Minimum Mtech

8/8/2019 Minimum Mtech

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Topics covered: Summary

1. Pre requisites; thevenin, super position, transistor characteristics, simple ac amplifier.

2. Diode equation;Id = Is*{exp(Vd/VT)-1} where VT is kT/q,

Is, the saturation current is about 10^-14 amp.

Diode characteristics; 60mV/decade, 18mV/octave. Current increases exponentially as

voltage is increases.

Diode voltage versus current is best plotted in semilog graph, where V in x axis is in linear

sacle and I in log scale is in Y asix. The plot is a straight line.

Dynamic resistance = .026/Id. At 1mA, it is 26 ohms.

3. Transistor: relationship between Ic, IB and IEIc/IB =

IE = IB*(+1)

Ic/IE = /(+1)

4. Three states:ON, OFF and linear. In ON state, base emitter is fwd biased

State Base emitter Collector base

ON FWD FWD

Linear FWD Rev

OFF Rev Rev

5. Three configurationsCE, CB and CC.

6. Equivalent circuit for transistor. Input resistance r. As input. Current source withparallel resistance rc at output. Rc = early voltage VA / Ic, collector current.


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7. Fixed bias, dc conditions: In CE, a fixed base current can be used to bias the transistor.Depending on supply and Rc, the transistor may be in linear region or ON.

8. Collector current increases by 10% if base emitter voltage is increased by 2.6mV.9. Min VCB for operation in linear mode is zero volts.

10.gm, Input resistance, output resistance, VG for common emittergm = Ic/. r= /gm; rE = 1/gm, rc = VA/Ic. When Ic is 1mA, gm is 38.5mA/volt, and rE is

26 ohms.

Output resistance = Rc||rc where Rc is resistance connected from collector to supply.

Draw circuit!!

11.Early voltage =VA.Affects current in collector even if base current is kept constant. If VCE is increased by

10% of VA, collector current increases by 10%.

12.Emitter degeneration.When a resistor RE is placed in emitter, and not bypassed, the voltage gain decreases

from Rc/rE to Rc/(rE +RE)

13.Common base characteristics: Rin, VG, Rout:In common base, input resistance = rE. VG is same as that for CE, except that there is

no reversal. Output res is Rc

14.Common collector characteristics: Rin, VG, RoutIn CC, input resistance is very high = *(rE+RE).

Output resistance is very low, = rE if Rs is zero. Else it is rE+(Rs/)

Voltage gain is close to but less than 1.

15.Current sources: Bipolar.The VBE of a transiode is applied as VBE for output transistor. Due to base current,

ratio is 1/{1+(2/)}. If is 50, ratio is 0.96.

16.Current source: beta helper


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A transistor (with collector connected to supply) is used to provide base current for

transdiode as well as output transistor. Advantage is that ratio is closer to 1. It is

1/{1+(2/^2)}. If is 50, ratio is 0.9984. Draw circuit!!

17

.Wilson current source.Here acurrent sense a transdiode senses the current in output transistor and

controls conduction oftransdiode. Draw circuit!!

18.Widlar current sourceUsed to get low current without having to use large resistor. Reduces cost of IC as cost

of IC is higher depending on area, and large resistances require larger areas. 6k

dropping 60mV in emitter reduces ref current of 100ua to 10uA.

Note 6k*10uA = 60mV if Ic = IE.

19.Differential amplifierTwo transistors with emitters connected together and returned to negative supply through

a resistance REE = R. Collector of each transistor connected to +ve supply through R.

Rin = 2*r. VG is half of what it is for CE, when output is taken between one collector and

ground.

Common mode gain is Rc/2REE. Common mode rej ratio is REE/rE.

Acm, Adm, Rdm, Rcm, CMRR

20.Active loadThe gain of an amplifier depends on load resistance. To get effectively high resistance

without increasing the drop across it, a current source is used. One can use a bipolar

current source as active load. In differential amplifier this is particularly attractive as

two collectors are present, and one can conveniently connect a transiode in one and the

output transistor in the other. Remember to take output from collector connected

to transistor .

21.Class A amplifierHas low efficiency. Dissipation is 3 times output power even when output power is max.

Draws current even when ac input is zero.

22.Class B amplifier; advantagesHigher efficiency, of78% max. Low dissipation when input is near zero.


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23.Class B cross over distortionDue to necessity to overcome threshold of pnp and npn, no output results when input is0.6V peak. When input is 1.4V peak, output peak is 0.6V lesser, with severe distortion,

called cross over distortion. Draw circuit, waveforms.

