Analog basics:

References:

1. Lessons in electric circuits: Various volumes: Kuphaldt. (See Discover

circuits: Kuphaldt: lessons in electric circuits)
2. Basic Electronics by Grob.
3. Various basic electronics books: snippet view available through “Google
books”
4. Various web sites on basic electronics, Videos, PDF files etc etc.

Waveforms, ac and dc.

1. Signals: These are of two types. Ac , dc and ac+dc
2. The voltages in ac are represented by rms. Some points regarding sinusoid
/ triangular waves are
(i) The peak is sqrt(2) times rms value. The mains voltage is 230V That
means rms. Unless qualified, voltages in ac mean rms values. The
peak is 230*sqrt(2) = 325V
(ii) The peak to peak value is twice the peak value. You have to qualify
the voltage statement by specifying whether it is peak or peak peak
. If not qualified, it can be taken as rms, and is hence prone to
misinterpretation!!
(iii) The average value of ac by definition, is zero. Sometimes average is
used to indicate rectified average!!
(iv) When you do a full wave rectification, rms value remains same. The
average dc of full wave rectified output is 0.6366 times peak value,
and 0.9 times rms value. See examples at end.
(v) The average for half wave is 0.3183 times peak value and 0.45 times
rms value.
(vi) The ripple factor for half wave is 121% as compared with 48.3% for
full wave. It is hence easier to filter full wave, or for same filtering,
the full wave will give lesser ripple than half wave.
 (vii) Rms to full wave rectified ratio is called from factor. It is 1.11 for
sine, and 1 for square wave.
(viii) Peak value for triangular waveform is sqrt(3) times its rms value. A
10V peak triangular waveform will show.

3. A true rms meter will enable measurement of rms value independent of

waveform. In some cases such a chopped sine waves, such a meter is
essential.
4. Many low cost meters are not “true rms” meters. They measure the ac by
full wave rectification, and the dc obtained is calibrated to show rms
value of ac applied. They are called “average reading rms calibrated”
meters. When used with waveforms other than sine, such as square and
triangle, their readings require “slope” correction (scale factor correction).
5. A square wave having 50% duty ratio has a form factor of 1, as against
1.11 for sine. If a square wave is measured as 230V in such a meter, its
true rms value will be 207V.

Circuit analysis methods:

1. When analyzing circuits, an equivalent circuit is very useful. The Thevenin

equivalent circuit has a voltage source VTH in series with a resistance
RTH. Look up text books to understand and become familiar with
thevenin equivalent circuits.
2. The Norton equivalent is another form used in some cases. It consists of a
Norton current source In in shunt with Rn, a Norton resistance. It is
related to VTH, and RTH as : (a) In = VTH/RTH and (b) Rn = RTH.
3. The super position theorem is super theorem that comes in handy when
more than one sources are present. In a differential amplifier, the output
is difference of inputs applied. This can be derived using superposition
theorem.
4. Resistive adders are a neglected lot. In many cases, one can add a voltage
using simple resistances instead of complicating the matter with “op
amps” and the supplies that they require.
 5. When measuring the current in an ac circuit or when summing two ac
voltages beware of the fact that the voltages can be complex. They have a
real part and imaginary part. These have to be separately added to get
the complex value for the “sum”. The magnitudes of the individual ac
voltages cannot be simply added to get the magnitude of the “sum”. The
sum is the quadrature sum of the voltages. [sum of A and B is
=sqrt(A^2+B^2)]
6. In finding transfer functions of networks, one can easily calculate these
when a C is replaced by (1/sC), and an L is replaced by sL. Inverse
transform of output will give the time functions.. as example, you should
identify the transfer function of a low pas filter and identify how the
expression for the output across capacitor can be obtained using the
transfer function of the low pass and Laplace transform of a step (= 1/s).
7. When using circuit maker for simulation and making measurements of ac
ensure that stop time minus start time is an integer times the period of
the ac, to avoid errors. When using 50Hz, one can use 40mS as start time
and 140ms as stop time. The step time should be about 0.1% of stop time.

DB SCALE

1. Often factors such as gain or attenuation are spoken of in dB. The dB

value for voltage gain is 20*log(gain). Note that you cannot place a
voltage or power inside the bracket. It has to be a ratio of voltage (or
power) with respect to another voltage (or power). dbM is a measure of
power with respect to the power of 1mW in 50 ohms. We can speak of a
gain of 1 as 0dB, 100 as 40dB and 0.01 as -40dB. Note that dB value for 0
gain is minus infinity.
2. Some standard values are as follows for voltage gain

gain 0.01 0.1 0.316 1 2 3.16 5 10 31.6 100

dB -40 -20 -10 0 6 10 14 20 30 40
 value

First order Filters:

1. Simple low pass filter is made using a series resistor and a shunt
capacitor.
2. The time constant of the second filter is RC = 159uS if R= 15.9k and C =
10nF. Filters with other time constants are shown below.
V1 V2 V3
-1/1V R1 -1/1V R2 -1/1V R3
159k 15.9k 1.59k

100 Hz C11kHz C2 10kHz C3

10n 10n 10n

100Hz LPF 1kHz LPF 10kHz LPF

3. If Time constant is Rc, cut off frequency fc is calculated as 1/(2.πRC). With

Rc as 159uS, fc is 1 khz.
4. fc has a big role. It indicates that drom dc to fc, the “gain” is between 0
dB and -3dB. At fc, the gain is -3dB = 0.707. It is 30% less than dc gain.
5. At frequencies above fc, gain keeps falling down. At 10*fc, gain is 1/10.
At 23*fc, gain is 1/23 and at k*fc, gain is 1/k. [For this to be true, choose
k as at least 3 or more.]

