comparators are used in | comparators are used in circuits such as | comparator definition meaning

comparators are used in | comparators are used in circuits such as | comparator definition meaning in digital electronics ?
Comparators
An analog comparator has two inputs one is usually a constant reference voltage V_R and other is a time varing signal V_I and one output V₀. the basic circuit of a comparator is shown in fig. 54.
When the non-inverting voltage is large than the inverting voltage, the comparator produces a high output voltage (+V_sat). When the non-inverting output is less than the inverting input, the output is low (-V_sat).Fig. 54 (b) also shows the output of a comparator for a sinusoidal.
V₀ = – V_sat, if V_i > V_R
= + V_sat if V_I < V_R
If V_R = 0, then slightest input voltage (in mV) is enough to saturate the op-amp and the circuit acts as zero crossing detector as shown in fig. 54 (c). if the supply voltages are + 15 V, then the output compliance is from approximate -13 V to + 13 V. the more the open-loop gain of op-amp, the smaller the voltage requires to saturate the output. if V_d required is very small then the characteristic is a vertical line as shown in fig. 54(d) and (e).
If we want to limit the output voltage of the comparator two voltage (one positive and other negative) then a resistor R and two zener diodes are added to clamp the output of the comparator. the circuit of such comparator is shown in fig. 67(f) the transfer characteristics of the circuit is also shown in fig. 54(g).
The resistance is chosen, so that the zener operates in zener region when V_R = 0, then output changes rapidly from one state to other very rapidly every time that the input passes through zero as shown in fig. 54(h) and (i).
Such a configuration is called zero crossing detector. if we want pulses at zero crossing then a differentiator and a series diode is connected at the output. it produces single pulses at the zero crossing point in every cycle.
If the input to a comparator contains noise, the output may be erractive when V_in is near a trip point. for instance, with a zero crossing, the output is low, when V_in instance, with a zero crossing, the output is low, when V_IN is positive and high when V_IN is negative. if the input contains a noise voltage with a peak of 1 mV or more, then the comparator will detect the zero crossing produced by the noise. fig. 55(a) shown the output of zero crossing detection, if the input contains noise.
This can be avoided by using a schmitt trigger, circuit which is basically a comparator with positive feedback.
Fig. 55(b) shows an inverting schmitt trigger circuit using op-amp.
Because of the voltage divider circuit, there is a positive feedback voltage. when op-amp is positively saturated a positive voltage is feedback to the non-inverting input, this positive voltage holds the output in high stage (V_IN < V_F). When the output voltage is negatively saturated, a negative voltage feedback to the inverting input, holding the output in low state.
When the output is +V_sat, then reference V_ref is given by
V_ref = R₁ / (R₁ + R₂) * V_sat = (+BV_sat)
If V_in is less than V_ref output will remain +V_sat.
When input V_in exceeds V_ref = +V_sat, the output switches from + V_sat to -V_sat. Then, the reference voltage is given by
V_ref = -R₂ / (R₁ + R₂) * V_sat = (-BV_sat)
The output will remain -V_sat as long as V_in > V_ref.
If V_in < V_ref, i.e,. V_in becomes more negative than -V_sat, then again output switches to +V_sat and so on. the transfer characteristic of schmitt trigger circuit is shown in fig. 55(c). the output is also shown in fig. 55(d) for a sinusoidal wave. if the input is different than sine even then the output will be determined in a same way.
Positive feedback has an unusual effect on the circuit. if forces the reference voltage to have the same polarity as the output voltage. the reference voltage is positive when the output voltage is high (+V_sat) and negative when the output is low (-V_sat).
In a schmitt trigger, the voltage at which the output switches from +V_sat to -V_sat or vice-versa are called upper trigger point (UTP) and lower trigger point (LTP). the difference between the two trip points is called hysteresis.
UTP = R₂ / R₁ + R₂ = V_set
LTP = R₂ / R₁ + R₂ = (-V_sat)
                   V_hys = UTP – LTP
= R₂ / R₁ + R₂ = V_sat – R₂ / R₁ + R₂ = (-V_sat)
= 2 (R₂ / R₁ + R₂) V_sat
= 2BV_sat
The hysteresis loop can be shifted to either side of zero point by connecting a voltage source as shown in fig. 55(e) and (f).
