Friday, July 9, 2010

Full Wave Rectifier

In a Full-wave rectifier circuit two diodes are used, together with a transformer whose secondary winding is split equally into two and has a common centre tapped connection, (C). Now each diode conducts in turn when its Anode terminal is positive with respect to the centre pointC.

The circuit consists of two Half-wave rectifiers connected to a single load resistance with each diode taking it in turn to supply current to the load. When point A is positive with respect to point B, diodeD1conducts in the forward direction as indicated by the arrows. When point B is positive (in the negative half of the cycle) with respect to point A, diode D2 conducts in the forward direction and the current flowing through resistor R is in the same direction for both circuits. As the output voltage across the resistor R is the sum of the two waveforms, this type of circuit is also known as a "bi-phase" circuit.

As the spaces between each half-wave developed by each diode is now being filled in by the other diode the average DC output voltage across the load resistor is now double that of the single half-wave rectifier circuit and is about 0.637Vmax of the peak voltage, assuming no losses.
The peak voltage of the output waveform is the same as before for the half-wave rectifier provided each half of the transformer windings have the same rms voltage value. To obtain a different d.c. voltage output different transformer ratios can be used, but one main disadvantage of this type of rectifier is that having a larger transformer for a given power output with two separate windings makes this type of circuit costly compared to a "Bridge Rectifier" circuit equivalent.

Full-wave Bridge Rectifier

The output of the half-wave rectifier is DC (it is all positive), but it is not a good a power supply for a circuit as the output voltage continually varies between 0V and Vs-0.7V. Further, half the time there is no output at all.

The circuit in figure 3 addresses the second of these problems. Here, output voltage does not remain 0V at any time. Here, four diodes are arranged in such a way that both the positive and negative parts of the AC waveform are converted to DC. The resulting waveform is shown in figure 4.


Figure 3: A full-wave rectifier When the AC input is positive, diodes A and B are forward-biased, while diodes C and D are reverse-biased. When the AC input is negative, the opposite is true - diodes C and D are forward-biased, while diodes A and B are reverse-biased.

While the full-wave rectifier is an improvement on the half-wave rectifier, its output still isn't suitable as a power supply for most circuits since the output voltage still varies between 0V and Vs-1.4V. So, if you put 12V AC in, you will 10.6V DC out.

Figure 4: Full-wave rectification

Half wave Rectifier

The purpose of a rectifier is to convert an AC waveform into a DC waveform. Rectifier use components called diodes to convert AC into DC. A diode is a device which only allows current to flow through it in one direction. In this direction, the diode is said to be 'forward-biased' and the only effect on the signal is that there will be a voltage loss of around 0.7V. In the opposite direction, the diode is said to be 'reverse-biased' and no current will flow though it.

Figure 2 shows the AC input waveform to this circuit and the resulting output. As you can see, when the AC input is positive, the diode is forward-biased and lets the current through. When the AC input is negative, the diode is reverse-biased and the diode does not let any current through, meaning the output is 0V. Because there is a 0.7V voltage loss across the diode, the peak output voltage will be 0.7V less than Vs.





Thursday, July 8, 2010

Astable Multivibrator

The output remains at each voltage level for a definite period of time. If you looked at this output on an oscilloscope, you would see continuous square or rectangular waveforms. The astable multivibrator has two outputs, but NO inputs.

Let's look at the multivibrator in figure 3-3 again. This is an astable multivibrator. The astable multivibrator is said to oscillate. To understand why the astable multivibrator oscillates, assume that transistor Q1 saturates and transistor Q2 cuts off when the circuit is energized. This situation is shown in figure 3-4. We assume Q1 saturates and Q2 is in cutoff because the circuit is symmetrical; that is, R1 = R4, R2 = R3, C1 = C2, and Q1 = Q2. It is impossible to tell which transistor will actually conduct when the circuit is energized. For this reason, either of the transistors may be assumed to conduct for circuit analysis purposes.

Figure 3-4. - Astable multivibrator (Q1 saturated).

Essentially, all the current in the circuit flows through Q1; Q1 offers almost no resistance to current flow. Notice that capacitor C1 is charging. Since Q1 offers almost no resistance in its saturated state, the rate of charge of C1 depends only on the time constant of R2 and C1 (recall that TC = RC). Notice that the right-hand side of capacitor C1 is connected to the base of transistor Q2, which is now at cutoff.

Let's analyze what is happening. The right-hand side of capacitor C1 is becoming increasingly negative. If the base of Q2 becomes sufficiently negative, Q2 will conduct. After a certain period of time, the base of Q2 will become sufficiently negative to cause Q2 to change states from cutoff to conduction. The time necessary for Q2 to become saturated is determined by the time constant R2C1.

