mV signal

[Pitt123]

Can someone explain to me how a 5mV singnal works and how the source output current is adjusted with voltage?

Or is the 5mV signal just outputed and the current just results from the resistane in the cable and load?

How is the mV signal kept from loosing accuracy from small voltage drops along the length of the cable?

 

[StephenSDH]

You will only have voltage drop if there is current. Your mV input will be a high impedance input. It will only sample the voltage and not serve as a load.

 

[Smart $]

5mV signal???

Equipment, please?

 

[jumper]

5mV signal???

Equipment, please?

I am a bit confused also.

I only know about a 4-20mA signal system.

http://www.allaboutcircuits.com/vol_1/chpt_9/3.html

 

[gar]

101113-1442 EST

Pitt123:

I do not know what is your signal source.

Suppose is was a bonded straingage transducer. The output rating was 2 MV/V at full scale and the bridge source resistance was 350 ohms. If you excite the bridge with 2.5 V, then at full scale the bridge output would be 5 MV.

It is quite likely that such a transducer would have the minus excitation connected to chassis and indirectly to earth ground. This means the output is floating off of ground. The receiver for this would be a differential amplifier and probably a moderately high input resistance.

Next suppose the signal wires from the transducer have a total resistance of 500 ohms and the input resistance of the amplifier is 99,150 ohms. Then the voltage at the amplifier is 991.5/1000 of the equivalent internal voltage at the transducer. There will be a small variation as a function of temperature of signal wires.

A bigger problem is the excitation to the transducer. Suppose the bridge input resistance is 500 ohms, and the wires supplying the bridge excitation have 500 ohms resistance. The source voltage back at the receiver (input amplifier) needs to be 5 V to get 2.5 V input to the transducer. Excitation wire temperature will have a much greater effect on calibration, than the signal wires. How would you solve this problem?

.

 

[BJ Conner]

Had an EKG lately?

Had an EKG lately?

Can someone explain to me how a 5mV singnal works and how the source output current is adjusted with voltage?

Or is the 5mV signal just outputed and the current just results from the resistane in the cable and load?

How is the mV signal kept from loosing accuracy from small voltage drops along the length of the cable?

I can generate 3-7mV signals. I did it about 6 months ago duringa semi-annual check up. EKG machines do it pretty well.
ECGs are much harder (micro-volts).

 

[Pitt123]

Next suppose the signal wires from the transducer have a total resistance of 500 ohms and the input resistance of the amplifier is 99,150 ohms. Then the voltage at the amplifier is 991.5/1000 of the equivalent internal voltage at the transducer. There will be a small variation as a function of temperature of signal wires.

.

It looks like you are are showing a voltage divider here but I am not following your math. If I understand your point correctly however you are saying that typically the reciever impedance will be very high an much higher than the source impedance and line impedance thus there will be very little current with the mV signal and there will also be a very small voltage drop. The current in your example would be very small in the order of 5mV/99,150ohms = 504micro amps?

Is there a typical maximum length of cable for these type of signals?

A bigger problem is the excitation to the transducer. Suppose the bridge input resistance is 500 ohms, and the wires supplying the bridge excitation have 500 ohms resistance. The source voltage back at the receiver (input amplifier) needs to be 5 V to get 2.5 V input to the transducer. Excitation wire temperature will have a much greater effect on calibration, than the signal wires. How would you solve this problem?

I'm assuming that the excitation voltage will always be the same which in this case lets say its 5V. Are you then saying that in order to get the full scale ouput from the transducer we need to ensure that we are supplying the full excitation voltage to the transducer and anything less will result in a less than full scale output?

To solve the problem of voltage drop with the excitation signal would it simply be a matter of using a larger size signal wire?

 

[gar]

101115-0919 EST

Pitt123:

Any real world voltage, signal, or power source can be approximated as a perfect voltage source with an internal resistance (impedance). Nonlinearities or other variations can modify this statement. But in many cases over a reasonable range this is a good approximation, even for a storage battery.