When V₀ = +V_sat, the refernce voltage (UTP) is given by,
UTP = (V_sat  -V_R)R₂ / R₁ + R₂ + V_R
BV_sat + R₁V_R / R₁ + R₂
When V₀ = – V_sat, the reference voltage (UTP) is given by,
LTP = (-V_sat – V_R) R₂ / R₁ + R₂ = V_R
= – BV_sat + R₁V_R / R₁ + R₂
If V_R is positive, the loop is shifted to right side; if V_R is negative, the loop is shifted to left side. the hysteresis voltage V_hys remains the same.
Non-Inverting Schmitt Trigger
In this case, again the feedback is given at non-inverting termilnal. the inverting terminal is grounded and the input voltage is connected to non-inverting input. fig. 56 shows an non-inverting schmitt trigger circuit.
To analyze the circuit behaviour, let us assume the output is negatively saturated. then, the feedback voltage is also negative (-V_sat). This feedback voltage will hold the output in negative saturation untill the input voltage becomes positive enough to make voltage positive.
V₊ = (-V_sat – V_in) /R₁ + R₂ = R₂ + V_IN
= -V_satR₂ / R_{1 +} R₂ + V_INR₁ /R₁ + R₂
= R₁ / R₁ + R₂ (-R₂V_sat + R₁V_IN)
= R₁ /R₁ + R₂ (R₂V_sat / R₁ = V_in)
When V_IN becomes positive and its magnitude is greater than (R₂ / R₁)V_sat, then the output switches to +V_sat. therefore, the UTP at which the output switches to +V_sat is given by.
UTP = (R₂V_sat /R₁)
Similarly, when the output is in positive saturation, feedback voltage is positive. to switch output state, the input voltage has to become negative enough to make. when it happens the output changes to the negative state from positive saturation to negative saturation voltage negative.
V₊ = (V_sat – V_IN)/ R₁ + R₂ = R₂ + V_in
= V_sat R₂ / R₁ + R₂ + V_in R₁ / R₁ + R₂
= R₁ / R₁ + R₂ (R₂V_sat + R₁V_in)
= R₁ / R₁ + R₂ (R₂ V_sat / R₁ + V_in)
When, V_in becomes negative and its magnitude is greater than R₂/R₁ V_sat, then the output switches to -V_sat.
Therefore,
LTP = (-R₂V_sat / R₁)
The difference of UTP and LTP given the hystersesis of the schmitt trigger.
V_bys = UTP – LTP
= 2 (R₂/R₁) V_sat
= 2BV_sat
In non-inverting schmitt trigger circuit, the B is defined as
B = R₂/R₁
Example 14. Design a voltage level detector with noise immunity that indicates when an input signal crosses the nominal threshold of -2.5 V. the output is to switch from high to low, when the signal crosses the threshold in the positive direction and vice-versa. noise level expected is 0.2 V_PP maximum assume the output levels are V_H = 10 V and V_L = 0 V.
Sol. for the triggering action required an inverting configuration is required. let the hysteresis voltage be 20% larger than the maximum pp noise voltage that is V_hys = 0.24 v.
Thus, the upper and lower trigger voltages are – 2.5 + 0.12 or UTP = 2.39 V and LTP = – 2.62 V
Since, the output levels are V_H and V_L instead of + V_sat and -V_sat, therefore, hysteresis voltage is given by
V_hys = R₂ / R₁ + R₂ (V_H – V_L)
Or                             V_H – V_L / V_hys = 1 + R₁ / R₂
and                       R₁/R₂ = 10 – 0/0.24 – 1 = 40.7
The reference voltage V_ref can be obtained from the expression of LTP.
LTP = BV_L + R₁V_ref / R₁ + R₂
Given that V_L = 0 and LTP = – 262, we obtain
V_R = (1 + R₂/R₁) LTP = (1 + 1/40.7) (-2.62) = – 2.68 V
We can select any values for R₂ and R₁ that satisfy the ratio of 40.7. it is a good practice to have more than 100 k for the sum of R₁ and R₂ and 1 k to 3 k for the pull-up resistor on the output. the circuit shown in figure above shows a possible final design. the potentiometer serves as a fine adjustment for V_R while the voltage follower makes V_R to appear as an almost ideal voltage source.