V4
-1.41/1.41V R4 240.3mV
15.9k AC V V5
-1.41/1.41V R5 166.4mV
15.9k AC V
4kHz C4
10n
5.8kHz C5
10n
1kHz LPF
1kHz LPF

In the figures above, input is 1V rms. Output can be approximated

to be ¼ = 0.25V for first circuit, and 1/5.8 = 0.172V for second
circuit. The readings of the meter show that this is more or less
true. The meter readings are “actuals”. (no approximation..)
6. The gain falls off as 1/f at higher frequencies. If frequency is doubled, the
gain becomes half. This is called 6dB/octave. (Note that octave is
doubling by a factor of 2). This is also equivalent to 20dB/decade.
 7. Filtering is always associated with phase shift. Low pass filter gives 0 deg
lag at dc, and 45 deg lag at fc. This lag increases to 90 deg as frequency
approaches infinity. At kfc, phase shift is arctan(k). k can be 0 to infinity.
8. The response of a first order to a step is of the type 1- exp(-t/RC). It takes
output 1 time constant to rise to 63% of the change, and 2.3*RC to rise to
90%. Response reaches 99% in 4.6RC. This is referred to as settling time
for 1%. Settling time for 0.1% is 6.9RC. Settling time to 100*(1/65536)% is
{ln(5536)}*RC, or about 11.1*RC

Diodes:

1. Current in diodes are exponentially related to voltage applied. Doubling

of voltage does not just double the current. The current can increase to
destructive values.
2. A silicon diode normally requires 0.6 to 0.7V to result in a current of 1mA.
This depends on the type of diode, but this voltage will be in the range
0.35V to 0.9V at 25 degC.
3. When current is doubled, voltage required just increases by 18mV. In
transdiodes (collector shorted to base, in a transistor), an increase in Vd
of 60mV will result in current becoming 10 times, such as from 120uA to
1.2 mA. This is called 60mV/decade behavior. Some diodes will have a
100mV/decade behavior.
4. The diode has a negative temperature coefficient. As temperature
increases, the voltage across the forward biased diode decreases at a rate
of 2 mV/degC. A 30 deg increase can mean reduction of Vd by 60mV.
5. Using the 60mV.decade rule, if Vd at 1ma is 680mV, that at 1uA is 680-
3*60=500mV (180mV lesser since 1uA is 3 decades down below 1mA) and
at 10na, Vd will be 500-2*60=380mV (since 10nA is 2 decades below 1uA)
6. At a constant applied voltage of 500mV, the current can be 1uA at 25
degC, but can increase to 1ma at 115degC and decrease to 10na at
-35degC.
 7. Diode characteristics are “straight lines” when log of Id is drawn against
Vd (lin). At currents in excess of 10mA, these will deviate from a straight
line, due to bulk resistance of diodes.

Zeners and LED’s

1. A zener diode provides a nearly constant voltage over a wide range of

currents, and is an attractive solution for stabilizing voltages requires as
“supplies” for electronic circuits.
2. In some cases, the voltage across the zener can also be used as a
reference source in a circuit. In many situations what one needs is a
voltage that is allowed to vary by a small amount (say 1%) but this effect
can be cancelled out in subsequent circuits. In such cases, a zener can be
used advantageously.
3. The zener has a dynamic resistance of about 10 to 15 ohms at 5 ma. This
could be substantially higher at lower currents such as 1mA.
4. The zener voltage may vary by 100 to 150mV for a 10mA change in zener
current.
5. Zener voltage is temperature dependent. At voltages less than 5V, the
temperature coefficient is negative, while for >5V, it is positive.
6. Zeners are available at 400mW, 1W and higher power ratings.
7. They have typically tolerance of +/- 5% and +/- 10%.
8. Just as diode, the voltage across a zener has a smooth variation as current
is lowered down to zero. (See text books to see the zener characteristics.
In many cases, the graphs are drawn with different scales for positive and
negative voltages. Take note of this.)
9. An LED is available in various colors. While the red Led has a typical drop
of 1.8V at 10mA, the white and blue ones may have drops of 3.4V.
10.Handle LEDs with care. ESD can affect them and result in failure.
11.Some of the blue LEDs border on principles of LASER. Never look at them
directly lest it should be the last time..
12.The LEDs also have logarithmic characteristics, just as a diode. Their
voltage also decrease generally as temperature is increased.
Transistors:

1. Two types are available: PNP and NPN.

2. In ON mode both junctions, base emitter and base collector are forward
biased.
3. In OFF condition both of these are reverse biased.
4. In “Linear” mode where the transistor can amplify, the BE junction is
forward biased, and CB junction is reverse biased. If this condition is not
maintained, amplifiers will cease to operate properly.
5. Proper biasing methods are hence essential in amplifiers and establish
desired dc conditions. No ac amplification is possible without proper dc
operating conditions.
6. The operating point of a transistor specifies Ic the collector current and
VCE the collector emitter voltage.
7. Fixed bias, self bias and potential divider bias are three methods usually
adopted for biasing the transistor.
8. In linear mode, common emitter operation is popular. Common collector
stages can be used to bring down high source resistances to low values.
The voltage gain here is less than 1.
9. 3 parameters are of high importance in amplifiers. They are Rin, Rout and
VG, the voltage gain.
10.Two amplifiers having gains of 40 and 100 may not provide a total gain of
4000 when cascaded, due to interactions between Rin, and Rout of the
stages.
11.A transistor with β of 100, operating at 12V with a 10V drop across its
resistor Rc in collector gives a gain of about -385, and has Rin of 2600
ohms when operated at 1mA. Since drop across rc is 10V, Rc is 10k, and
Rout is about 10k. In the figure below can you identify why the output has
halved ?? And can you estimate the voltage across C2 and C1??
 V4 V1
12 12
+V +V
993.4uA 993.4uA
DC~A DC~A

R5 R6 R2 R1
1130k 10k 1130k 10k
C2 R3 C1
1u 370.6mV 1u 185.5mV
NPN 2.6k NPN
V3 AC V V2 AC V
-1.41m/1.41mV Q2 -1.41m/1.41mV Q1

1kHz 1kHz

Voltage across C2 is about 58uV!!!

12.A transistor with β of 100, operating at 12V with a 10V drop across its
resistor Rc in collector
gives a gain of about -385, and has Rin of 26k ohms when operated at
100uA. Since drop across rc is 10V, Rc is 100k, and Rout is about 100k.
13.Input resistance is that resistance of source for which gain becomes half
of the gain
when Rs = 0.
14.How do you measure Rout?? Verify the method using a simulated
amplifier.
15.Emitter degeneration can increase Rin and decrease VG. A small resistor
in emitter does this job. To avoid emitter degeneration, resistances in
emitters should be decoupled.
16.RC coupled amplifiers using a single supply employ capacitor coupling to
ensure that dc conditions are maintained.
17.When having to amplify dc, capacitors cannot be used. Using dual
supplies and a differential amplifier, one can amplify dc and all
frequencies up to a maximum limit. Dc amplifiers do not just amplify dc.
They also amplify ac!!
18.Using an active load (- a current source is used as load-), one can get gains
exceeding 1000 in a single stage, even with low voltage supplies.
 19.Differential amplifiers have two types of signals called DM (differential
Mode) and CM (common mode) signals. If V1 and V2 are inputs V2-v1 is
DM and (V2+V1)/2 is CM signal.
20.CMRR is a measure of how well the unwanted CM signals are rejected by
the differential amplifier. Typical values are 60dB to 120 dB.
21.One can use a current source in emitters of differential amplifiers to get
high CMRR.
22.An op amp is merely a dc amplifier with high gain. When inputs are zero,
output should be about zero volts.
23.These can be used with feedback to form an inverting amplifier or a
noninverting amplifier or a differential amplifier, and a host of other
functions.
24.Amplifiers can have nonlinear characteristics that can compensate for
reverse type non linearity of sensors. RTDs have a temperature coefficient
of 0.00385 %/degC in the range 0 to 100 degC and 0.00373%/degC in the
range 100 to 200 degC. This results in 0.4% nonlinearity when the 0 to 200
degC range is considered. This nonlinearity can be compensated for using
a non linear amplifier, which has a gain of 1/0.385 for 0 to 1V and 1/0.373
gain for 1 to 2V.
25.Typical diodes cannot be used to convert low level ac signals such as 0 to
1V to proportional dc. This is because of the threshold of the diode which
is typically 500mV to 600mV. Precision diodes using op amps can be used
for this purpose, with signals as low as 10mV.
26.Positive feedback will result in regeneration, and throw the output to a
full high level (close to V+) or low level (close to V-). This is used in Schmitt
triggers, that can take bold decisions!!. The hysteresis of Schmitt triggers
depends on the amount of positive feedback applied.

Power supplies:
1. Power supplies are extremely important in electronics, as the whole
operation depends on their ability to provide an adequately stable
voltage at the required current.
2. A simple source of supply is a battery. One has to know what the
internal resistance of the battery is to guess whether it can provide the
desired current and how long. Over a period the charge stored in the
battery depletes and the internal resistance of the battery increases.
(The voltage reduces, but that is less important!!). At some point this
higher internal resistance results in the battery not being able to
provide the desired minimum voltage on load.
3. The battery has to be tested with specified or equivalent load. Merely
measuring the battery voltage often will give erroneous confirmation
of its usefulness. Always check a battery with specified load.
4. Shunt and series regulators are used in practice to “stabilize” the
voltage output of a power supply. Load regulation refers to variation
observed in the supply output voltage at full load, when input voltage
is changed.
5. Load regulation refers to changes in output voltage due to load current
being changed from 0% to 100% of rated load current.
6. Apart from these changes, the supply also may possess some ripple
and noise. Study the specifications of commercial supplies available
from LAMDA, APLAB, Agilent, and other reputed manufacturers.
7. Zeners can be used with or without additional transistors to provide
regulated output at low currents. A +/- 6V supply using zeners and
transistors is shown below: This can supply currents upto 100mA and
has no short circuit protection..(Simulate and see what the short
circuit current is…)
 2N3904
Q1
R4
D3