Therefore, as soon as you place a load on the source you have created a voltage divider.

In the example I provided you the internal resistance of the transducer was 350 ohms. Added to this is the wire resistance of 500 ohms. Thus, at the receiving end the transducer and wire resistance combined are the internal resistance, 850 ohms, of the source as seen by the receiver. For convenience I made the receiver input resistance 99,150 ohms so that the total resistance of the voltage divider was 100,000 ohms. Thus, the divided voltage at the receiver is 99,150/100,000 = 0.9915 times the theoretical ideal voltage source in the transducer.

Suppose the wiring resistance changed by 100% from 500 to 1000 ohms, then the new divider ratio would be 99,150/100,500 = 0.9866 or not quite a 1% change in the measurement for the 100% change is wiring resistance.

This is because the receiver input resistance is large relative to the source resistance. With an infinite receiver input resistance there would be no error in the received voltage compared to the source voltage in the transducer.


Next to the excitation side. Roughly the same size wires would be used to supply excitation to the transducer as are used for the signal side. For example Belden 8723.

Assuming the bridge input resistance is 500 ohms and the wiring resistance is 500 ohms and the source voltage for excitation at the receiver end is 5 V, then 2.5 V is actually the excitation voltage at the transducer. A characteristic of this type of transducer is that the output voltage from the transducer for a fixed mechanical input to the transducer is proportional to the excitation voltage.

Now change the wiring resistance to 1000 ohms and the excitation drops to 5/3 = 1.67 V. Thus, transducer output has changed by (2.5-1.67)/2.5 = 0.333 or a drop of 33% for the 100% change in input wire resistance. This is very significant compare to the output side because of the differences in resistances.

You could excite the transducer with a constant current if its input resistance was constant. What is usually done is to provide a second set of signal wires. These are used to monitor the excitation voltage at the transducer. At the receiving end the ratio of the transducer output signal and the transducer excitation is obtained.


Generally speaking -- at the receiver end there will be an ability to adjust calibration by gain and offest adjustments.

.

 

[Pitt123]

101115-0919 EST

Pitt123:

Any real world voltage, signal, or power source can be approximated as a perfect voltage source with an internal resistance (impedance). Nonlinearities or other variations can modify this statement. But in many cases over a reasonable range this is a good approximation, even for a storage battery.

Therefore, as soon as you place a load on the source you have created a voltage divider.

In the example I provided you the internal resistance of the transducer was 350 ohms. Added to this is the wire resistance of 500 ohms. Thus, at the receiving end the transducer and wire resistance combined are the internal resistance, 850 ohms, of the source as seen by the receiver. For convenience I made the receiver input resistance 99,150 ohms so that the total resistance of the voltage divider was 100,000 ohms. Thus, the divided voltage at the receiver is 99,150/100,000 = 0.9915 times the theoretical ideal voltage source in the transducer.

Suppose the wiring resistance changed by 100% from 500 to 1000 ohms, then the new divider ratio would be 99,150/100,500 = 0.9866 or not quite a 1% change in the measurement for the 100% change is wiring resistance.

This is because the receiver input resistance is large relative to the source resistance. With an infinite receiver input resistance there would be no error in the received voltage compared to the source voltage in the transducer.


Next to the excitation side. Roughly the same size wires would be used to supply excitation to the transducer as are used for the signal side. For example Belden 8723.

Assuming the bridge input resistance is 500 ohms and the wiring resistance is 500 ohms and the source voltage for excitation at the receiver end is 5 V, then 2.5 V is actually the excitation voltage at the transducer. A characteristic of this type of transducer is that the output voltage from the transducer for a fixed mechanical input to the transducer is proportional to the excitation voltage.

Now change the wiring resistance to 1000 ohms and the excitation drops to 5/3 = 1.67 V. Thus, transducer output has changed by (2.5-1.67)/2.5 = 0.333 or a drop of 33% for the 100% change in input wire resistance. This is very significant compare to the output side because of the differences in resistances.