Relaxation Oscillator
With positive feedback, it is also possible to build relaxation oscillator which produces rectangular wave.
The circuit is shown in fig. 57(a).
In this circuit a fraction R₂/(R₁ + R₂) = B to the output is feedback to the non-inverting input terminal. the operation of the circuit can be explained as follows:
Assume that the output voltage is +V_sat. the capacitor will charge exponentially toward +V_sat. the feedback voltage is +bV_sat. when capacitor voltage exceeds + bV_sat the output switches from + V_sat to -V_sat. then feedback voltage becomes -V_sat and the output will remain -V_sat. now, the capacitor charges in the reverse direction. when capacitor voltage decreases below – bV_sat (more negative than -bV_sat) the output again switches to +V_sat. this process continues and it produces a square wave. under steady state conditions, the output voltage and capacitor voltage are shown in fig. 57 (b). the frequency of the output can be obtained as follows.
The capacitor charges from -BV_sat to +BV_sat during time period T/2. the capacitor charging voltage expression is given by
V_C (t) = {V_sat – (-BV_sat)} (1 – e^-t/RC) -BV_sat
At                               T = t/2, V_C (t) = +BV_sat
BV_sat = (1 + B) V_sat (1- e^-t/2RC) – BV_sat
2B = (1 + B) (1 – e^-t/2RC)
Or                        1 – e^-t/2RC = 1B/1 + B
Or                       e^-t/1RC = 1 – B/1 + B
Or                     e^-t/RC = 1 + B/1 – B
Or                      t/2RC = 1n 1 + B/1 – B
Or                         t = 2RC 1n (1 + B / 1 – B)
The frequency of the output is given by f = 1 / t
f = 1 / t = 1 / 2RC 1n (1 + B /1 – B)
This square wave generator is useful in the frequency range of 10 Hz to 10 KHz. at higher frequencies, the slew rate of the op-amp limits the slope of the output square wave.
The duty cycle of the output wave can be changed replacing the resistance R by another circuit consisting of variable resistance and two diodes D₁ and D₂ as shown in fig. 57 (c).
When the output is positive then D₂ conducts and D₁ is off. the total feedback resistance becomes R_b + R and charging voltage is reduced by V_D. During the interval when the output is negative than D₁ conducts and D₂ is off. the charging resistance becomes R + R_A. the total charging and reverse charging period is decided by total resistance (2R + R_A + R_B) = constant. therefore, frequency will remain constant but duty cycle changes. the capacitor voltage and output valtage of the oscillator are shown in fig. 57 (d).
t₁ = (R + R_B) C 1n (1 + B/1 – B)
t₂ = (R + R_A) C 1n (1 + B/1 – B)
t = t₁ + t₂ = (2R + R_A + R_B) C 1n (1 + B/1 – B)
The duty cycle is given by D = t₁ / t = R + R_B / 2R + R_A + R_B
BY varying R_A and R_B the duty cycle can be charged keeping frequency constant.
Triangular Wave Generator
In the relaxation oscillator, capacitor voltage V_C has a near triangular wave shape but the sides of the triangles are exponential rather than straight line. to linear size the triangles, if is required that C be charged with a constant current rather that the exponential current through R. the improved circuit is shown in fig. 58(a).
In this circuit an op-amp integrator is used to supply a constant current to C, so that the output is linear. because of inversion through the integrator, this voltage is feedback to the non-inverting terminal of the comparator rather than to the inverting terminal. the inverter behaves as a non-inverting schmitt trigger. the voltage V_R is used to shift the DC level of the triangular wave and voltage V_S is used to change the slopes of triangular wave form is shown in fig. 58(b).
To find the maximum value of the triangular waveform assume that the square wave voltage V_o is at its negative value = – V_sat. with a negative input, the output V_(t) of the integrator is an increasing ramp. the voltage at the non-inverting comparator input V₁ is given by.
