0-1V to 4-20 mA Converter

Ensure +5/-5 dual supply for chip TL062 IC3. Gnd is common ps ground, let grounds radiate from ground plane in one side of PCB. R3modified.

In output R23 is for protection from shorting of +5V supply, R23 can also go to an unregulated or external. supply upto 24V DC which is referenced to this circuits gnd. More voltage more distance. Q2 is the current control device, and R22 50E is the shun20mA in the output (provided suitable load is connected) means 200mVshunt. This is fed to close loop control system of IC3a inverting pin. An opamp on this type of feedback tries to drive the outputthe inputs at same level. If there is 1V at pin 3 and no current is flowing pin 2 is at 0V so output goes positive and drives Q2. this results in a flow of current till a 1V builds across shunt, if it exceeds then oopamp falls This reduces drive to transistor and hence current reduces. That is the part of V to I conversion with open collector output. Now we need 200mV to 1000mV to get 4are more noise prone. that is the reason 4mA and not 0mA. Now we need to convert 0-2 V to 0.2 offset set by R16 pot. the opamp IC3B adds both the input and this offset to get 200mV to

20 mA Converter

5 dual supply for chip TL062 IC3. Gnd is common ps ground, let grounds radiate from ground plane in one side of PCB. R3-R8 is an attenuator that may need to be designed or

In output R23 is for protection from shorting of +5V supply, R23 can also go to an unregulated or external. supply upto 24V DC which is referenced to this circuits gnd. More voltage more

Q2 is the current control device, and R22 50E is the shunt for taking a sample of current. 420mA in the output (provided suitable load is connected) means 200mV- 1000mV across 50E shunt. This is fed to close loop control system of IC3a inverting pin.

An opamp on this type of feedback tries to drive the output in such a way, so as to maintain both

If there is 1V at pin 3 and no current is flowing pin 2 is at 0V so output goes positive and drives Q2. this results in a flow of current till a 1V builds across shunt, if it exceeds then oopamp falls This reduces drive to transistor and hence current reduces. That is the part of V to I conversion with open collector output.

Now we need 200mV to 1000mV to get 4-20mA 4mA is good for 0 as low level measurements that is the reason 4mA and not 0mA.

2 V to 0.2 - 1.0 V using IC3B. R14 is a representation of that 200mV offset set by R16 pot. the opamp IC3B adds both the input and this offset to get 200mV to

5 dual supply for chip TL062 IC3. Gnd is common ps ground, let grounds radiate R8 is an attenuator that may need to be designed or

In output R23 is for protection from shorting of +5V supply, R23 can also go to an unregulated or external. supply upto 24V DC which is referenced to this circuits gnd. More voltage more

t for taking a sample of current. 4-1000mV across 50E

in such a way, so as to maintain both

If there is 1V at pin 3 and no current is flowing pin 2 is at 0V so output goes positive and drives Q2. this results in a flow of current till a 1V builds across shunt, if it exceeds then output of opamp falls This reduces drive to transistor and hence current reduces. That is the part of V to I

20mA 4mA is good for 0 as low level measurements

1.0 V using IC3B. R14 is a representation of that 200mV offset set by R16 pot. the opamp IC3B adds both the input and this offset to get 200mV to

1000mV. for that the opamp IC3B icalculate the values for that.

Voltage-to-current signal conversion

In instrumentation circuitry, DC signals are often used as analog representations of physical measurements such as temperature, pressucurrent signals are used in preference to equal in magnitude throughout the series circuit loop carrying current from the source (measuring device) to the load (indicator, recorder, or controller), whereas voltage signals in a parallel circuit may vary from one end to the other due to resistive wire losses. Furthermore, current-sensing instruments typically have low impedances (while voltagehave high impedances), which gives currentimmunity.

