Differential Amplifier : input resistance

transcript

Differential amplifiers are widely used in engineeringinstrumentation

Differential Amplifier : input resistance

iRiRiRvv

Differential Amplifier: why useful

Difference amplifier find application in many areas, in particular in the design of instrumentation systems.

ExampleConsider a transducer that produces between its two output terminals a relatively small signal around 1mV, But between each of the two wires (transducer terminals) and the ground there may be a large interference around 1V.

Differential Amplifier: why useful

The instrumentation amplifier must reject this large interference signal (a common-mode signal), and amplify the small difference (differential ) signal.

Vcm is the common mode signalVd is the differential signalThis Circuit has low input resistance and the gain cannot be easily varied

A difference Amplifier

To overcome the drawbacks of the difference amplifier the following instrumentation amplifier has been proposed

A Instrumentation Amplifier

The circuit consists of two stage 1- the first stage formed by op amp A1 and A2 with their associated resistors.2- the second stage is formed by op amp A3 together with its four resistors.

Find v0

Solution:

An ideal differential amplifier will not have any output that depends on the value of the common mode voltage; ( The circuit gain for common mode voltage, Acm will be zero.)

The common mode rejection ratio (CMRR) of a differential amplifier is defined as the ratio of the gain to the common modegain (Acm). The common mode rejection (CMR) is the CMRR expressed in dB.

Common Mode Rejection Ratio (CMRR)

Clearly, the larger these numbers, the better the differential amplifier. Typical values of CMR range from 80 to 100 dB.

Let, R1=R2 and Rf=R3, then

)(vand

gain aldifferenti theis

Common mode signal

CMRRCMRVV

log20AACMRRrejection modeCommon

gain modecommon

Any signal common to both inputs is effectively canceled and free to put the ground anywhere, the noise is common.

)A(V s CMRRVVAAVv

ncmsout

A typical example of an op amp is a 741 integrated circuit IC.

Op Amp Integrated Circuit IC

Compensation for input offset voltage can be provided as a variable resistor connected to two terminals (offset null).

Differentiator Amp

The output voltage is given by:

Integrator Amp

Example

Develop a circuit to realize the following equation

dtVVV ininout 410

Example

Develop a circuit to realize the following equation( Solution)

Example

The above circuit consists of three op-amp circuits. The first one at the top to left is used to realize the gain = (-10), while the second one is the integrator with gain =(-4). The last one atthe top to right is a summing amplifier with inverse sign to produce the final output.

Example

Develop a circuit to realize the following equation

dtVVV ininout 410

Example

Develop a circuit to realize the following equation( Solution)

Nonlinear (logarithmic) amplifier

The op-amp can also implement a nonlinear relationship, this is achieved by placing a nonlinear element in the feedback of the op-amp. For example, a diode can be used as shown figure.

The summation of currents provides

Were I is the current passes through R and at the same time in the diode. Note that the current in the diode has a nonlinear relation as function of Vout. In the diode we have the relation

Where Io= amplitude constant and α=exponential constant. The inverse of this relation is the logarithm, and thus

Were I is the current passes through R and at the same time in the diode. Note that the current in the diode has a nonlinear relation as function of Vout. In the diode we have the relation

Where Io= amplitude constant and α=exponential constant. The inverse of this relation is the logarithm, and thus

which constitutes a logarithmic amplifier.

Analog signal conditioning

Measurement systems are usually used for:

• Displaying data about some event or variable

• Inspection or testing, i.e to determine whether an item is to specification or not (calibration)

• Providing feedback information in the control loop

Example: temperature measurement

The measurement system consists of basic three components:• Sensor to transform the variations in the physical variable into a measured form(resistance, displacement, current or volt)• Signal conditioning to change the sensor output signal either in its form or range to met the control loop requirements (signal processing)• Display element to monitor the variations in the physical variable

Analog signal conditioning includes:

• Signal level change- Attenuation (gain<1)- Amplification (gain>1)

• Linearization, if the transducer gain is nonlinear, anon linearamplifier can be used to compensate this problem. Finally, the over all gain of transducer and amplifier is linear.

Note that, linearity is a very important characteristic in control loop that we have to maintain.• Conversion of the nature of signal as- Passive change (resistance, capacitance or inductance) to active change (volt or current)- Voltage to current- Current to voltage- Electric current to pneumatic signal• Filtering and impedance matching

Signal conditioning circuits could be implemented using:

• Passive circuits- Divider circuits- Bridges- RC filters• Op-amps

1- Voltage divider

Passive circuits

2- Bridge circuit (Wheatstone bridge)

1- Current to voltage converter

Active circuits

The current signal I (mA) is supplied from the transducer and R (KΩ) , so the output voltage is:

To avoid the negative sign in the above relation the circuit maybe modified as following

Active circuits

Develop a signal conditioning circuit to convert a transducer output (4 20 mA) into a voltage range (0 10 Volt) and draw the circuit diagram.

Example

Solution:First step we have to change the current signal into voltage signal. selecting R = 100 Ω.

Example

Therefore the current range will be converted into avoltage range (0.4 2 volt).

Example: solution

Second step, we will use an amplifier to obtain the requiredoutput voltage range in assuming linear relation as:

(Vout = A Vin + offset).

Substitute in the above relation by the values (Vout= 0 at Vin= 0.4 V) and (Vout = 10 V at Vin= 2 V). We will obtain the equations:

0 = 0.4 A +offset (1)10 = 2 A +offset (2)

Solving (1) and (2) we get:

Offset = -2.5 and A 6.25

solution

Differential Amplifier : input resistance

Documents