Wednesday, May 25, 2011
Third-Order High-Pass Filter
This report focuses on active third-order high-pass filter design using operational amplifiers. High-pass filters are commonly used to implement high frequency in a system. Design of Third-order filters is the main topic of consideration. To illustrate an actual circuit implementation, separated into two types of filters first-order and second-order.
A third-order high pass filter is formed by connecting in series first and second order high pass section and so on. As the order of the filter increased, the actual stop-band response of the filter approaches its ideal stop-band characteristic. The overall gain of the filter is fixed because all the frequency determining resistors and capacitor are equal.
A filter is a device that passes electric signals at certain frequencies or frequency ranges while preventing the passage of others. High Pass Filter (HPF), sometimes called a low-cut filter, is a filter that high frequencies can be transmitted well and frequencies lower than the cutoff frequency are attenuated or reduced. The actual amount of attenuation for a particular frequency varies from filter to filter.
A third-order high pass filter is formed by connecting in series first and second order high pass section and so on. In the stop-band the gain of the filter changes at the rate of 20 dB/decade for first-order filter and at 40 dB/decade for second-order filter. This means that, as the order of the filter is increased, the actual stop-band response of the filter approaches its ideal stop band characteristic.
Higher order filter are formed simply by using the first and second order filter. A third-order high pass filter is formed by connecting in series first and second order high pass section and so on. Although there is no limit to the order of the filter that can be formed, as the order of the filter increase, so does its size.
Also, its accuracy declines, in that the difference between the actual stop-band response and the theoretical stop-band response increase with an increase in the order of the filter. Note that in the third order filter the voltage gain of the first order section is one, and that of the second order section is two. This gain values are necessary to guarantee butter-worth response and must remain the same regardless of the filter’s cutoff frequency. Furthermore, the overall gain of the filter is fixed because all the frequency determining resistors and capacitor are equal.
Generally, the minimum-order filter required depends on the application specifications. Although a higher-order filter than necessary gives a better stop-band response, the higher-order type filter is more complex, occupies more space, and is more expensive. As the order of the filter increased, the actual stop-band response of the filter approaches its ideal stop-band characteristic.
555 Timer Modeling
The 555 IC timer chip is widely used as both a monostable and astable multivibrator. Originally introduced by Signetics using bipolar technology, the 555 is now available from several manufacturers in bipolar and CMOS technology.
First-Order Band-Pass Filter
Application of first order band pass filter to find out the output voltage gain magnitude of specific frequency. A specific range of frequency can pass through the amplifier which has a specific bandwidth of this band pass filter.
A band pass filter is a frequency selector. It allows one to select or pass only one particular band of frequencies from all other frequencies that may be present in a circuit. This type of filter has a maximum gain at a resonant frequency.
A band pass filter is the combination of high pass and low pass filter combination. It has a pass band between two cut off frequency fH and fL such that fH > fL. Any input frequency outside this pass band is attenuated. There are two types of band pass filters wide band pass and narrow band pass. If the quality factor Q < 10 and Q > 10 then it would be wide band pass and narrow band pass filter.
The relationship between Q, the 3-dB bandwidth and the center frequency fc is given by,
According to Figure 01 of band pass filter circuit, there are two sections. One is first order high pass section and other is low pass section.
For first order high pass section the output voltage equation is,
For first order low pass section the output voltage equation is,
Putting the value of first order high pass section output voltage from equation (1),
So the final voltage equation of first order band pass filter is,
The voltage gain magnitude of the band pass filter is equal to the product of the voltage gain magnitudes of the high pass and low pass filters.
Therefore the equation (2) is,
Where, = Total band pass gain
f = frequency of the input signal (Hz)
= low cut off frequency (Hz)
= high cut off frequency (Hz)
The above first order low pass band width filter was designed by taking following precautions,
a.
Limited magnitude of input voltage was applied at the input, so that the op-amp must not be driven to saturation.
Limited magnitude of input voltage was applied at the input, so that the op-amp must not be driven to saturation.
b. Only selected frequency can pass through the filter.
c. Overall gain of the first order band pass filter is the multiplication of high pass filter gain and low pass filter gain.
Electric Analog Computation
Electronic analog computation is such kind of electronic computation in which basic analog computing elements such as adders ,integrators ,multipliers, comparators etc are used to solve any desired equation such as differential equations etc .It is the basic concepts of analog computer.
In this a differential equation is solved by electronic analog computation. Let a differential equation be :
D2v+k1Dv+k2v-v1=0……………….(1) Where, k1 and k2 are constant terms.
In the starting I assumed that D2v is available in the form of a voltage .Then by means of an integrator I will get the voltage proportional to Dv. A second integrator gives the voltage proportional to v .Then an adder gives –( k1Dv+k2v-v1) From the equation it is equal to D2v
and hence the output of this summing amplifier is fed to the input terminal ,where I had assumed that D2v was available in the first place.
The integrator 1 has a time constant RC=1s, and hence its output at terminal 1 is –Dv .This voltage is fed to a similar integrator 2 and the voltage at terminal 2 is +v. The voltage at terminal 1 is fed to summing amplifier 1 which gain is 1 and in the output terminal 3 I get + k1Dv- v1.
where k1=(R/R1).At the end the output of terminal 2 and 3 are fed to summing amplifier 2,from where I will get D2v= - (k1Dv+k2v-v1) at terminal 4.