In order to use current as an analog representation of a physical quantity, we have to have some way of generating a precise amount of current within a precise current signal when we might not know the resistance of the loop? The answer is to use an amplifier designed to hold current to a prescribed value, applying as much or as little voltage as necessary to the load circuit to maintain that value. Such an amplifier performs the function of a current source. An op-amp with negative feedback is a perfect candidate for such a task:

The input voltage to this circuit is assumed to be coming from some type of phytransducer/amplifier arrangement, calibrated to produce 1 volt at 0 percent of physical measurement, and 5 volts at 100 percent of physical measurement. The standard analog current signal range is 4 mA to 20 mA, signifying 0% to 100% of measurement rvolts input, the 250 Ω (precision) resistor will have 5 volts applied across it, resulting in 20 mA of current in the large loop circuit (with Rhow much wire resistance is present in that large loop, so long as the oppower supply voltage to output the voltage necessary to get 20 mA flowing through R250 Ω resistor establishes the relationship between input voltage and output current, in this

1000mV. for that the opamp IC3B is an analog computer, summer, subtracter. Try to now

current signal conversion

In instrumentation circuitry, DC signals are often used as analog representations of physical measurements such as temperature, pressure, flow, weight, and motion. Most commonly,

signals are used in preference to DC voltage signals, because current signals are exactly equal in magnitude throughout the series circuit loop carrying current from the source

he load (indicator, recorder, or controller), whereas voltage signals in a parallel circuit may vary from one end to the other due to resistive wire losses. Furthermore,

sensing instruments typically have low impedances (while voltage-have high impedances), which gives current-sensing instruments greater electrical noise

In order to use current as an analog representation of a physical quantity, we have to have some way of generating a precise amount of current within the signal circuit. But how do we generate a precise current signal when we might not know the resistance of the loop? The answer is to use an amplifier designed to hold current to a prescribed value, applying as much or as little voltage

he load circuit to maintain that value. Such an amplifier performs the function of amp with negative feedback is a perfect candidate for such a task:

The input voltage to this circuit is assumed to be coming from some type of phytransducer/amplifier arrangement, calibrated to produce 1 volt at 0 percent of physical measurement, and 5 volts at 100 percent of physical measurement. The standard analog current signal range is 4 mA to 20 mA, signifying 0% to 100% of measurement range, respectively. At 5

Ω (precision) resistor will have 5 volts applied across it, resulting in 20 mA of current in the large loop circuit (with Rload). It does not matter what resistance value R

s present in that large loop, so long as the op-amp has a high enough power supply voltage to output the voltage necessary to get 20 mA flowing through R

Ω resistor establishes the relationship between input voltage and output current, in this

s an analog computer, summer, subtracter. Try to now

In instrumentation circuitry, DC signals are often used as analog representations of physical re, flow, weight, and motion. Most commonly, DC

signals, because current signals are exactly equal in magnitude throughout the series circuit loop carrying current from the source

he load (indicator, recorder, or controller), whereas voltage signals in a parallel circuit may vary from one end to the other due to resistive wire losses. Furthermore,

-sensing instruments sensing instruments greater electrical noise

In order to use current as an analog representation of a physical quantity, we have to have some the signal circuit. But how do we generate

a precise current signal when we might not know the resistance of the loop? The answer is to use an amplifier designed to hold current to a prescribed value, applying as much or as little voltage

he load circuit to maintain that value. Such an amplifier performs the function of amp with negative feedback is a perfect candidate for such a task:

The input voltage to this circuit is assumed to be coming from some type of physical transducer/amplifier arrangement, calibrated to produce 1 volt at 0 percent of physical measurement, and 5 volts at 100 percent of physical measurement. The standard analog current

ange, respectively. At 5 Ω (precision) resistor will have 5 volts applied across it, resulting in 20 mA

). It does not matter what resistance value Rload is, or amp has a high enough

power supply voltage to output the voltage necessary to get 20 mA flowing through Rload. The Ω resistor establishes the relationship between input voltage and output current, in this case

creating the equivalence of 1-5 V in / 4-20 mA out. If we were converting the 1-5 volt input signal to a 10-50 mA output signal (an older, obsolete instrumentation standard for industry), we'd use a 100 Ω precision resistor instead.