+
680

+
BRIDGE C3 C1 R3
V1 100uF 100uF 500
5TO1CT D2
1N4736

60 Hz D1
R2 1N4736

+
C2 680

+
100uF C4 R1
100uF 500

Q2
2N3906

The power dissipated in the zeners are less than 100mW. The
transistors dissipate about 800mW when load current is 100mA.
8. In comparison, on short circuit, the power dissipated in the transistors
rise to very high levels (3W) as a result of which these transistors can
even fail.
9. 7805 and 7905 are popular series regulators often used to get +/- 5V.
While 7805 is a positive regulator requiring an input higher than the 5V
output (7 to 25V is common), the 7905 will operate from inputs in the
range -7 to -25V and provide a -5V output.
10.These regulators have special internal current protection apart from
thermal shutdown to safeguard the device when load becomes a short
or when load exceeds specified rating.
11.TL431 is a programmable precision reference and can also be used as a
shunt regulator for low currents. The current can be increased to more
than 200mA using additional external transistors.
12.Switching regulators provide much higher efficiency than linear
regulators and hence have much lesser power dissipation. They can
also be designed to cater for wide ranges of input voltage such as 80V
to 270V ac.
13.The choice of an appropriate heat sink is essential for reliable
operation of power supply regulator chips. Do not ever touch them
and burn your hands. Some of them are comfortable operating at case
temperature exceeding 100 degC.
 14.With knowledge of ѲHS and ѲJC, one can estimate the case
temperature and junction temperature of an IC at given ambient
temperature, for a specific power dissipated.

Op amps:

1. Inverting amplifier has an input resistance = Ri. For example A 10k / 100k
amplifier has input resistance of 10k at low frequencies.

R2
100k

-1.117mV
V1 DC~V
5V
+V
V3
0/0V U1
R3 R1
10k +
source resistance
1kHz LT1006A
V2
+V -5

2. A resistor may be used in positive lead to reduce errors due to base

current.
 R5
100k

V6 -298.7uV
5V DC~V
+V
V4
0/0V R4 R6 U2
10k +
source resistance
1kHz LT1006A
R7 reduces offset at output due to bias current R7
9.1k
V5
+V -5

The bias current will cause large offset if resistors are chosen as ten times
higher.
3. Non inverting amplifiers do not have positive feedback!! The feedback
remains negative. They provide very high input resistance. However direct
ac coupling to +ve input without a ground return resistor can result in
large offset at output and often saturated output.
R5
100k

V6 3.998 V
5V DC~V
R6 +V
10k U2
V4 +
0/0V C1
1uF
LT1006A

1kHz V5
+V -5

BAD circuit!!

4. The amplifier gives high CMRR provided input is within allowable “input
range”. In some cases, this can include negative supply. Thus if negative
supply is -5V, inputs can go down to this value, and amplifiers will still
operate. Example LM 324, LM 358. In some cases, “input voltage” range
might include positive supply. E.g, LF 353.
5. Rail to rail input op amps allow “input voltage range” from positive supply
to negative supply. The output voltage range may not include supply
voltages. In other words, output can swing only from V+ - 1.2V to V- +
1.2V.
 6. If amplifiers are Rail to rail output, input cannot include supply voltages.
7. Rail to Rail input / output amplifiers allow inputs up and down to supply
voltages and can also swing from +ve supply to –ve supply. Refer to data
sheets if there is any doubt.
8. In general FET op amps will be preferable for use with large value resistors.
However, their offset/ bias current at 125 degC can be 100 to 1000 times
its value at 25 degC. Look for graphs giving bias current variation with
temperature in data sheets.
9. Instrumentation amplifiers have high CMRR but more importantly they
permit source unbalance. Even if source resistances are not equal (i.e they
are unbalanced), the high input resistance enables high CMRR to be
obtained. They also have high gain stability and linearity.
10.Instrumentation amplifiers need a return to ground through a resistance, if
source is totally floating. If no such return is provided, the first stage can
go into saturation, and measurements on second stage may not reveal
that!!