V₁ = (-V_sat – V_out / R₁ + R₂) R₂ + V_out
Or          V₁ = R₁ / R₁ + R₂ = V_out – R₂ / R₁ + R₂ = V_out
When V₁ rises to V_R, the comparator changes state from -V_sat to +V_sat and V_(t) starts decreasing linearly similarly,when V₁ falls below V_R, the comparator output changes from +V_sat to -V_sat.
Hence, the minimum value of triangular V_min occurs for V₁ = V_R. hence, the pecak value V_max of the triangular waveform occurs for V₁ = V_R.
Therefore,       V_max = R₁ + R₂ / R₁ = V_R + R₂ / R₁ = V_sat
And                  V_min = R₁ + R₂ / R₁ = V_R – R₂/ R₁ = V_sat
The peak to peak swing is given by
V_max – V_min = 2V_sat R₂/R₁
The average output voltage is given by V_R (R₁ + R₂ / R₁).
If V_R = 0, the waveform extends between -V_sat (R₂/R₁) and +V_sat (R₂/R₁).
The sweep times t₁ and t₂ for V_S = 0 can be calculated as follows :
The capacitor charging current is given by
I = C dV_C / dt = – C dV_out / dt
Where, V_C = – V_out is the capacitor voltage.
For                      V_out = – V_sat,       I = – V_sat / R
Therefore,            -V_sat / R = – C dV_out / dt
dV_out / dt = V_sat / RC
Or                              V_{max –} V_min / t₁ = V_sat / RC
Therefore,                t₁ = V_max – V_min / V_sat / RC
When the output voltage of first op-amp is +V_sat, then the voltage V₁ is given by
V₁ = (-V_sat – V_out / R₁ + R₂ = R₂ + V_out
As triangular voltage increases, the voltage V₁ also increases. At t = t₁, when the voltage V_out become V_max the voltage V₁ becomes equal to V_R and switching takes place. Therefore,
V_R = -V_satR₂ / R₁ + R₂ + R₁ / R₁ + R₂ =  V_max
Or                         V_max = R₁ + R₂ / R₁ = V_R + R₂ / R₁ = V_sat
Similarly, when output voltage is +V_sat, then voltage V₁ is given by
V₁ = (V_sat – v_out ) / R₁ + R₂ = R₂ + V_out
As triangular voltage decreases, the voltage V₁ also decreases. At t = t₂, when the voltage V_out becomes V_min, the voltage V₁ becomes equal to V_g and switching takes place. therefore,
V_R = V_sat R₂ / R₁ + R₂ + R₁ / R₁ + R₂ = V_max
Or                    V_max = R₁ + R₂ / R₁ V_R – R₂ / R₁ V_sat
Therefore,       V_max – V_min = 2R₂ / R₁ V_sat
and                  V_av = R₁ + R₂ / R₁ = V_R
Solving the above equations, we get
t₁ = 2R₂RC / R₁ = V_sat
t₂ = t₁ = 1/2f
f = R₁ / 4R₂RC
The frequnency f is independent of V_O, maximum frequency is limited either by slew rate or its maximum output current which determines the charging rate of C. slowest speed is limited by the bias current of op-amp.
If unequal sweep intervals t₁ = t₂ are desried, then V_S can be changed. the positive sweep speed is given by (V_sat + V_S )/ RC and the negative sweep speed is given by (V_sat – V_S) / RC . the peak to peak triangular amplitude is unaffeeted by the voltage V_S.
When                 V_S = 0,  I = C dV_C / dt
-V_sat – V_S /R = – C dV_Out /dt = – C. (V_{max –} V_min) /t₁
Therefore,                   t₁ = V_max – V_min / (V_sat + V_S ) / RC
Or                                  t₁ = 2V_satR₂/R₂ / (V_sat + V_S) / RC
Similarly, when the output voltage of first op-amp -V_sat, then
V_sat – V_S / R = – C (V_max – V_min / t₂
Therefore,               t₂ = V_max – V_min / (V_sat – V_S) / RC
Or                              t₂ = 2V_satR₂/R₁ / (V_sat – V_S) / RC
Therefore,                 t_{1 /} t₂ = V_sat – V_S / V_sat + V_S
and                            t = t₁ + t₂
2V_sat  R₂RC / R₁ [2V_sat / V²_sat -V²_S]
The oscillation frequency is given by
f = 1 / t = R₁ / 4R₂RC [1 – (V_S / V_sat)²]
and the duty cycle is given by
Duty cycle t₁/t = V²_sat – V²_S / 2V_sat (V_sat + V_S)
= V_sat – V_S / 2V_sat = 1/2 (1 – V_S / V_sat)
Example 15. (a) Consideer the pulse generator shown in figure below. in the quiescent state (before a trigger pulse is applied), find V₂ , V₀ and V₁.