You could excite the transducer with a constant current if its input resistance was constant. What is usually done is to provide a second set of signal wires. These are used to monitor the excitation voltage at the transducer. At the receiving end the ratio of the transducer output signal and the transducer excitation is obtained.


Generally speaking -- at the receiver end there will be an ability to adjust calibration by gain and offest adjustments.

.

Great explanation Gar this is clear now.

If I'm correct then any voltage drop on the wire of a 4-20mA signal is taken care of by the transducer adjusting its output impedance and therefore supplying more voltage in order to supply the appropriate amount of voltage to drive the 4-20ma signal through the external resistance?

 

[gar]

101115-1751 EST

Pitt123:

In a current loop system the transmitter operates to adjust the current in the loop to a value proportional to its sensed measurement. The transmitter has limits as does the loop.

The major limits on the loop and transmitter are:
1. Maximum voltage rating across the transmitter. This rating must be greater than the source voltage of the loop.
2. The transmitter must have capability to pass current somewhat greater than the maximum loop current.
3.The total loop resistance external to the transmitter, this includes the receiver resistance, must not exceed a value that would prevent the transmitter from supplying the maximum current. This is related to source voltage and the voltage drop across the transmitter at maximum current.

In a 4-20 MA system 4 MA will often correspond to an input measurement of the ZERO value of whatever is being measured. On the other hand it might correspond to 90 deg C and 20 MA could represent 110 deg C. In this case the measurement range would be 20 deg C referenced to 90 deg C.

Fundamentally the transmitter is an electronically controlled variable resistor. At 20 MA is has to have its lowest resistance. Any higher resistance causes the loop current to be less than 20 MA. At 4 MA the transmitter sees the greatest voltage across its terminals. All this assumes the source voltage is physically separate from the transmitter. But the theory of the loop doesn't have anything do do with where in the series loop the source voltage is present.

.

 

[Smart $]

...

In a 4-20 MA system 4 MA will often correspond to an input measurement of the ZERO value of whatever is being measured. On the other hand it might correspond to 90 deg C and 20 MA could represent 110 deg C. In this case the measurement range would be 20 deg C referenced to 90 deg C.

...

In Instrumentation & Control speak (read: terminology) what you noted as measurement range is actually the measurement span. Range always has two values associated with it—Lower Range Value (LRV) and Upper Range Value (URV)—while the Span is the absolute value of URV minus LRV. For your example, the value of the preceding terms are:

LRV: 90?C
URV: 110?C
Span: 20?C
​

Range may also be stated in the fashion: 90?C to 110?C, or 100?C ? 10?C (though this latter form is typically viewed as a nominal value and tolerance rather than range).

 

[Pitt123]

Fundamentally the transmitter is an electronically controlled variable resistor. At 20 MA is has to have its lowest resistance. Any higher resistance causes the loop current to be less than 20 MA. At 4 MA the transmitter sees the greatest voltage across its terminals.
.

So at 4mA the transmitter sees most of the voltage (lets say 24V in this case) across its internal impedance or terminals and supplies very little voltage to the external of the circuit?

So at 4mA with an external resistance of 250ohms we might only meausure 1V between the transmitter output and some common point int the circuit, or in other words 1V across the external 250ohm load?

All this assumes the source voltage is physically separate from the transmitter. But the theory of the loop doesn't have anything do do with where in the series loop the source voltage is present.
.

Does the analog output card on a PLC behave the same as a transmitter in that the output adjusts its internal impedance to supply the correct mA signal to the circuit? It will therefore adjust the voltage output as well. So even though the output card has a 24V supply, this ouput voltage will be adjusted to meet the current requirements of the external circuit.

 

[gar]

101116-1027 EST

Pitt123:

On your first question:

So at 4mA the transmitter sees most of the voltage (lets say 24V in this case) across its internal impedance or terminals and supplies very little voltage to the external of the circuit?