The next state is shown in figure 3-5. The negative voltage accumulated on the right side on capacitor C1 has caused Q2 to conduct. Now the following sequence of events takes place almost instantaneously. Q2 starts conducting and quickly saturates, and the voltage at output 2 changes from approximately -VCC to approximately 0 volts. This change in voltage is coupled through C2 to the base of Q1, forcing Q1 to cutoff. Now Q1 is in cutoff and Q2 is in saturation. This is the circuit situation shown in figure 3-6.

Figure 3-6. - Astable multivibrator. (Q2 saturated).

Notice that figure 3-6 is the mirror image of figure 3-4. In figure 3-6 the left side of capacitor C2 becomes more negative at a rate determined by the time constant R3C2. As the left side of C2 becomes more negative, the base of Q1 also becomes more negative. When the base of Q1 becomes negative enough to allow Q1 to conduct, Q1 will again go into saturation. The resulting change in voltage at output 1 will cause Q2 to return to the cutoff state.

Look at the output waveform from transistor Q2, as shown in figure 3-7. The output voltage (from either output of the multivibrator) alternates from approximately 0 volts to approximately -VCC, remaining in each state for a definite period of time. The time may range from a microsecond to as much as a second or two. In some applications, the time period of higher voltage (-VCC) and the time period of lower voltage (0 volts) will be equal. Other applications require differing higher- and lower-voltage times. For example, timing and gating circuits often have different pulse widths as shown in figure 3-8.

Figure 3-7. - Square wave output from Q2.

Figure 3-8. - Rectangular waves.

Wednesday, July 7, 2010

555-Monostable

The circuit diagram of the 555 monostable circuit is given in figure 2. The value of R and C determine the length of time that it's output is in the high state. It can be calculated using the equation below...

[Equation]

...where T is the time period in seconds, and R and C are the component values in Ohms (Ω) and Farads (F).

Figure 2: The 555 monostable circuit


Calculations

  1. Firstly, decide the time period T that you require. This can be very small (milliseconds) or large (minutes), but it must be expressed in seconds. {I choose T = 10 seconds}
  2. Next, guess a value for the capacitor C, expressed in Farads. For starters, try 100μF. {I choose C = 100μF}
  3. Put the values of T and C into the equation below and calculator resistor R...
    [Equation]

If the resistor value you calculated is smaller than 1kΩ or larger than 1MΩ, you should re-do the calculation with a different value for capacitor C until you get a resistor value within the acceptable range.

Varying the Time Period

If you will need to adjust the time period of the monostable circuit in use, you can use a linear variable resistor for R, as shown in figure 3.

Because the resistance of a variable resistor goes down to around 0Ω at one end of its range, a 1kΩ resistor is placed in series with it so that the value of R will never fall below 1kΩ. As the shaft of the variable resistor is turned from its lowest setting to its highest, T will become longer.

If your chosen variable resistor has three connections, it is a potentiometer, and you should connect to the centre connection and either of the end connections.

Further InformationResistors - Includes information about variable resistors

Figure 3: Varying the time period with a variable resistor

The Trigger Input

As you can see from figure 1, the 555's Trigger input must be taken low to trigger the monostable. This is achieved in figure 2 by placing a button in series with a resistor across the power supply. Normally, the 10kΩ resistor keeps the Trigger input high, at the voltage Vs, and the monostable is in its steady state. When the button is pushed, the Trigger input is directly connected to 0V and the time period T starts.

The Reset Input

If you want to make the monostable output go low before the time period has elapsed, simply take the 555's Reset input briefly low. This can be achieved with a push button in exactly the same way as with the Trigger input.

Astable Operation


555 astable operationWith the output high (+Vs) the capacitor C1 is charged by current flowing through R1 and R2. The threshold and trigger inputs monitor the capacitor voltage and when it reaches 2/3Vs (threshold voltage) the output becomes low and the discharge pin is connected to 0V.

The capacitor now discharges with current flowing through R2 into the discharge pin. When the voltage falls to 1/3Vs (trigger voltage) the output becomes high again and the discharge pin is disconnected, allowing the capacitor to start charging again.

This cycle repeats continuously unless the reset input is connected to 0V which forces the output low while reset is 0V.

An astable can be used to provide the clock signal for circuits such as counters.

A low frequency astable (<>

An audio frequency astable (20Hz to 20kHz) can be used to produce a sound from a loudspeaker or piezo transducer. The sound is suitable for buzzes and beeps. The natural (resonant) frequency of most piezo transducers is about 3kHz and this will make them produce a particularly loud sound.

Duty cycles

Duty cycle

The duty cycle of an astable circuit is the proportion of the complete cycle for which the output is high (the mark time). It is usually given as a percentage.