Another name for this circuit is transconductance amplifier. In electronics, transconductance is the mathematical ratio of current change divided by voltage change (∆I / ∆ V), and it is measured in the unit of Siemens, the same unit used to express conductance (the mathematical reciprocal of resistance: current/voltage). In this circuit, the transconductance ratio is fixed by the value of the 250 Ω resistor, giving a linear current-out/voltage-in relationship.

• REVIEW:

• In industry, DC current signals are often used in preference to DC voltage signals as analog

representations of physical quantities. Current in a series circuit is absolutely equal at all points

in that circuit regardless of wiring resistance, whereas voltage in a parallel-connected circuit

may vary from end to end because of wire resistance, making current-signaling more accurate

from the "transmitting" to the "receiving" instrument.

• Voltage signals are relatively easy to produce directly from transducer devices, whereas

accurate current signals are not. Op-amps can be used to "convert" a voltage signal into a

current signal quite easily. In this mode, the op-amp will output whatever voltage is necessary to

maintain current through the signaling circuit at the proper value.

Howland Current Source for Grounded Load

Idea: Keeping a constant current by adding a current.

As we already know, the simple current source V-R is imperfect as the load affects the current. The voltage drop VL across the load (suppose again that it is a varying resistor RL) is harmful as it enervates the excitation voltage V. As a result, the current I decreases with the value VL/R.

So far, we aided the exciting voltage source by using an additional supplementary voltage source. Now, we will try to aid the voltage source injecting an additional current Is = VL/R by another "helping" current source. We may build it by connecting a "helping" voltage source Bs through another resistor Rs = R to the load.

Implementation: Using an additional op-amp helping current source.

Тhis genius idea is implemented in the famous Howland current source. In this clever circuit, the excitation voltage V and the resistor R form a basic current source. It produces a current I = V/R - VL/R, which passes from the left hand side through the load RL.

The non-inverting amplifier R1-R2-OA and the resistor R3 form a helping current source. It produces a current Is = VL/R, which passes from the right hand side through the load RL. As a result, a steady total current IL = V/R - VL/R + VL/R = V/R flows through the varying load RL. It is a genius idea, isn't it?

Variations Among Current-Sink/Source Circuits with a Single Op Amp Aug 21, 2006

Abstract: The design differences among five constant-current circuits are analyzed. Performance data are

discussed, and illustrate the different compromises between precision, dynamic impedance, and

compliance range for some of the circuit designs.

A controlled constant-current source or sink is a useful circuit block to use in designing electronic

applications such as sensor bias, amplifier bias, and special waveform generation. These types of circuits

exhibit high-output dynamic impedance, while delivering relatively large currents within the specified

voltage range.

Constant-current circuits are usually implemented with an op amp and a discrete external transistor.

Source, sink, and bipolar versions can also be designed with a single op amp and some resistors (Figure

1). The Figure 1 circuits A, B, and C are constant-current sinks offering different compromises among

precision, dynamic impedance, and compliance range. Circuit D is a bipolar current source whose

feedback connection is simpler than that of the usual Howland current pump (no positive feedback,

constant input impedance). Circuit E is a constant-current source. All circuits in Figure 1 exhibit excellent

linearity of output current with respect to input voltage.

Figure 1. Constant-current circuits: current sink (A); current sink (B); current sink (C); bipolar current

(sink/source) circuit (D); and current source (E).

The output of Circuit A is uncertain because of the op amp's quiescent current, which adds to the

calculated output. (The 25µA quiescent current for the op amp shown, however, is negligible in most

applications.) Circuit B behaves similarly to Circuit A, but subtracts quiescent current from the ideal

output-current value. Circuit C is a sink with no quiescent current error, and Circuit D sinks or sources

current depending on the polarity of the input voltage (i.e., a bipolar output). The performance of Circuit

D depends on good resistor matching for the pairs R1-R2 and R3-R4, and good tracking between the +V

and -V power supplies. Any difference between the absolute values of positive and negative supply

voltage appears as an offset at zero input voltage.