I want to change the viewpoint a little. The transmitter, when it does not include the voltage source, simply modulates the current flow in the loop. It does not supply voltage to the loop, but by its modulation of the current adjusts the distribution of voltage between itself and the loop resistance external to the transmitter.

Consider the receiver to be 250 ohms. Typically the receiver will have an internal resistor (current shunt) used to convert current to voltage to measure current in the loop. This is the reason the receiver has a substantial input resistance. Thus, at 4 MA loop current the receiver has a voltage drop of 1 V, as you said. At 20 MA this voltage drop is 5 V. This sensing resistance could be smaller, but for normal systems at reasonable voltages 250 ohms would be a maximum. If this resistance was dropped to 25 ohms and thus 0.5 V for full scale there would be no real problem because DC amplifiers today are readily available with noise levels in the 1 microvolt range. Even 2.5 ohms or 50 MV full scale would work.

Back to your question. Consider wiring from the transmitter to the receiver to have zero resistance. Thus, resistance external to the transmitter is 250 ohms. In other words put the transmitter close to the receiver. Assume a source voltage for the loop of 25 V and this source is separate from the transmitter. To produce 4 MA in the loop the voltage drop across the transmitter has to be 25-1 = 24 V. At 20 MA this becomes 25-5 = 20 V across the transmitter. The power dissipation in the transmitter is 20*0.02 = 0.4 W at 20 MA. This would be the worst case. Any additional resistance in the loop will reduce the power dissipation in the transmitter.

If the transmitter was to adjusted to 0 current, meaning it ceases to let current flow, then the maximum voltage across the transmitter terminals becomes 25 V. The internal output transistor in the transmitter in this current loop has to withstand this voltage. Probably a transistor with at least a 50 V rating would be chosen.

At the maximum current extreme, 20 MA, the lowest voltage drop across the transmitter will be determined by the sum of the lowest voltage drop across the output transistor and internal current sensing resistor. Suppose the current sense resistor is 1 ohm and an FET (field effect transistor) of 4 ohms drain-source resistance is used. Then the lowest possible resistance of the transmitter is 5 ohms. The transmitter can have a resistance anywhere from 5 to near infinity ohms. The minimum resistance is important in determining the maximum wiring resistance. With a 25 V source and 255 ohms minimum between the transmitter and receiver, and the need to provide 20 MA the result for maximum wiring resistance is (25/0.02)-255 = 1250-255 = 995 ohms. But then you need some margin and thus maximum wiring resistance has to be somewhat lower than 900 ohms. The purpose of using a current loop system is to eliminate errors from variation of resistance in the loop. For example temperature of the wiring.

So at 4mA with an external resistance of 250ohms we might only meausure 1V between the transmitter output and some common point int the circuit, or in other words 1V across the external 250ohm load?

I am not sure what is your question. In the external circuit (I have been including the voltage source in the external circuit) you have to define the points between which voltage is measured. So depending upon the points measured the voltage could be the 1 V across the 250 ohms or something else.

Does the analog output card on a PLC behave the same as a transmitter in that the output adjusts its internal impedance to supply the correct mA signal to the circuit? It will therefore adjust the voltage output as well. So even though the output card has a 24V supply, this ouput voltage will be adjusted to meet the current requirements of the external circuit.

If you are describing a PLC current loop transmitter with an internal voltage source, then yes. But I would rather say the device controls the current rather than voltage. Indirectly it adjusts the output voltage, but it does not measure output voltage and use this measurement for control. Rather it adjusts its internal resistance to achieve a certain current flow.

Note: This type of device can be described as an adjustable current source. Being a constant current device is only valid over a limited range of parameters. Some physical devices are inherently somewhat of a constant current source. Such as: a pentode vacuum tube, a bipolar transistor, and some designed circuits with many components in an integrated circuit.

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