For a standard 555/556 astable circuit the mark time (Tm) must be greater than the space time (Ts), so the duty cycle must be at least 50%:

Duty cycle = Tm = R1 + R2
Tm + TsR1 + 2R2

555 astable circuit with diode across R2

555 Astable Multivibrator


555 astable output
555 astable output, a square wave
(Tm and Ts may be different)
555 astable circuit
555 astable circuit
An astable circuit using 555 produces a square wave. Figure shows a digital waveform with sharp transitions between low 0V and high. The durations of the low and high states may be different. The time period (T) of the square wave is the time for one complete cycle, but it is usually better to consider frequency(f) which is the number of cycles per second.

T = 0.7 × (R1 + 2R2) × C1 and f = 1.4
(R1 + 2R2) × C1

T = time period in seconds (s)
f = frequency in hertz (Hz)
R1 = resistance in ohms (ohm)
R2 = resistance in ohms (ohm)
C1 = capacitance in farads (F)

The time period can be split into two parts: T = Tm + Ts
Mark time (output high): Tm = 0.7 × (R1 + R2) × C1
Space time (output low): Ts = 0.7 × R2 × C1

Choosing R1, R2 and C1

555 astable frequencies
C1R2 = 10kohm
R1 = 1kohm
R2 = 100kohm
R1 = 10kohm
R2 = 1Mohm
R1 = 100kohm
0.001µF68kHz6.8kHz680Hz
0.01µF6.8kHz680Hz68Hz
0.1µF680Hz68Hz6.8Hz
1µF68Hz6.8Hz0.68Hz
10µF6.8Hz0.68Hz
(41 per min.)
0.068Hz
(4 per min.)
R1 and R2 should be in the range 1kohm to 1Mohm. It is best to choose C1 first because capacitors are available in just a few values.
  • Choose C1 to suit the frequency range you require (use the table as a guide).
  • Choose R2 to give the frequency (f) you require. Assume that R1 is much smaller than R2 (so that Tm and Ts are almost equal), then you can use:
    R2 = 0.7
    f × C1
  • Choose R1 to be about a tenth of R2 (1kohm min.) unless you want the mark time Tm to be significantly longer than the space time Ts.
  • If you wish to use a variable resistor it is best to make it R2.
  • If R1 is variable it must have a fixed resistor of at least 1kohm in series
    (this is not required for R2 if it is variable).


Tuesday, July 6, 2010

Decade Counter

The 7490A monolithic counter contains four masterslave flip-flops and additional gating to provide a divide-by two counter and a three-stage binary counter for which the count cycle length is divide-by-five. The counter has a gated zero reset and also has gated setto- nine inputs for use in BCD nine’s complement applications.

To use the maximum count length (decade or four-bit binary), the B input is connected to the QA output. The input count pulses are applied to input A and the outputs are as described in the appropriate Function Table. A symmetrical divide-by-ten count can be obtained from the counters by connecting the QD output to the A input and applying the input count to the B input which gives a divideby- ten square wave at output QA.

Decade Counter

A decade counter is a binary counter that is designed to count from 00002 to 10102. An ordinary four-stage counter can be easily modified to a decade counter by adding a NAND gate as shown in figure 3-25. Notice that FF2 and FF4 provide the inputs to the NAND gate. The NAND gate outputs are connected to the CLR input of each of the FFs.

The counter operates as a normal counter until it reaches a count of 10102, or 1010. At that time, both inputs to the NAND gate are HIGH, and the output goes LOW. This LOW applied to the CLR input of the FFs causes them to reset to 0. Remember from the discussion of J-K FFs that CLR and PS or PR override any existing condition of the FF. Once the FFs are reset, the count may begin again. The following table shows the binary count and the inputs and outputs of the NAND gate for each count of the decade counter:

BINARY COUNTNAND GATE INPUTSNAND GATE OUTPUT
*******AB*******
0000001
0001001
0010101
0011101
0100001
0101001
0110101
0111101
1000011
1001011

Changing the inputs to the NAND gate can cause the maximum count to be changed. For instance, if FF4 and FF3 were wired to the NAND gate, the counter would count to 11002 (1210), and then reset.








7 Segment Counter Circuit


Here is a seven segment counter that uses counter IC CD 4033. NE 555 is used as an astable multivibrator for triggering 4033. For each pulse the out put of CD 4033 advances by one count. The output of CD 4033 is displayed by the seven segment LED display LT543. Switch S1 is used to initiate the counting.Diode D1 prevents the risk of accidental polarity reversal.