(b)   At t = 0, a narrow, positive trggering pulse V, whose magnitude exceeds V_R is applied. At t = 0 +, find V₀ and V₁.
(c) Verify that the pulse width t = RC in (2V₀)/ V_R.
Sol. (a) Before a trigger pulse is applied, the circuit is in stable state with the output at V₀ = + V₀ (~V_z + 0.7). the capacitor C is charged with the polarity shown in figure below.
Thus,               V₁ ~ 0.7 V
And                  V₂ = – V_R
(b) At t = 0, a narrow positive triggering pulse of higher magnitude is applied. the capacitor C voltage can not charge instantaneously. therefore, V₂ becomes positiven and greater than V₁ (~ 0.7 V) the comparator output changes.
Thus,                         V_O = – (V_Z + 0.7 V) = – V₀
Since, capacitor C voltage cannot change instantaneously, therefore,
V₁ = 2V₀
(c) The input trigger pulse in of very short duration therefore, after the short duration pulse the voltage V₂ returns to (-V_R). But the output remains -V₀ because V₁ is – 2V₀.
The capacitor now starts charging exponentially with a time constant t = RC through R towards V₀, because diode is reverse biased.
V_C = (-V₀ – V₀) (1 – e^-1/RC) – V₀
The voltage at point V₁ is given by,
V₁ = – V₀ – V_C
= – V₀ + 2V₀(1 -e^{-1 / RC}) – V₀
= 2V_0e^{-1 / RC} – t /RC
When V₁ voltage becomes more than V_R. the comparator output switches back to + V_O let at t = T. the voltage V₁ becomes – V_R
– V_R = – 2V₀ e^-T/RC
T = RC In 2V_O / V_R
The capacitor now starts charging towards +V₀ through R untill V_C reaches +V₀ and V₁ becomes 0.7 V.The waveform at different points are shown in figure above.
Voltage Regulators
An ideal power supply maintains a constant voltage at is output terminals under all operating condititons. the output voltage of a practical power supply changes with load generally dropping as load current increase as shown in fig. 59(a).
The terminal voltage when full load current is drawn is called full load voltage (V_FL) The no load voltage is the terminal voltage when zero current is drawn from the supply, that is the open circuit terminal voltage.
Power supply performance is measured in terms of per cent voltage regulation, which indicates its ability to maintain a constant voltage it is defined as
V_R = V_NL – V_FL / V_FL x 100%
The thevenin’s equivalent of a power supply is shown in fig. 59 (b). the thevenin voltage is the no load voltage V_NL and the thevenin resistance is called the output resistance R₀ let the full load current be I_fl. therefore, the full load resistance R_fl is given by,
R_fl = V_fl / I_fl
From the equivalent circuit, we have
V_fl = (R_fl / R_fl + R₀) V_nl
and the voltage regulation is given by
V_nl –  (R_fl / R_fl + R₀)  V_nl
V_R = (R_fl / R_fl + R₀)  V_nl  x 100%
V_R = R₀ / R_fl x 100%
or           V_R = R₀ (I_fl / V_fl) x 100%
it is clear that the ideal power supply has zero output resistance.
Example 16. A power supply having output resistance 1.5 supplies a full load current of 500 mA to a 50 load determine,
(a) Percentage voltage regulation of the supply
(b) no load output voltage.
Sol. (a) full load output voltage
V_FL = (500mA) (50) = 25 V
Therefore,   V_R = R₀ (I_FL/V_FL) x 100%
= 1. 5 (0.5 /25)x 100%
= 3.0%
(b) the no load voltage V_nl = (R_fl + R₀ / R_fl) = V_fl
= 25 (50 + 1.5) / 50 = 25.75 V