24.Class B: prebias is used to reduce cross over distortion. Base voltage of npn is held1.2V higher using two diode drops, with constant current through them. Draw circuit.

When input is applied to base of pnp. Output is always higher by an offset dc of 0.6V.

25.Class B efficiency, power dissipation.If power input is Pin, Power output Po is 78% and power dissipation Pd is 22%.

Current drawn from supply is pulsating. Looks like half wave waveform. If peakcurrent in load is Ip, current in power supply is halfwave sine with peak value of Ip.

Average of this is Ip/.

With supply as Ip*RL in ideal case, Power output is (Ip*RL)*(Ip)/2, while Pin =

2*(Ip*RL)*(Ip/.). Thus efficiency = /4 = 78%, in ideal case.

Power dissipation is not max when power output is max!! Pd is higher at output power

of 50% rated power. This occurs when peak voltage across load is 70.7% of supply

voltage, in ideal case. Pd at half rated power output is 50% higher than Pd at rated

power output.

Calculate for 8W into 8 ohms. Supply is +/-11.3V. Supply current at 8W output = .45

amp. Pin = 10.17, of which Po is 8W leaving Pd as 2.17W.

At half rated power, supply current is 0.318amp. Pin = 2*11.3*.318= 7.19W, of which

Po is 4W. Thus Pd is 3.19W. This is 47% higher than 2.17W. This is in ideal case. In

practice the max power can be 30 to 40% higher than Pd at rated power output.

26.Choice of heat sinkWhen power is dissipated, it has to be taken off to ensure than junction temperature

does not rise to a value higher than max allowed value. Heat sinks may be required.

HS = (T/P)-CA, where T is max junction temperature minus max ambient. If

CA is 4degC/Watt, P= 5W, TJmax = 125 deg, TA max is 65 deg, HS = 12-4 =8degC/watt. It is good practice to derate, and take TAmax as 10deg higher and TJ max

as 10 deg lesser. In that case, HS will have to be 4 degC/W or lesser. (A 3degC/W heat

sink is bigger than a 4degC/W heat sink!!)

27.Class B protection: current limiting


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741: 3 stages.

First stage simplified schematic

2nd

stage: simplified schematic

Final stage. Simplified schematic.

Calculation of currents in various branches.

Calculation of open loop gain

Equivalent circuit representation of three stages.

Filters: First order: amplitude and phase, step response

Second order, Butterworth: realisation. Gain/phase response.

Step response: overshoot of butterworth second order is 4.2%

Zeta for overshootless response is 1. There will be overshoot is zeta is less than 1.

Zeta for maximally flat response is 0.707. For lesser zeta, there will be a peak in frequency

response.

Third order LP, Butterworth: TF.

Feedback and its effects.

723 and its branch currents.

723: estimation of output resistance, and regulation

General op amp configurations:

Inv amplifier, Input resistance, output resistance and gain

Noninv amplifier, 3 parameters.

Input resistance of non inv amplifier,

Diff amplifier, with large input res and high gain

Inv and non inv Schmitt trigger

Precision rectifier, ac to dc conversion. Form factor

Average of half wave recified, full wave rectified.


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Filtering of output to get low ripple

Log amplifier

Analog multiplier:

Gilberts cell

Transducers: RTD, and thermocouple

Cold junction compensation: Required in thermocouples as it is convenient to keep cold

junction at ambient temperature. Errors are then corrected by measuring the cold junction

temperature (- this can be done using a solid state temp sensor-) and adding that to

temperature interpreted from thermocouple. RTD has nocold junction compensation

requirement!!

3 wire RTD: Eliminates need for recalibration when cable to RTD is changed.

RTD linearization: RTD is nonlinear and gives 0.4% error in 0 to 200 degC range. A two

slope amplifier can bring this down to 0.1% error. Instead of a fixed gain of 2000/75.8, one

can have two slopes, viz 100/38.5 from 0 to 100 degC and 100/37.3 from 100 degC to 200

degC. Note that 37.3 = 75.8-38.5

LVDT, signal conditioning.

Communication topics: RS 232, RS485. Driver requirements, length , data rate. Signal at

receiver

Digital topics: Synchronous and asynchronous flip flops, problem ofracing.

Combinational and sequential circuits.

Set up time, hold time and propagation delay.

Karnoughs map.


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Minimum Mtech

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