======================== ========================= =============

Worked examples.
1. What is the rms value of a dc of 100V added to ac of 70.7V?
Ans: sqrt*(100^2+70.7^2)=122.5V
2. An ac of 230V is converted to dc using a bridge rectifier. What is the rms
value of the output?
Ans: 230V if diode drops are negligible. Simulation shows this as 228.3V.
D1
BRIDGE 205.6 V
V1
-325/325V DC~V

50 Hz 99.32 V
R1 AC V
100k

sqrt[205.6^2+99.32^2]=228.3
 Note that the ac meter in the simulator shows only the ac rms. The dc meter
shows ony the dc content. Sum is not 205.6+99.32 = 304.9!!! The sum is
quadrature sum meaning sqrt(A^2+B^2), and is 228.3V
3. What is the average of 230V when full wave rectified?
The average is 2*peakvalue/pi.= 0.6366 of peak. In terms of rms, it is 90% of
rms. Thus 230V will yield 0.9*230=207V. Simulation shown above has some
small drops in diodes and that is the reason it shows 205.6V
4. Is the output of the full wave rectifier, ac, dc or ac+dc?
Asn: The output is ac+dc. The dc value is 205.6, while the ac content is
99.32V. One can talk about the ripple factor as the ratio of these, and is
99.32/205.6 = 48.3%.
5. What is ripple factor for a half wave rectifier?
In the case of half wave, the rectified average is peak value / pi =
0.3183*peak. In terms of rms it is 45%. The ac is pretty high as seen in the
following simulation. The rms value is ideally 0.707*rms of input. For 230V
input rms of output is 230*0.707 = 162.6V. since average is 103.5, ac rms can
be calculated as sqrt(162.6^2-103.5^2)= 125.4volts, and ripple factor is 121%

V1 103.1 V
-325/325V D1
DIODE DC~V

50 Hz 124.8 V
R1 AC V
100k

6. How is the ripple in the full wave rectifier reduced?

Often it is reduced by resorting to peak hold using a capacitor.

7. What would be the dc at output of a (i) half wave rectifier (ii) full wave
rectifier when a capacitor is used to filter the ac off?

The 230V has a peak value of 325V. Thus filtered dc will be about 325V.
Surprised??
 V1 323.9 V
-325/325V D1 DC~V D2 323.3 V
DIODE V2
-325/325V BRIDGE DC~V

50 Hz 100u 185.6mV
C1 AC V 50 Hz 100u 91.89mV
R1 C2 AC V
100k R2
100k

8. So a step down transformer with 20V secondary voltage will give dc of

20*sqrt(2), when it is rectified and filtered??
Yes. See the simulation below: In the case of halfwave, the dc is 0.7V less
than peak. In the case of bridge, dc is 2*Vd = 1.4V less than peak value of ac.
Note that the source is set to peak value of 20Vrms sine.
V1 27.54 V
-28.2/28.2V D1 DC~V D2 26.86 V
DIODE V2
-28.2/28.2V BRIDGE DC~V

50 Hz 100u 15.66mV
C1 AC V 50 Hz 100u 7.419mV
R1 C2 AC V
100k R2
100k

9. What is the rms value of a triangular waveform of peak value 10V?

Ans: 10/sqrt(3)= 5.7735V
V3
-10/10V 5.774 V
AC V

100 Hz

10.What is the rms value of a 90V peak square wave with 50% duty ratio? Ans:
90V.
 Final meter Final Meter
V1 D1 88.73 V V2 D2 88.70 V
-141/141V 99.98 V BRIDGE DC~V -90/90V 90.00 V BRIDGE DC~V
AC V AC V

50 Hz 50 Hz
R1 R2
100k 100k

Though the input applied has differing rms values, the final meter shows same reading..
This is the problem with average reading rms calibrated meters

Though the input applied has differing rms values, the final meter shows
same reading!! This is the problem with average reading rms calibrated
meters.

11.Find the thevenin equivalent of the following potential divider with input of
10V, when set to 5V and 2.5V.
5.000 V 4.000 V
DC~V DC~V

+ V1 S1 + V2 S2
R1 R3
10V 10k 50% 10V 10k 50%
R2 R4
10k 10k

5.000 V 4.000 V
DC~V DC~V
R9 R11
2.5k S5 2.5k S6
+ R10 +
Vs1 Vs2 R12
5 10k 10k
5
- -

The equivalent circuit as shown is a voltage source in series with a

resistance. When setting is 50%, Rth = 5k||5k = 2.5k. See that the
voltages are same when load is connected in actual circuit and
equivalent circuit. For the 25% setting, the following is the
simulation:
 2.500 V 2.105 V
DC~V DC~V

+ V4 R6 S4 + V3 R5 S3
10V 10k 25% 10V 10k 25%
R8 R7
10k 10k

2.500 V 2.105 V
DC~V DC~V
R9 R11
1.875k S5 1.875k S6
+ R10 +
Vs1 Vs2 R12
2.5 10k 10k
2.5
- -

12.A 5V peak sine wave at 40Hz has to be added to a dc so that the voltage
swings from 2 to 4V. This is to be fed to an ADC. Show a simple arrangement
using the 5V dc to do this offsetting.
Ans: The change required is 4-2 = 2V. the change available = peak to peak ac
= 10V. Thus the “gain” required is 40%. A 6k in series and a 4k in shunt will
do this. The rms value will be 2/sqrt(2) = 1.414V.
The average value of output = (4+2)/2 = 3V. When the 4k is not grounded but
connected to 5V, the voltage developed across 6k = 3V. Thus the circuit is as
follows:
V2 3.000 V
5V DC~V
+V

V1 R2
4k 1.414 V
-5/5V R1
6k AC V

40 Hz
Resistive summer

13.Do as above, with dc value = 2.5V and ac peak = 1.5V. Output should swing
between 1 and 4V.