Monday, April 19, 2010

Differenciator-Application



Applications for this, besides representing the derivative calculus function inside of an analog computer, include rate-of-change indicators for process instrumentation. One such rate-of-change signal application might be for monitoring (or controlling) the rate of temperature change in a furnace, where too high or too low of a temperature rise rate could be detrimental. The DC voltage produced by the differentiator circuit could be used to drive a comparator, which would signal an alarm or activate a control if the rate of change exceeded a pre-set level.
In process control, the derivative function is used to make control decisions for maintaining a process at setpoint, by monitoring the rate of process change over time and taking action to prevent excessive rates of change, which can lead to an unstable condition. Analog electronic controllers use variations of this circuitry to perform the derivative function.

Sunday, April 18, 2010

Integrator-Application


There are applications where we need precisely the opposite function, called integration in calculus. Here, the op-amp circuit would generate an output voltage proportional to the magnitude and duration that an input voltage signal has deviated from 0 volts. Stated differently, a constant input signal would generate a certain rate of change in the output voltage: differentiation in reverse. To do this, all we have to do is swap the capacitor and resistor in the previous circuit:


One application for this device would be to keep a "running total" of radiation exposure, or dosage, if the input voltage was a proportional signal supplied by an electronic radiation detector. Nuclear radiation can be just as damaging at low intensities for long periods of time as it is at high intensities for short periods of time. An integrator circuit would take both the intensity (input voltage magnitude) and time into account, generating an output voltage representing total radiation dosage.

Another application would be to integrate a signal representing water flow, producing a signal representing total quantity of water that has passed by the flowmeter. This application of an integrator is sometimes called a totalizer in the industrial instrumentation trade.

Saturday, January 2, 2010

Integrator




The negative feedback of the op-amp ensures that the inverting input will be held at 0 volts (the virtual ground). If the input voltage is exactly 0 volts, there will be no current through the resistor, therefore no charging of the capacitor, and therefore the output voltage will not change. We cannot guarantee what voltage will be at the output with respect to ground in this condition, but we can say that the output voltage will be constant.

However, if we apply a constant, positive voltage to the input, the op-amp output will fall negative at a linear rate, in an attempt to produce the changing voltage across the capacitor necessary to maintain the current established by the voltage difference across the resistor. Conversely, a constant, negative voltage at the input results in a linear, rising (positive) voltage at the output. The output voltage rate-of-change will be proportional to the value of the input voltage.
The formula for determining voltage output for the integrator is as follows:










Differenciator


We can build an op-amp circuit which measures change in voltage by measuring current through a capacitor, and outputs a voltage proportional to that current:

The right-hand side of the capacitor is held to a voltage of 0 volts, due to the "virtual ground" effect. Therefore, current "through" the capacitor is solely due to change in the input voltage. A steady input voltage won't cause a current through C, but a changing input voltage will.
Capacitor current moves through the feedback resistor, producing a drop across it, which is the same as the output voltage. A linear, positive rate of input voltage change will result in a steady negative voltage at the output of the op-amp. Conversely, a linear, negative rate of input voltage change will result in a steady positive voltage at the output of the op-amp. This polarity inversion from input to output is due to the fact that the input signal is being sent (essentially) to the inverting input of the op-amp, so it acts like the inverting amplifier mentioned previously. The faster the rate of voltage change at the input (either positive or negative), the greater the voltage at the output.

The formula for determining voltage output for the differentiator is as follows:
Vout = -RC dVin/dt

Differentiator and integrator circuits



By introducing electrical reactance into the feedback loops of op-amp amplifier circuits, we can cause the output to respond to changes in the input voltage over time. Drawing their names from their respective calculus functions, the integrator produces a voltage output proportional to the product (multiplication) of the input voltage and time; and the differentiator (not to be confused with differential) produces a voltage output proportional to the input voltage's rate of change. Capacitance can be defined as the measure of a capacitor's opposition to changes in voltage. The greater the capacitance, the more the opposition. Capacitors oppose voltage change by creating current in the circuit: that is, they either charge or discharge in response to a change in applied voltage. So, the more capacitance a capacitor has, the greater its charge or discharge current will be for any given rate of voltage change across it. The equation for this is quite simple:

The dv/dt fraction is a calculus expression representing the rate of voltage change over time. If the DC supply in the above circuit were steadily increased from a voltage of 15 volts to a voltage of 16 volts over a time span of 1 hour, the current through the capacitor would most likely be very small, because of the very low rate of voltage change (dv/dt = 1 volt / 3600 seconds). However, if we steadily increased the DC supply from 15 volts to 16 volts over a shorter time span of 1 second, the rate of voltage change would be much higher, and thus the charging current would be much higher (3600 times higher, to be exact). Same amount of change in voltage, but vastly different rates of change, resulting in vastly different amounts of current in the circuit.

To put some definite numbers to this formula, if the voltage across a 47 µF capacitor was changing at a linear rate of 3 volts per second, the current "through" the capacitor would be (47 µF)(3 V/s) = 141 µA.