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what benefit using 2 diodes back to back in analog sinals 4- 20 ma 4
- Thread starter ddsf
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1
- #4
Skogsgurra
Electrical
Are they really connected back to back? That would mean that no signal passes through.
A more common arrangement is to connect them in parallel with one diode forward and one diode the other way. Usually, there is just one diode, but when there are long cable runs and a noisy environment, using one diode can result in noise being rectified and added to the original 4-20 mA signal. Using two diodes eliminates that risk.
Why? Because it is a convenient way to measure actual loop current. Kust connect a milliampmeter across the diodes. Most meters have a voltage drop less than the diode's threshold voltage and will therefore take over all the current so it can be read on the meter.
It is a safe way of checking mA signals without breaking the loop.
Gunnar Englund
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Half full - Half empty? I don't mind. It's what in it that counts.
A more common arrangement is to connect them in parallel with one diode forward and one diode the other way. Usually, there is just one diode, but when there are long cable runs and a noisy environment, using one diode can result in noise being rectified and added to the original 4-20 mA signal. Using two diodes eliminates that risk.
Why? Because it is a convenient way to measure actual loop current. Kust connect a milliampmeter across the diodes. Most meters have a voltage drop less than the diode's threshold voltage and will therefore take over all the current so it can be read on the meter.
It is a safe way of checking mA signals without breaking the loop.
Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
- Thread starter
- #5
- Thread starter
- #6
another question
if the diode putted in this way to measure the milliampere across it , then it will not measure the actual current in the current loop by the transmitter bcz the diode will drop 0.7 volt and the transmitter will have 24 - 0.7 = 23.3 v only across it and will not give the same milliampere that i have measured when i shorted the diode with the ammeter
is it right??
if the diode putted in this way to measure the milliampere across it , then it will not measure the actual current in the current loop by the transmitter bcz the diode will drop 0.7 volt and the transmitter will have 24 - 0.7 = 23.3 v only across it and will not give the same milliampere that i have measured when i shorted the diode with the ammeter
is it right??
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1
- #7
Skogsgurra
Electrical
ddsf
I explained that in my answer. If you put one diode there, you will risk rectification of noise and that adds to the mA signal. If you put two diodes, as shown in your picture, there will not be any net rectification. It may be a little hard to understnd this, but you can trust me. Been there lots of times.
Next: It doesn't matter at all if your voltage is 24V or 23.3 V. It may, in fact, vary from probably 12 or 15 V up to typically 30 V without affecting the signal. This is because the transducer outputs a current signal and its internal (dynamic) resistance is usually in the Megohms region.
Also, At the typical voltage drop of a multimeter (around 100 mV), there is no current (OK, a few microamps - perhaps - but no more) in the diodes. Those few microamps are typically .1 percent of the actual signal and will not be significant at all if you want to do a quick check of the signal.
If you want to calibrate the thing, you should do it out of circuit. Not that it is needed, but the routine is such.
Gunnar Englund
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Half full - Half empty? I don't mind. It's what in it that counts.
I explained that in my answer. If you put one diode there, you will risk rectification of noise and that adds to the mA signal. If you put two diodes, as shown in your picture, there will not be any net rectification. It may be a little hard to understnd this, but you can trust me. Been there lots of times.
Next: It doesn't matter at all if your voltage is 24V or 23.3 V. It may, in fact, vary from probably 12 or 15 V up to typically 30 V without affecting the signal. This is because the transducer outputs a current signal and its internal (dynamic) resistance is usually in the Megohms region.
Also, At the typical voltage drop of a multimeter (around 100 mV), there is no current (OK, a few microamps - perhaps - but no more) in the diodes. Those few microamps are typically .1 percent of the actual signal and will not be significant at all if you want to do a quick check of the signal.
If you want to calibrate the thing, you should do it out of circuit. Not that it is needed, but the routine is such.
Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
- Thread starter
- #8
mmm i appreciate your logic answer, but kindly let me imagine another situation . u said that the transducer have high megaohms resistance , then what about putting a small resistor 10 ohm for example rather than the diode and try to put the ammeter across it ..in this case also i think the whole milliampere will go through the zero ohm ammeter and will measure the milliampere in the loop and at the same tome the 10 ohm will drop a very small voltage in normal opearation and will not affect on the actuall current
is it right??
is it right??
Skogsgurra
Electrical
No, you will not have all current go through the ammeter if you do that. Typically, the internal resistance of a DMM in mA mode and ranges between 10 and 100 mA will be anything from 1 to 10 ohms, with a Fluke 87 having around 2 ohms.
To calculate the correct current, you first need to know the internal resistance of the meter and then calculate how the total current is divided between resistor and instrument. I do not find that practical in any way. Nor is it more exact than the diode method.
Just accept that this is a thechnique that has been used for a long time (probably before you were born) and that the accuracy is just as good as the instrument you use to measure current. The diodes introduce no error whatsoever.
Also, if you want to get anywhere in life - please try to use normal punctuation, skip texting and use capital letters in the beginning of sentences. That is all part of engineering. The sooner you get used to it the better.
Gunnar Englund
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Half full - Half empty? I don't mind. It's what in it that counts.
To calculate the correct current, you first need to know the internal resistance of the meter and then calculate how the total current is divided between resistor and instrument. I do not find that practical in any way. Nor is it more exact than the diode method.
Just accept that this is a thechnique that has been used for a long time (probably before you were born) and that the accuracy is just as good as the instrument you use to measure current. The diodes introduce no error whatsoever.
Also, if you want to get anywhere in life - please try to use normal punctuation, skip texting and use capital letters in the beginning of sentences. That is all part of engineering. The sooner you get used to it the better.
Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
- Thread starter
- #10
Skogsgurra
Electrical
Appreciate the "humor". But do not expect much help next time if you do not change your manners.
Gunnar Englund
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Half full - Half empty? I don't mind. It's what in it that counts.
Gunnar Englund
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
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