14.In the circuit below, calculate the voltages across R. Prove that the sum of
voltages across R and L is 100V, the applied voltage. (Hint: the voltages are
vectors.)
 V1
-141/141V R1 70.72 V
314 AC V

50 Hz L1
1H

15.If gain is 29.35, 10^5 and 0.632, find the dB equivalent.

Ans: 29.35 corresponds to a dB value of 29.35dB

10^5 corresponds to a dB value of 100dB

0.632 corresponds to a dB value of 6-10=4dB.

16.In the low pass filter circuit shown below, what is the transfer function?
What is the equation governing the build up of voltage across C when input
applied is a step.
V1
-100/100V R1 49.99 V
15.9k AC V

1kHz C1
10n

Ans: The capacitance C1 can be replaced by 1/(sC1) to find transfer function.

The voltage ratio is then, {1/(sC1)} /{R1+ 1/(sC1)} = 1/(1+sR1C1).
The application of a step will result in a voltage governed by 1- exp(-t/R1C1).
Proof:
Step has a Laplace transform = 1/s. The Laplace transform of output
= (1/s)*{1/(sR1C1)}
Expressing this as A/s + B/(1+sR1 C1), and equating coefficients of s,
one gets
A = 1 and B = -R1C1.
Thus (1/s)*{1/(sR1C1)}
 = 1/s – R1C1/(1+sR1C1)
= (1/s) – 1/{s+1/(R1C1)}
Take inverse Laplace transform to get, 1- exp(-1/R1C1)
Exercise: Plot this with R1C1 as 159uS, for t=0 to t=954us. Tabulate your
results for t = 0, 15.9us, 159uS, 365.7uS and 731.4uS against voltages
obtained.
17.An oven acts as a first order system with a time constant of 30 mins. If the
temperature is set to 125 degC, when will the temperature reach 115 degC if
initial temperature is 25 degC?
Ans: The equation governing the build up of temperature is 25 +100*exp(-
t/RC) where Rc is the time constant of the oven. This becomes equal to 115
degC when time is 2.3RC = 1.15Hrs= 1Hr and 9 mins.
18.If the oven which is at 115 degC is turned off, when will in reach 30 deg, if
ambient temperature is 25 degC and time constant of oven is 30mins?
Ans: The equation governing the temperature is 25+(115-25)*exp(-t/RC).
This becomes 30 after a time lapse of 1 hour and 26.7 mins. (Verify.. by
simulation. Have a capacitor of 1F across 1800 ohms. Connect this junction
to 25V.If capacitor has an initial value of 115V across it, when it will it drop
to 30V? )
19.Using the 60mV/decade rule (and 18 mV/octave rule) estimate the voltage
across a diode at 0.2mA and 10uA.Given: At 1mA, voltage is 640mV
Ans: At 100uA, voltage is 60mv lesser at 580mV. At 200uA, it is 18mv higher
at 598mV.
At 10ua, it is 580-60=520mV.
20.A diode has a max voltage of 600mV and minimum of 400mv at 100uA
@25degC. Estimate what these would be at -25degC. What will be the limits
at 1mA?
Ans: The limits will be 50*2 = 100mV higher, at 0.7Vmax and 0.5Vmin. Limits
for 1mA will be 60mV higher at 0.76V max and 0.56V min.
See the table giving Vd values in mV.

Temperatur 25deg100u 0deg100u - 0deg,1m -

e A A 25deg100u A 25deg,1m
 A A
Vd min 400 450 500 510 560
Vdmax 600 650 700 710 760

21.Two transdiodes in series are operated from a 5v supply at a current of 2ma.

The voltage is found to be 1.32V. If the 5V has a ripple of 100mv peak to
peak, what is the ripple of 1.32V?
Ans: The dynamic resistance of a transdiode at 1ma is 26 ohms, and at 2 ma,
it is 13 ohms. When two of these are in series, the dynamic resistance is 26
ohms.
The diodes are provided a current from 5V through a series resistor R. Since
current through the diodes is 2mA, and voltage across series resistor is 5-
1.32 = 3.68V. Since current is 2ma, R = 1.84k
The ripple voltage hence divides as follows across R and the pair of diodes.
Ripple current = 100mV/(1840+26), and ripple voltage across the diodes is
(26/1866)*100mVpp=1.39mV pp.[Note that this is better than what you
would get if you used two resistors of 1.84k and 660 ohms to drive a voltage
of 1.32V from the 5V. In such a case, the ripple would be 1.32/5*100=
26.4mV, about 20 times higher!!!
22.A zener of 5.1V operated at 5ma has a voltage of 5.087V at 5mA and 5.132V
at 10mA. What is its dynamic resistance?
Ans: The change in voltage is 45mV and change in current is 5mA. Hence rz =
9 ohms.
23.Draw the circuit of a common emitter amplifier and identify the important
parameters. The operating point desired is 2mA, 4V, and supply is 12V.Use
2N2222A. Draw the equivalent circuit. Use potential divider bias, with the
biasing resistors having Ic/10 as current.
The circuit is as shown below. The emitter voltage is taken as 0.5V. The VCE
is 12-4.15*2.022 = 3.6 V, and ic is 2.023mA. The input resistance is calculated
as β/gm where gm = Ic/26mv = 2.023/26 = 77.8 mA/volt. Β for 2N2222A is
about 170 and Rin = 170/.0778=2.18k. This is in shunt with 5.6k and 51k.
hence actual input resistance is 1.52k
Voltage gain = gm*Rc = 0.0778*3900*= 303. The output resistance Ro = 3.9k