Circuit E is a current source, which requires close matching of the R2-R3 and R4-R5 resistor pairs to

ensure insensitivity to changes in the supply voltage. Output currents for the different circuits can be

calculated using the following formulas:

Circuit (A) IOUT = -[VIN/RLOAD + (25µA)]

Circuit (B) IOUT = -[VIN/RLOAD - (25µA)]

Circuit (C) IOUT = -(VIN/RLOAD)

Circuit (D) IOUT = -2VIN/RLOAD

Circuit (E) IOUT = VIN/RLOAD

The formula for Circuit D assumes perfect matches: R3 = R4, R1 = R2 and +V = -V. It also assumes that R3

>> R1.

The data in Figure 2 reference the circuits of Figure 1. Figure 2 shows the dynamic impedance and range

of useful output voltage (compliance) for a fixed value of output current. The output current for all plots

is 5mA. (A high value was chosen to show the higher end of the current amplitude range.) Depending on

the application, you may be able to improve the dynamic impedance and range of these circuits with a

judicious choice of op amp and resistor values.

Figure 2. Output current vs. output voltage is illustrated for the circuits of Figure 1. Note that for Circuits

B and C, the dynamic output impedance closely resembles an ideal current source: ΔVOUT/ΔIOUT = ∞

Computer Controlled 100ma Current Source (July 11, 2008) Often in industrial control systems a constant current source is needed, which is controlled

by a computer and referenced to circuit ground. The circuit below converts a zero to 5v

signal from a computer’s analog output into a current, with a full scale of 100ma. The circuit

shown requires a 9v DC supply but any voltage from 9v to 12v will work.

Click on Drawing Below to view PDF version of Schematic

Constant current source circuit

We can use a voltage reference turn into a Constant current source circuit and use pnp-transistors for current-boosting

By op amp action ,the voltage across R is away Vref , give

Ie = Vref / R Ie = Ib + Ic = (Ic/B) + Ic Ic = [B/(B+1)] *Ie Io = Ic = [B/(B+1)] * Vref / R about Vref / R VL = Io*RL Where maximum voltage of Vl(max) <>

We can adjust output current by adjust current setting resistance R

ADJUSTABLE CURRENT SOURCE Circuit

The HIP5600 can supply a 450 A (20%) constant current. It makes use of the internal bias network.

See Figure 27 for bias current versus input voltage. With the addition of a potentiometer and a 10 F capacitor the HIP5600 will provide a constant current source. IOUT is given by Equation 13 in Figure 16.

FIGURE 16. ADJUSTABLE CURRENT SOURCE 1.5A ADJUSTABLE CURRENT SOURCE Circuit

The LM317 is an adjustable 3−terminal positive voltage regulator capable of supplying in excess of 1.5 A over an output voltage range of 1.2 V to 37 V. This voltage regulator is exceptionally easy to use and requires only two external resistors to set the output voltage. Further, it employs internal current limiting, thermal shutdown and safe area compensation, making it essentially blow−out proof. The LM317 serves a wide variety of applications including local, on card regulation. This device can also be used to make a programmable output regulator, or by connecting a fixed resistor between the adjustment and output, the LM317 can be used as a precision current regulator. Features -Output Current in Excess of 1.5 A -Output Adjustable between 1.2 V and 37 V -Internal Thermal Overload Protection -Internal Short Circuit Current Limiting Constant with Temperature -Output Transistor Safe−Area Compensation -Floating Operation for High Voltage Applications -Available in Surface Mount D2PAK−3, and Standard 3−Lead Transistor Package -Eliminates Stocking many Fixed Voltages -Pb−Free Packages are Available

