Opamp-Circuits (Page 3)

his Circuit converts a voltage control output from a Process Controller to be converted into a Current Control if the AC-Drive or Valve needs a Current Control Signal.

Significance of Current Loop 4 to 20 mA Standard

Voltage to Current Convertor using LM723

This is a three wire voltage to current loop converter. The 1-5 V DC is attenuated and fed to pin 5 LM723 opamp section which tries to maintain the same voltage at pin 10 across the 10 E, thereby producing a open collector constant current sink proportional to the 1-5V input. By trimming the attenuator you can scale-calibrate 1-5V input to 4-20mA output for looping many instruments in series, like a controller, recorder or PLC. With a supply voltage upto 24V, three instruments can be looped. The connection to pin 6 is required to convert 0-1 input to 4-20mA.

All the transmitter circuits can be seen here. Industrial Process Control Circuits

This circuit was designed by me in the eighties, the 555 was for negative supply, The whole thing went into the anodized cast aluminuim head of a sensor.

How 4-20mA Works

In the circuit, use only metal film resistors (MFR) of 1 per cent tolerance, as this is an instrumentation application. Power supply should be a stable +5V, -5V supply, for which one can use 7805 and 7905 regulators.

The inputs TC+ and TC- terminals should go to a 4-way barrier terminal block, the 2 extra terminals are used to mount TH1 Cu thermistor. This forms an isothermal block, which is good enough.

A simple way to make a TH1 Cu thermistor, is to take a 1 Meg-ohm 2W resistor as a former and wind 2 meters of 46 SWG enameled copper (Cu) wire (5.91 ohm/meter) over it. This gives a 12-ohm value. Terminate wire ends on resistor leads.

Thermocouple Temperature using DPM or DMM

Test and Calibration –

For calibration, you will need a DMM-DPM and a milli-volt source (as shown in the Fig.). First connect source to terminals TC+ and TC-, then set source to 0.00 mV (verify with DMM for zero). The output across +out and -out (use DMM) terminals must be mV representing the room temperature (RT). For example, if RT is 30° C (use a glass thermometer) then +out should be 30mV at 0mV input. Adjust VR1 till 30mV is read at +out terminal. This is ‘zero cal’.

A sample and hold is like an analog memory. If The digital control A is low 4066 switch is open, and when A is high switch is closed. U2B is a buffer so as to ensure quick charging of C1 thru 4066 on resistance of 100E.

Simple Sample and Hold with CD4066

Mixed and Interface Circuits

U2A is a FET input opamp buffer which does not load or drain the cap C1. When A goes high the input analog sample is stored in C1. A has to be high for say 10*1uF*100E = 1mS, so that a proper stable sample is stored. When A is low C1 undergoes very slow discharge as opamp input resistance and 4066 off resistance is in giga ohms. The accuracy of reading Vout falls with respect to time due to leakage currents.

Learn how to use the AD590 to measure environment temperatures for display, logging or cold junction compensation.

The voltage at the point 1 of R4 will be :Vo=( 1+ ( 10K/22K)) * Vref = 3.63V as nominal Vref is
2.5V.AD590 is a current source which gives 1 uA / kelvin, It is independent of the voltage across the device. you can treat it like a current source or sink or impedance. total voltage across AD590 is 5V as opamp pin 2 is at virtual ground.

Analog Circuits – OpAmp, Signal Condition, Mixed Signal.

AD590 based Temperature Sensor

This is the way you try to understand the design.

The AD590, here is a constant current sink as cathode goes to -5. The current it sucks away or drains from node pin 2 of OP07 is 1uA/ kelvin. at 0 deg C the current drained is 273 uA at 26 deg C it is 300uA.

You know according to theory that the amount of current entering the node, is equal to the amount of current leaving the node. do not look at voltages now, look at the currents. the AD590 drinks 273uA from Node pin 2 of OP07 at 0 deg C. Now no current can come from opamp OP07 pin 2 as resistance is in giga ohms and leakage in pico amps. now the pot R5 and resistor R4 are just in series and connected to 3.63 V as established earlier. The TL431 is a shunt regulator with reference and has a low impedence. Now the R5 + R4 combination should not load the TL431, that is not the case as 3.6 / 10K = 360uA .

By varying R5 pot you can pump 3.6 / 10K = 360uA down to 130uA when R5 is max into node pin 2 of OP07. This pot will be calibrated with AD590 in ICE to give a 0 mV output of the Op07. When calibrated R5+R4 pump 273 uA into node pin 2 of op07. this is sucked away by the AD590 which is draining 273uA at 0 deg C. This leaves the pin 2 at zero potential as currents leaving = currents entering.

Now to understand the opamp functioning.

