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University Year 1 > Electrical- Diodes, Transistors and Amplifiers > Flashcards

Electrical- Diodes, Transistors and Amplifiers Flashcards

1
Q

What is a diode made from?

A

Silicon. It has a P-type material next to a N-type material.

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2
Q

Current against voltage graph for a diode

A

Line along positive x-axis until just before 0.7V when there is a steep exponential increase. Line along negative x-axis until 100s of V when it goes pretty much straight down.

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3
Q

Equation of a diode

A 
i=I0e^(v/nkT)
Where I0 is small
n is a constant between 1 and 2
T is temperature in K
k is Boltzmann’s constant
i is current and v is voltage
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4
Q

What is the ideal diode model?

A

In forwards mode it conducts like a wire and there is no voltage drop across it. When it is not conducting, no current goes through it and it acts like an open circuit (no voltage drop across it).

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5
Q

What is the simplified (0.7V) model for a diode?

A

It only conducts in forwards mode and acts like a 0.7V voltage source opposing the direction of current through it. When it isn’t conducting, it acts like an open circuit and the voltage drop across it is less than 0.7V

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6
Q

How to decide if a diode is conducting or not

A

Assume it is or isn’t conducting, then check the circuit to see it makes sense. If you assumed it is conducting, using the 0.7V model, the current going into the diode in forwards mode must be greater than 0 for this assumption to be correct. If you assumed it isn’t conducting, the voltage drop across it must be less than 0.7V for this assumption to be correct.

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7
Q

How to analyse a circuit with a diode and a varying source?

A

Identify the state of the diode when it’s at the cusp of conduction, i.e If=0 and Vf=0 or 0.7 (depends on model) and f means forwards. Solve the circuit to find the value of the source under this condition (threshold when diode state changes). Slightly increase (or decrease) the input from the source to see if the diode conducts above or below this value.

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8
Q

How to get equations to draw the output waveform for a circuit with a diode and varying source.

A

Need equations for vout both for when diode is and isn’t conducting. Write the input voltage as v(t). For conducting state, solve the circuit using diode as 0.7V source or wire (model) to get equation for vout in terms of v(t). For non-conducting state, solve the circuit using diode as open circuit to get another equation for vout.

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9
Q

How to draw the output waveform for a circuit with a diode and a varying source

A

Have the equations for vout in terms of v(t). Draw voltage time graph for v(t) (using information in the question). Draw vertical dotted lines for when cusp of conduction requirement is met. This is when you switch equations. Draw graph for vout in comparison with v(t).

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10
Q

What does a peak detector do?

A

Finds the tip of a sinusoidal input voltage.

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11
Q

What is a peak detector?

A

A circuit that has an output voltage that is (ideally) equal to the peak value of an input signal.

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12
Q

Example of a peak detector circuit

A

Varying voltage source in series with a diode (such that it conducts when Vin is positive) and a capacitor. The capacitor is in parallel with a resistor.

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13
Q

Describe the voltage-time graph for Vin and Vout for a peak detector

A

This is for a sinusoidal input voltage that initially goes positive. The line for Vout follows the sine wave up to the peak, then it is an almost straight diagonal line down to the next positive part of a wave cycle. It follows that wave up to the peak and the process repeats. The diagonal line is less steep for a greater value of R as the time constant (RC) will be larger. This is for the ideal model of diode.

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14
Q

How to find percentage ripple for a peak detector

A

Assume Vout against time graph has straight diagonal lines that go to the centre of each Vin wave. Voltage across capacitor is
(1/C)Sidt. This is across one time period T so
ΔV=IT/C. The current (I) is Vp/R so
ΔV=VpT/RC
%ripple=100xΔV/Vp=100xT/RC

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15
Q

What is the point of the resistor in a peak detector circuit?

A

If it wasn’t there, there would be no mechanism by which the capacitor could lose charge and the voltage across it would be the highest value of source voltage that ever occurred (ideal model). Sometimes the source voltage has different sized peaks that would then be missed.

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16
Q

What does a rectifier do?

A

Converts an AC voltage into a DC voltage.

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17
Q

Describe a half wave rectifier and what it does

A

AC voltage source in series with a diode and resistor (voltage across which is Vout). When the Vin is such to make the diode conducts in forwards mode, positive half cycles are transmitted to the resistor. Otherwise, the negative half cycles are blocked. The Vout against time graph is basically the Vin graph but negative regions replace by horizontal lines along x-axis.

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18
Q

What does a half wave rectifier with smoothing look like?