Fig1.1: Electronic analog computing circuit for calculating a differential equation .
Design a Temperature Indicator
This circuit is a temperature indicator circuit or differential instrumentation amplifier using a transducer bridge. This circuit is calibrate in degrees Celsius or Fahrenheit. In the circuit used buffer in and points for exact voltage of and points. Because gain is always 1 of buffer circuit. Then the output voltage of buffers is input voltage of differential amplifier.
The differential amplifier is difference voltage of and points using 741 Op Amp. When temperature is increased then resistance is also decreased and output voltage is decreases and when temperature is decreased then resistance is also increased and output voltage is increases.
The temperature indicator is a circuit that indicates of temperature in degrees Celsius or Fahrenheit. The temperature is inversely proportional to the resistance or transducer.
Fig.01 - Temperature indicator
In the circuit used as the transducer in the bridge circuit is a thermistor and replaced output voltmeter to temperature indicating meter. Then temperature indicating meter is calibrate in degrees Celsius or Fahrenheit.
The bridge can be balanced at a desired reference condition, for instance 250C. As the temperature varies from its reference value, the resistance of the thermistor changes and the bridge become unbalanced. This unbalance bridge in turn produces the meter movement.
Analog weight Scale
The most common weight-scale implementation is to use a transducer bridge, with voltage output directly proportional to the weight placed on it. The trend in weight scales towards higher accuracy and lower cost has produced an increased demand for high-performance analog signal processing at low cost.By connecting a strain Gage in the bridge ,the differential instrumentation amplifier can be converted in to a simple analog weight scale.
In the analog weight scale, strain Gage elements are connected in all four arms of the bridge. The elements are mounted on the base of the weight platform, so that, when an external force or weight is applied to the platform, one pair of elements in the opposite arms elongates, whereas the other pair of elements in the opposite arms compresses.
On the other hand, When no weight is placed on the platform, the bridge is unbalanced, RT1=RT2=RT3=RT4=R, and the output voltage of the weight scale can be zero. When a weight is placed on the scale platform, the bridge becomes unbalanced. In other words, when the weight is placed on the platform, RT1 and RT3 both decrease in resistance and RT2 and RT4 both increase in resistance (Figure 01).
The analog weighing measuring scales need to be hung properly before we start measuring anything using the same. They never allow measuring the weight of anything that we want as they have some limitations. If we think about measuring anything using an analog scale we need to check the analog lines to get the nearby accurate measurement.
One thing we need to keep in our mind while using such a scale is the limitation, if we load anything which weights more than the mentioned limit by the manufacturer of the scale we may end up damaging the same. Analog scales were never capable of providing accurate weight of any item and at the same time, it involves human effort to measure anything where the chances of human error are quite normal
One thing we need to keep in our mind while using such a scale is the limitation, if we load anything which weights more than the mentioned limit by the manufacturer of the scale we may end up damaging the same. Analog scales were never capable of providing accurate weight of any item and at the same time, it involves human effort to measure anything where the chances of human error are quite normal
For better accuracy, a microprocessor-based digital weight scale may be constructed.But it is much more complex and expensive than the analog scale.
The Differential Input and Differential Output
For getting balanced differential output we use this circuit. In this circuit two source is present, so the superposition theory is applied to get the output. In this circuit both the inverting and non-inverting terminal is working.It rejects the common-mode voltages, so it is very useful in noisy environments.
A differential input and differential output amplifier using two identical Op amp. It is most commonly used as a preamplifier and driving push-pull arrangement. The differential input and output are inphase or the same polarity provided Vin =Vx – Vy and
Vo =Vox – Voy
When we want to find out the 1st op-amps output VOX , we will use the superposition theory.
When we get VX is active , VY is inactive then ,
In non inverting terminal
V1= (1+ )VX
When we get Vy is active, Vx is inactive then,
In inverting terminal,
V1 = - Vy
So, Vox = V1+V1
=(1+ )VX - Vy
Fig: The circuit diagram of the differential input and output amplifier.
When we want to find out the 2nd op-amps output VOy ,we will use superposition theory.
When we get VX is active , VY is inactive then
In inverting terminal ,
V2= - Vx
Again, when we get Vy is active , Vx is inactive then
In non inverting terminal we get,
V2= (1+ )Vy
So, Voy = V2+V2
=(1+ )Vy - Vx
So the output result
Vo = Vox – Voy
= (1+
)VX - Vy –[(1+ )Vy - Vx ]
)VX - Vy –[(1+ )Vy - Vx ] = (1+ ) ( VX - Vy ) +
( VX - Vy )
) ( VX - Vy ) +( VX - Vy ) = ( VX - Vy ) (1+
)
)Design
To design a input and differential output amplifier, taking a differential output of at least 3.7V and the differential input Vin =10V.
We know,
Vo = ( VX - Vy ) (1+ )
) Or, 3.7 = (0.1) (1+ )
) Or, 37 = (1+ )
) Or, 36 =
Or, Rf = 18 R1
Let, R1 = 100Ώ, then Rf = 1.8 KΏ .
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