12
+V
12
999.3uV 2.023mA +V
AC V DC~A 275.2mV
AC V 623.6uV 2.023mA 171.8mV
R7 R1 AC V DC~A AC V
51k 3.9k C2
10u R5
-1.41m/1.41mV C1 R9
10u 51k 3.9k C6
C5 10u
-1.41m/1.41mV R11
2N2222A 1k 10u
1kHz R2
R8 R3 1meg
5.6k 2N2222A
250 C3 1kHz R4
1000u R10 R6 1meg
5.6k 250 C4
1000u

The Rin for the simulated amplifier cannot be ascertained from circuit on
left. It can be ascertained from circuit on right. The voltage at base is
623.6mv and 999.3mV-623.6mV drops across the resistance R11,1k. Thus
input current is 376/1= 376uA. Using this, one can now estimate the Rin as
0.6236/.376= 1.66k. This compares well with estimated value of 1.52k. the
voltage gain is seen to be 275.2 from circuit on left as one gets 275.2mV as
output when input is 1mV.This figure for gain compares reasonably well with
estimated value of 303. [There will be better match if the internal resistance
rc of the transistor is also considered].
How will you identify the output resistance?? This can be done by reducing
R2 till the voltage 275.2mv drops down to half its value = 137.6mV. See
what happens when it is made 3.6k. the output voltage reduces to almost
half its value of 275.2mV. Thus the output resistance of this amplifier is 3.6k.
This matches well with the estimate of 3.9k (Noting that rc has not been
considered. Rc = VA/Ic=113/2.023=55.6k, and 3.9k||55.8k= 3.65k!!!!)
 12
+V

999.5uV 2.023mA
AC V DC~A 137.1mV
AC V
R7 R1
51k 3.9k C2
10u
-1.41m/1.41mV C1
10u

2N2222A
1kHz R2
R8 R3 3.6k
5.6k 250 C3
1000u

24.For the circuit considered, what happens when the resistance in emitter is
bypassed partially?
This is shown in next figure.
12 12
+V +V

2.023mA 776.9uV 2.023mA 46.86mV

999.7uV 60.31mV
DC~A AC V DC~A AC V
AC V AC V
R13 R9 R5
R7 3.9k C6
51k 3.9k C1 51k
10u C5 10u
-1.41m/1.41mV R11
-1.41m/1.41mV C2 1k 10u
10u

2N2222A 2N2222A
1kHz R4
1kHz R14 R10
R3 1meg R6 1meg
R8 5.6k
5.6k 50 50

R1 R12 C4
200 C3 200
1000u 1000u

The gain has dropped from 275 to 60.3. The input resistance has increased
as the base voltage is 776.9mV as against 623.6mV earlier. The estimation of
the input resistance is left to you.
25.If a 5.1V zener with rz of 10 ohms is used to stabilize a 12V having a ripple of
100mV peak to peak, find the ripple in the zener output if current in zener is
5ma .
Ans: Assuming that voltage across zener is 5.1V, (-this can be 10% off if
tolerance of zener is 10%), the series resistance is 6.9/5= 1.38k. the ripple
across zener is (10/1390)*100mV= 0.72mV pp.
26.Draw the circuit of a zener and transistor based series regulator providing
output of 12V for inputs from 20 to 25V. Current desired is 200mA. Max.
What will be the worst case dissipation in the transistor?
Ans: The circuit is as shown below. The worst case dissipation will be about
2.6 watts, since drop across the transistor will be about 25-12 = 13V and
current through the transistor will be 200mA.
Caution: (No internal protection!! Shorting the output can kill the transistor)
11.47 V
11.71 V DC~V
DC~V
V2
V1 25 Q1
20 Q2 +V MPSA28
+V MPSA28
R4
R1 3.9k
3.9k
D2 R3
D1 R2 1N4743 60
1N4743 1k

27.What is the load regulation in the above case?

Ans: If one assumes that output at 12mA remains same even when V1 is
made 25V, then change in voltage is 240mV in 11.47V. This is equivalent of
2.1% load regulation. Much better regulation is possible. (Better means
lower %age of regulation).
28.Explain a method of protection used to limit current in regulator to a desired
value on short circuiting the load.
Ans: The idea is to merely have an additional sense resistor that senses the
load current. If this exceeds say 140% of rated maximum current, the base
current supplied by R1 is diverted directly to emitter. The following circuit
exemplifies this. In feedback regulators, the feedback can be taken directly
from voltage across the load. This reduces the effect of RSC on load
regulation.
29.The circuit of a feedback controlled series regulator is shown below. Explain
the function of the subsections:

The resistors R2 and R1 provide a fixed fraction of the output Vout to a

differential amplifier where it is compared with a stable reference provided
by Vref. The output of the differential amplifier controls the “pass” transistor
between pins 1 and 5, so that any change in Vout is corrected for. The
resistance R3 senses the load current and “diverts” the current fed to base of
“pass” transistor directly to its emitter and provides protection when load
resistance connected to Vout is too small.
30.Show the various blocks provided by the 723 regulator.
 The block diagram is shown below.