The pin 2 of opamp is a 0 potential as calculated above and pin 3 also is at zero pulled down by R7. Now as both inputs are at same potential the output of opamp also is zero. The feedback resistors R1 and R2 will carry no current as both their ends are at 0. the Vout is now 0 mV and AD590 is on a block of ICE and opamp is stable.

If pin 2 (-) becomes more dominant or positive than pin 3 (+) the output swings negative. If pin 3 (+) becomes more dominant or positive than pin 2 (-) the output swings positive. The opamp on feedback tries to maintain both the inputs at the same potential. This thumb rule can be used to make opamp oscillate, amplify or compute.

Now what happens when the AD590 is removed from the block of ICE. It comes to room temperature say 26 deg C which means 300uA. Now the AD590 demands to draw 300uA from node pin 2 of OP07. The R4 + R5 from 3.6 V can give 273uA as it is fixed, not a uA more. The rest which is 300 – 273 = 27uA leads to a drop in potential at pin 2 and it turns negative. as demand is greater than supply. which makes pin 3 which is at zero more positive than pin 2. ( theory : 0 is positive compared to -1) as pin 3 is more dominant opamp swings positive as per thumb rule. and a current starts flowing thru R1 + R2 till the current reaches 27uA. at this point the extra current 27uA drawn by AD590 is supplied by opamp thru R1+R2. The Pin 2 now comes to 0 as currents leaving = currents entering.

Test & Measurement, Instrumentation

At this point the voltage at opamp output is given by ( R1 + R2 ) * 27uA = 270mV (assume R1+R2 is 10K after calibration) now opamp gives 10mV per deg C.as opamp now is a closed loop control the rise and fall in temperature, results in AD590 current variation which produces a proportional OP07 output.

Now the explanation above is in steps but all that happens in real time in an instant.

This is the best Instrumentation OpAmp, Great CMRR, ensure supply has no ripple and keep analog and digital grounds separate. Ri can be replaced with a trimpot and resistor to alter gain. Connect a preset ends to pins 1 and 8 and preset wiper to VCC for Offset Null when high gains are configured.

Thermocouple and Pt-100 RTD

The Input zeners and diodes form a protective clamp for all voltages above VCC-VDD. If supply is changed to +12 -12 change zeners to 12V zeners. Use similar Zeners at output to protect Output from being zapped by overvoltages or high energy – voltage*frequency transients. Add plastic capacitors across Rf for damping AC operation or ripple. Also avoid floating inputs by providing a bias.

3 Op-Amp Differential Instrumentation Amp

Vout = (Vp – Vn) * (2Rf+Ri)/Ri

Related Reading

The Input Impedance of this module is very high and is symmetric. This circuit can be used for strain gauges and for four wire measurements. If inputs are in mV use OP07. The merit is that it uses only 2 OpAmps yet has high differential Input Impedance.

Dual Differential Amp – Interactive Simulation

The Outputs of Opamps are low impedance but still have limits they cannot drive more than a few mA of Current into the Load. If low ohmic value loads are to be applied use external transistors as amplifiers. If inputs Vn-Vp are floating Outputs may be random or Oscillating, it is good to have a bias network of 10M resistors to a potential even zero or COM this enables Vout when input floats.

Two Op-Amp Differential Amplifier

Vout = (Vp – Vn) * (Rf+Ri)/Ri

Related Reading

Precision Instrumentation Amplifiers

If output impedance of a point is a high value then connecting another circuit at that point will load it resulting in malfunction or error. Buffers are used as interface between circuits. Low impedance of an output means it can source sink lot of current, when you need 2 opamps use LF353 or TL072 which are dual opamps.

Blind Dial Proportional Temperature Controller

A non-inv FET input is the best buffer, for inverting buffer use high R values Using very high R values like 2.2M or higher requires a glass epoxy PCB and guard rings around pin 2, 3 to prevent leakage currents on the PCB reaching the PINs. Also moisture and dust has to be prevented by using RTV coating or Varnish. Use 78L05 79L05 for the dual supply required by this circuit.

Buffer or Unity Gain Op-Amp

Vout = -(Vin) for inverting

Vout = Vin for non-inverting

 

This amplifies the difference between two inputs Vp and Vn the low impedance of this configuration is a drawback, but can be used in analog computing. Optimum VCC VDD can be +12/-12. AC signals common to Vp and Vn are canceled by this configuration.

Use a capacitor like 10nF plastic from pin 2 to 3 or across R2 to make circuit stable. For AC applications use LF351 TLO71 as they have good slew rate and also are FET inputs. For AC applications use a capacitor (1uF) in series with Ri to block DC Components. The Inputs have asymmetrical input impedance this affects CMRR, also use 1% tolerance MFR resistors for Rf and Ri.

Differential Amplifier - Op-Amp Circuits

Vout = (Vp – Vn) * (Rf/Ri)

Instrumentation and Measurement Circuits