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 dB value 0.01 -40 0.1 -20 0.316 1 -10 0 2 6 3.16 10 5 14 10 20 31.6 30 100 40

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 -1/1V R1 159k V2 -1/1V R2 15.9k C2 10n 1kHz LPF V3 -1/1V R3 1.59k C3 10n 10kHz LPF

100 Hz

C11kHz 10n 100Hz LPF

10kHz

3. If Time constant is Rc, cut off frequency fc is calculated as 1/(2. RC). 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 240.3mV AC V C4 10n

R4 15.9k

V5 -1.41/1.41V

R5 15.9k

166.4mV AC V C5 10n

4kHz

5.8kHz

1kHz LPF 1kHz LPF

In the figures above, input is 1V rms. Output can be approximated to be = 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 6803*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 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 12 +V

993.4uA DC~A R5 1130k R6 10k 370.6mV AC V V2 -1.41m/1.41mV R3 2.6k

V1 12 +V

993.4uA DC~A R2 1130k R1 10k 185.5mV AC V

C2 1u V3 -1.41m/1.41mV

NPN Q2

C1 1u

NPN Q1

1kHz

1kHz

Voltage across C2 is about 58uV!!! 12.A transistor with 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 D3 BRIDGE 5TO1CT 60 Hz C2 100uF R2 680 + C3 100uF R4 680 D2 1N4736 D1 1N4736 + C4 100uF R1 500 + + R3 500

V1

C1 100uF

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 HS and JC, 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 DC~V

V3 0/0V

V1 5V +V R3 source resistance R1 10k +

U1

1kHz

LT1006A V2 +V-5

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

R5 100k

V4 0/0V

V6 5V +V R4 source resistance R6 10k +

-298.7uV DC~V U2

1kHz

LT1006A R7 9.1k V5 +V-5

R7 reduces offset at output due to bias current

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

R6 10k V4 0/0V C1 1uF

V6 5V +V U2 +

3.998 V DC~V

LT1006A V5 +V-5

1kHz

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.V1 -325/325V D1 BRIDGE 205.6 V DC~V

50 Hz R1 100k

99.32 V AC V

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 -325/325V 103.1 V DC~V

D1 DIODE

50 Hz R1 100k

124.8 V AC V

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 -325/325V

D1 DIODE 100u C1 R1 100k

323.9 V DC~V

V2 -325/325V

D2 BRIDGE

323.3 V DC~V

50 Hz

185.6mV AC V

50 Hz

100u C2 R2 100k

91.89mV AC V

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 -28.2/28.2V D1 DIODE 100u C1 R1 100k 27.54 V DC~V V2 -28.2/28.2V D2 BRIDGE 26.86 V DC~V

50 Hz

15.66mV AC V

50 Hz

100u C2 R2 100k

7.419mV AC V

9. What is the rms value of a triangular waveform of peak value 10V? Ans: 10/sqrt(3)= 5.7735VV3 -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.

V1 -141/141V

99.98 V AC V

D1 BRIDGE

Final meter 88.73 V V2 DC~V -90/90V

90.00 V AC V

D2 BRIDGE

Final Meter 88.70 V DC~V

50 Hz R1 100k

50 Hz R2 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 DC~V + V1 10V S1 R2 10k 5.000 V DC~V R9 2.5k + Vs1 5 S5 R10 10k R11 2.5k + Vs2 5 + V2 10V 4.000 V DC~V S2 R4 10k

R1 10k 50%

R3 10k 50%

4.000 V DC~V S6 R12 10k

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 DC~V R6 10k 25% S4 R8 10k + V 10V

2.105 V DC~V

2.500 V DC~V R9 1.875k + Vs1 2.5 S5 R10 10k R11 1.875k + Vs2 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 5V +V V1 -5 5V R2 4k 3.000 V DC~V

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.)

40

z

R s s

v

summer

R1 6k

1.414 V C V

+ V4 10V

R5 10k 25%

S3 R7 10k

2.105 V DC~V S6 R12 10k

V1 -141/141V

R1 314

70.72 V AC V L1 1H

50 Hz

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 15.9k 49.99 V AC V C1 10n

1kHz

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 Vdmax

400 600

450 650

500 700

510 710

560 760

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 51.32 = 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 /gm where gm = Ic/26mv = 2.023/26 = 77.8 mA/volt. 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 = 3.9k12 +V 999.3uV AC V 2.023mA DC~A R7 51k R1 3.9k 12 +V 275.2mV AC V C2 10u -1.41m/1.41mV R11 1k R2 1meg 1kHz C5 10u 623.6uV AC V 2.023mA DC~A R9 51k R5 3.9k 171.8mV AC V C6 10u

-1.41m/1.41mV C1 10u 1kHz

2N2222A R8 5.6k R3 250 C3 1000u

2N2222A R10 5.6k R6 250 R4 1meg 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 AC V 2.023mA DC~A R7 51k R1 3.9k

137.1mV AC V C2 10u

-1.41m/1.41mV C1 10u 1kHz

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

24.For the circuit considered, what happens when the resistance in emitter is bypassed partially? This is shown in next figure.12 +V 999.7uV AC V 2.023mA DC~A R7 51k -1.41m/1.41mV C2 10u 1kHz R13 3.9k 60.31mV AC V C1 10u 776.9uV AC V 12 +V 2.023mA DC~A R9 51k R5 3.9k 46.86mV AC V C6 10u

-1.41m/1.41mV R11 1k 1kHz R14 1meg

C5 10u

2N2222A R3 5.6k R8 50

2N2222A R10 5.6k R6 50 R4 1meg

R1 200

C3 1000u

R12 200

C4 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)

11.71 V DC~V V1 20 +V R1 3.9k D1 1N4743 Q2 MPSA28

11.47 V DC~V V2 25 +V R4 3.9k D2 1N4743 R3 60 Q1 MPSA28

R2 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 5V +V U3 LM358 D3 DIODE R12 10k

R17 1k

+

Q2 NPN

R11 10k + R18 10k 100k R20

LM358 U4

D4 DIODE R14 9.979 18.35 V DC~V

R4 6k

R19 97.6k

R10 4k

R16 250

5.000 V DC~V

V1 20 +V V2 5V +V U1 LM358 D1 DIODE R1 10k

R5 1k

+

Q1 NPN

R6 10k + R7 10k 100k R9 R15 6k

LM358 U2

D2 DIODE R2 9.979 5.384 V DC~V

R8 97.6k

R13 4k

R3 249

4.992 V 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.R1 100k V1 5V +V R2 20k V3 500m U1 LM358 + R5 90k V4 + R3 20k R6 10k 775.7mV DC~V R4 100k V6 5V +V UB LM358 +

+

5V 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 12 +V + Is1 10u

V3 0/3V

166.mHz

V2 +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:

DIODE D2 D1 R1 DIODE 10k

U1 OPAMP

D3 DIODE D4 DIODE

1.072 V DC~V

Xa: 6.000 Yc: 1.173 Offsets A B b

Xb: 0.000 Yd:-800.0m X: 0.000

a-b: 6.000 c-d: 1.973 Y: 0.000

freq: 166.7m

a

Y=voltage

c

0

1

2 3 Ref=Ground

4 X=1/Div

5

6

d

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 V1 12 +V + R3 100k V6 -12 +V

V3 -3/3V

R1 10k

U1 LM358 V4 -3/3V

R4 10k

U2 LM358

166.m z

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!!

166.m z

V2 +V -12

+

V +V 12