The 723 has a separate reference generator that outputs the voltage Vref. An
uncommitted amplifier compares the voltage at +ve input with that at –ve
input and drives the “series pass transistor”. Typically for outputs lesser than
Vref, one connects the Vref through a voltage divider to +ve input and the
output Vout is fed back to negative input of amplifier as shown below.

The resistance RSC connected between Vout and “regulated output” ensures
that when current exceeds a certain set value, the current limit transistor
comes into action and diverts excess current fed to base of ‘series pass”
transistor directly to its emitter thus providing current limit action.
31.Draw the circuit of a CVCC supply and explain its action.
The circuit in simplified form has two diodes, one of which conducts in CV
mode and the other conducts in CC mode. If one conducts, the other turns
off !!. So when load is 250 ohms or above, the diode D3 in first circuit
conducts and diode D4 is Off as indicated by the meter reading of 17.85V.
[The base of Q2 is obviously at a much lesser voltage as decided by U3 and
the conducting D3]. The operation is in CV mode.
When load resistance is decreased to 249 ohms, the supply moves to
constant current (CC) mode, and diode D2 now conducts throwing D1 to cut
off. The control is now from U2 (and not U1). The load current as sensed by
R2 is converted to a voltage and is compared with fraction of the reference
voltage V2 derived through R13 and R15. This ensures that for all load
resistances lower than 249 ohms, the supply will operate in constant current
mode. Serious learners should simulate this circuit and convince one self as
to what happens at various load resistance values. One should check what
the load current is for lesser load resistances such as 100 ohms and 25 ohms.
One can modify to give a different constant current of say 100mA (as against
the setting of 20ma, as shown) and make necessary changes to get proper
CVCC action.
V3
20
+V
V4 U3
5V LM358 D3 R12
+V R17 + DIODE 10k
1k Q2
NPN
R11 LM358
10k
U4 D4
+ DIODE
R18
10k R14
9.979
18.35 V
100k
DC~V
R20

R19
R4 97.6k R10 R16 5.000 V
6k 4k 250 DC~V
 V1
20
+V
V2 U1
5V LM358 D1 R1
+V R5 + DIODE 10k
1k Q1
NPN
R6 LM358
10k
U2 D2
+ DIODE
R7
10k R2
9.979
5.384 V
100k DC~V
R9
R8
R15 97.6k R13 R3 4.992 V
6k 4k 249 DC~V

32. What is the gain of an inverting amplifier? How does it get affected if any by
presence of source resistance.
To examine the effect of source resistance, let us take a source that has anon
zero source resistance. Explain why the meter shows 776mV though the
voltages applied to the amplifier are same. [5V*R6/(R6+R5)=500mV].
The reason is that in the first case, the source resistance is zero while in
second case, it is 10k||90k=9k. Thus though the gain determining resistors
have a ratio of 10 in second case, the gain is actually -100/(20k+9k) = -100/29
= -3.448, as compared to gain of -5 in first case.
775.7mV
R1 DC~V
100k
R4
V1 100k
5V
+V V6
5V
R2 U1 +V
20k LM358 R5 R3 UB
+ LM358
90k 20k +
V3
+ 500m V4 R6
5V 10k
+

V2
+V-5 V5
+V-5

33. For the nonlinear amplifier of fig below, estimate the gain for 0 to 100mV
and 1 to 1.1V.
 R2
5k

V1 Is1
DIODE 12 10u
1.072 V
D2 D1 R1 +V U1
DIODE OPAMP5 D3 DC~V
V3 10k +
0/3V DIODE
D4
DIODE
V2
166.mHz +V-12

The negative input has same voltage as input from source V3. When voltage
at negative input is about 1V, the diodes start conducting. Once they
conduct gain is not unity, but 1+5k/10k = 1.5. Thus gain for 0 to 100mV is 1.
Gain for 1V to 1.1V is 1.5.
See the comparison of input with output shown below:
Xa: 6.000 Xb: 0.000 a-b: 6.000 freq: 166.7m
Yc: 1.173 Yd:-800.0m c-d: 1.973
Offsets X: 0.000 Y: 0.000
b a
A
B
Y=voltage

d
0 1 2 3 4 5 6
Ref=Ground X=1/Div

34.For the schmitt trigger shown below what is the type of feedback used? How
is that different from feedback used for a non inverting amplifier?
 R2
20k R3
100k

V1
12 V6
+V U1 -12
R1 +V U2
V3 10k LM358 R4 LM358
+ V4 10k
-3/3V
-3/3V
V2 +
+V-12 V5
166.mHz +V12
166.mHz

The non inverting amplifier has negative feedback while the Schmitt trigger
has positive feedback, as can be seen from the feedback connection from
output to input. The Schmitt trigger will swing between two levels, and
cannot stay in linear region due to regenerative feedback. Recall Japan
disaster!!