4-20mA Pressure Transducer Circuit

Complete 4-20mA Pressure Transducer Solution with PGA309 and XTR117 The XTR117 is a precision current output converter designed to transmit analog 4-20mA signals over an industry-standard current loop. It provides accurate current scaling and output current limit functions. XTR117 datasheet pdf The PGA309 is a programmable analog signal conditioner designed for bridge sensors. The analog signal path amplifies the sensor signal and provides digital calibration for zero, span, zero drift, span drift, and sensor linearization errors with applied stress (pressure, strain, etc.). The calibration is done via a One-Wire digital serial interface or through a Two-Wire industry-standard connection. The calibration parameters are stored in external nonvolatile memory (typically SOT23-5) to eliminate manual trimming and achieve long-term stability. PGA309 datasheet pdf

4-20mA Current-Loop Transmitter Circuit

The XTR117 is a precision current output converter designed to transmit analog 4-20mA signals over an industry-standard current loop. It provides accurate current scaling and output current limit functions. The on-chip voltage regulator (5V) can be used to power external circuitry. A current return pin (IRET) senses any current used in external circuitry to assure an accurate control of the output current. FEATURES _ LOW QUIESCENT CURRENT: 130 uA _ 5V REGULATOR FOR EXTERNAL CIRCUITS _ LOW SPAN ERROR: 0.05% _ LOW NONLINEARITY ERROR: 0.003% _ WIDE-LOOP SUPPLY RANGE: 7.5V to 40V _ MSOP-8 AND DFN-8 PACKAGES XTR117 datasheet pdf

0-5V To 4-20mA Current-Loop Transmitter Circuit

The AM422 is a low cost monolithic voltage– to–current converter specially designed for analog signal transmission. The AM422 is available in a 3– or 2–wire version, which allows applications with flexible input voltage ranges to be used for a standard output current. Output current range and current offset level are freely adjustable by external resistors. The IC consists of three basic sections: an operational amplifier input stage for single ended input signals (0.5–4.5V, 0–10V, or other), a programmable 4.5 to 10V reference for transducer excitation, and a current output, freely adjustable in a wide current range (4–20mA, 0–20mA, other). With the broad spectrum of possible input signals the AM422 is a flexible and multipurpose voltage–to–current converter for single ended transducers or voltage transmission. FEATURES - Wide Supply Voltage Range: 6...35V - Wide Operating Temperature Range: –40°C...+85°C - Adjustable Voltage Reference:4.5 to 10V

- Operational Amplifier Input:0.5...4.5V, 0...5V, other - Adjustable Offset Current - Available as Three– (0/4...20mA) or Two–Wire Version (4...20mA) - Adjustable Output Current Range - Protection Against Reverse Polarity - Protected Current Output AM422 datasheet pdf

4-20mA Current Loop Receiver Circuit

4-20mA Current Loop Receiver with Input Overload Protection circuit

The RCV420 is a precision current-loop receiver designed to convert a 4–20mA input signal into a 0–5V output signal. As a monolithic circuit, it offers high reliability at low cost. The circuit consists of a premium grade operational amplifier, an on-chip precision resistor network, and a precision 10V reference. The RCV420 features 0.1% overall conversion accuracy, 86dB CMR, and ±40V common-mode input range. FEATURES -COMPLETE 4-20mA TO 0-5V CONVERSION

- INTERNAL SENSE RESISTORS -PRECISION 10V REFERENCE - BUILT-IN LEVEL-SHIFTING - ±40V COMMON-MODE INPUT RANGE - 0.1% OVERALL CONVERSION ACCURACY - HIGH NOISE IMMUNITY: 86dB CMR A current-sensing circuit derives its power from the 4-20-mA current loop.

4-20mA Current Loop Receiver with fault protection and digital-signal recovery circuit

Figure shows one form of flexible fault protection for the 24VDC power supply of a 4-20mA loop. Also included is circuitry for recovering a digital signal superimposed on that loop. U1 (a high-side current- sense amplifier with comparator and reference) senses the loop current in R1 as an 8-40mV voltage and amplifies it by 100, producing an output-voltage range of 0.8V to 4V. That output (VOUT) can directly drive external meters, strip-chart recorders, and A/D converter inputs. More pdf