 1/2 of the voltage applied between R1 and R2.
If R2 is not known, and R1 is a variable resistor, adjusting R1 until the voltage across the unknown resistor is 1/2 of the total voltage applied, R1 is equal to R2, and "Bob's your uncle", R2 goes from being unknown to known.
You observed when you applied about 0.5 volts across the meter the needle deflected to its maximum. This would imply that the resistance of the meter is around 500 ohms if it is a 1ma meter movement. You need to find out what it actually is.
If you put a 1000 ohm variable resistor in series with the meter and apply 5.0v across the whole thing (MAKE SURE THE VARIABLE RESISTOR IS SET TO 1000 ohms) and decrease the variable resistor until you measure 2.5 V across it, the resistance of the variable resistor will be equal to the internal resistance of the meter.
Lets ASSUME (the actual value that you discover may be different) that the value you measure on the variable resistor 500 ohms, how then, do you cause the meter to fully deflect when 5.0 volts are applied?
Place a 4.2K in series with 300 ohms resistor (4.5 k is not a standard value, but 4.2k and 300 ohms are). 5 Volts /(4500 + 500 ohms) would as Mr Ohm has proved many times would allow 1ma of current to flow thru the meter.
Anyway, the point is to place a resistor in series with the meter movement that will allow a current sufficient to deflect the meter to its maximum without pegging it. All other voltages between 0v and 5v should cause the meter to deflect in a linear manner.
If you want to measure voltages corresponding to the numbers printed on the dial, a much more sophisticated network would have to be attached. I would suggest googling "logarithmic amplifiers"
Again, do you have any information on the history of this device? I would really like to know any details. Your picture of it was not clear enough to read all of its markings, could you note them for me, and or point me to any links?
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