help needed for this circuit

[udit]

i have to design a telemetry system usig lm324 and bjts etc and other analog devices .the objective is:
1.we have to generate 4ma of current when we have a input of 0V and 20ma when input is 5V.
this is transmitter.
next
2.we have a receiver it is supposed that receiver will not be using any power supply and there are two recievers and we can simulate them by using resistors of 500 ohms each
At the receiver i have to convert 4ma to 0V and 20 ma to 5 V. (no use of power supply)
and both the transmitter and receiver are connected in closed loop.
if u have got any schematic of such problem then pls post it.i think this is some of the basic exp in instrumentation.
thanks

 

[seadolphine2000]

As I understood, you want to convert current to voltage at the TX and vice versa at the RX side, Right.???

If I'm right, I can help with a schematic.

 

[IanP]

To generate 4-20mA signal you can use the attached circuit.
Iload = [Vsupply-V(+)]/R1
As you will need "voltage" at the receiver end I suggest that you select Vsupply=30V and R1=500Ω.
In this case, for 4mA the V(+) volage has to be 28V, and for 20mA the V(+) = 20V.
All what you need is additional differential amplifier (in classic configuration) based on 1/4 324 which will shift voltage from 0V to 28V (for 4mA) and 5V to 20V (for 20mA).

At the receiver end you will need different type of opam, as 324 voltage swing is not full 0-to something. Also, you need an opamp that cen work from very low supply voltage generated by a 500Ω resistor and 4-20mA.
Here I would suggest the CA3140 (can work of 4V rail-to-rail).
So to generate supply voltage for this opamp you will use 500Ω resistor connected in series with a 250Ω which will be used as the 4-20mA current sensor. The other end of the 500Ω will be connected to 0V and 250Ω to 4-20mA current generator.
Again, this opamp, in classic differential opamp configuration, will convert voltage generated accross 250Ω resistor (as differential voltage this will be 1V to 5V) to output voltage 0-5V.

Regards,
IanP

 

[VVV]

This is a system used widely in industrial controls. The transmitters are referred to as "loop-powered transmitters" or "two-wire" transmitters. You can understand why by looking at the block diagram.

The power supply provides power to the "loop", formed by the wires, the transmitter and the sense resistor in the measurement system. The transmitter works as a current source. The advantage of the system are:
- you are dealing with current, so voltage drops across the loop wires do not matter (if they are low enough, so as not to prevent the transmitter from operating properly.
- it uses only two wires, which makes it simpler and less expensive, especially that transmitters are usually located hundreds of meters from the measurement system.
- the power supply does not have to be well regulated, since the transmitter regulates the current
- loss of the minimum current (4mA) indicates the loop wires are shorted or at least one of them is broken.

The measurement system generally powered from the power supply. I do not know why you said it had no power.
The current is measured across a 250Ω or 500Ω resistor, which produces a 1-5V or 2-10V voltage. Generally, an A/D measures directly this voltage. This resistor is generally a precision resistor and a low tempco one, such as RN60 or RN55. I prefer 250Ω, because it produces 5V max, which can be measured easily with many A/Ds.

The transmitter schematic I am showing is pretty much a classical one. Many variations are possible, I thought this was a simpler one.
The power is provided from the loop. The entire circuit is connected across the current regulating transistor. The current is sensed across the 49.9Ω resistor, again a precision, stable one. As you can see, the ENTIRE circuit current (except about 10µA through R6) flows through the sense resistor (R1). That allows you to use a BJT, without worrying about its base current and errors associated with it.

The total current of your circuit MUST be lower than the 4mA, so that the circuit can regulate it (usually, 3.6mA is the maximum allocated for all the circuitry, including sensor excitation).

This is how it works: the opamp has one input grounded and the voltage at the other input is summed resistively so as to obtain 0V with respect to "GROUND". In fact, all voltages in the transmitter are referred to "GROUND". The voltage at R6 (right side pin) will be negative with respect to "GROUND" and it will be 49.9Ω*Iout, where Iout is 4-20mA. So you get 0.2-1V across R1. You can make R1 smaller, if needed, but I think this is OK. The voltage at the (+) input of the opamp will be zero when the current develops a high enough voltage across R1, so as to bring the (+) input back to zero volts (remember, the voltage drop across R1 is negative with respect to "GROUND"). It is important to use an opamp that can work down to 0V common mode, else you need to prebias the (-) input with a voltage derived from the reference.

With the values shown, the input is 0-2V. For other values, this is how you calculate R8+R2: The sum of the currents at the (+) input of the opamp has to be zero. So at the maximum output current (20mA), 10µA will flow through R6 to the output. You have 2µA coming through R7 from the reference. The rest (8µA) must come from the input. So R8+R2=Vinmax/8µA. Allocate a portion to R2, to allow easy adjustment.

Example: Vinmax=1V
Then R2+R8=1V/8µA=125kΩ. Use 100kΩ in series with a 50kΩ pot, or 120kΩ in series with a 10kΩ pot, depending on other considerations (resistor tolerances and especially the tolerance on Vinmax).

Other sections of the opamp can be used for signal conditioning for the actual transducer used in the transmitter (could be a bridge, etc.)
Remember, the total circuit current must be less than 3.6mA.

To adjust, set Vin to 0V anad adjust R9 for 4mA out. Then set Vin to max and adjust R2 for 20mA output.

The minimum voltage the transmitter will operate at is the minimum voltage of the opamp, plus the voltage drop across the diode, plus the voltage drop across R1 (max 1V). With the parts shown, it will be about 5V (a pretty much usual value). Using a 250Ω resistor in the measurement system, the minimum loop voltage is about 10V, plus losses across loop wires. The maximum voltage across the transmitter is limited by the opamp, about 30V. That should not be a problem, since loop power is generally 12-28VDC (as you can see, it's wide range, not necessarily regulated). If in doubt, add a 28V or 30V Zener across C1.
I like including diode D1 (even if it loses 0.6V), since in the field wires can be easily miswired and you do not want the transmitter damaged just because of a small mistake.

 

[seadolphine2000]

Hi,..

You can try these circuits, they have been tested in my college labs, but you have to specify the resistors values yourself.

Hope this helps.