A

Basically a peak detector circuit but the capacitor used has a hollow rectangle instead of top line.

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19
Q

What does a full wave rectifier circuit look like?

A

Same as half wave one but connected to bottom of AC voltage source is another diode that takes current from when the original diode doesn’t conduct, to between the original diode and the resistor. Results in a Vout against time graph as the magnitude of the Vin graph (peak, peak, peak…)

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20
Q

What does a bridge rectifier look like?

A

Has an AC voltage source connected to top and bottom of a diamond of four diodes. The top two conducts clockwise, the bottom two conduct anti-clockwise. The right vertex goes to a capacitor and resistor in parallel with each other. This then goes back to the left vertex of the diamond. Means the current in C is in the same direction no matter which way the voltage source acts.

21
Q

Graph of Vout against time for bridge rectifier and formula for ripple

A

Without the capacitor it would be the same as a full wave rectifier. With capacitor, it is like peak detector line for the graph of full wave rectifier. This means formula for ripple is
ΔV/Vp=T/2RC

22
Q

What is regulation?

A

The process of making the output independent of lead current changes.

23
Q

Series regulator circuit

A

Basically a potential divider. But the regulator part is just that one is a variable resistor which can automatically be changed to control the output across the other resistor.

24
Q

Shunt regulator circuit

A

One resistor (Rs) in series with a another resistor (Rshunt, not actually a resistor) and the load resistor (for Vout) which are in parallel with each other. The input voltage is DC. The regulator part is Rs and Rshunt.

25
Q

Symbol for Zener diode

A

Normal diode but line has a flick on each end (in opposite directions)

26
Q

How do Zener diodes behave?

A

They have a voltage that, in reverse mode, above which the diode breaks down by design. Conducting state is when current goes into the point of the triangle and it is modelled as a voltage source with the breakdown voltage (Vz) opposing the current. Non-conducting state when voltage applied to it is less than Vz so it is modelled like an open circuit with the applied voltage across it. 0.7V drop ignored.

27
Q

What does a Zener shunt regulator look like?

A

Same as normal shunt regulator but the Rshunt is replaced with a Zener diode pointing upwards.

28
Q

What to do when solving a shunt regulator circuit when load is given in terms of current

A

Redraw the load resistance of a current source acting in the same direction as the input voltage with a value given in the question.

29
Q

How to solve a shunt regulator circuit when given a required output and minimum input voltage and maximum load current

A

Find maximum value for Rs when the Zener diode conducts. Know the potential difference across load resistor is output and is same as potential drop across Zener diode. Choose worst case (lowest Vin, highest load current). The lowest current going into diode for circuit to work is 0A. Now find potential across Rs using one loop. Use V=IR to find its resistance (maximum value).

30
Q

How to find efficiency of shunt regulator

A

Power in is Vin x Iin. Know load current but not current in diode.
Power out is Vout x Iout (both known).
Power loss is power loss from resistor and in diode (Iin^R and VzIz)
Can work out Iin from current in resistor (work out its pd)
Do power out over in for efficiency.

31
Q

What does a transistor do?

A

Controls a large current/voltage using a small current/voltage

32
Q

Circuit representation of bipolar junction transistor (BJT)

A

Has two forms. Both have wire (base) going into vertical line. Diagonal line up from this (not joining base) goes to vertical line (collector). Diagonal line down from near base goes to other vertical line (emitter). This diagonal line has an arrow pointing to and touching the emitter for NPN form. For PNP form the arrow points to base and touches the vertical line it goes into.

33
Q

How do the currents and voltages work in a NPN transistor?

A

Base and collector currents go into transistor. Emitter current goes out. There is a voltage drop (0.7V) between base and emitter as this part acts like (or is) a diode. The collector-emitter part acts as a current source. The collector current is equal to the current gain (β) times the base current. There is a minimum value of voltage between the collector and emitter (0.4V) meaning too high a base current can’t turn on the transistor.

34
Q

Properties of an ideal switch

A

When closed, there is a short circuit, its resistance is 0 and the pd across it is 0.
When open, there is an open circuit, it’s resistance is infinite and the current through it is 0.
In either case the power loss is 0.

35
Q

How does a BJT acts as a switch?

A

Voltage (Vcc) and resistor in series going into the collector. Vcc is equal to current in collector times resistance of resistor plus the pd across the collector and emitter (Vce). Rearrange to find
Ic=Vcc/RL-Vce/RL
L is subscript for load resistance

36
Q

Graph of Ic against Vce for BJT acting as a switch

A

For a fixed base current, line curves up from origin until 0.4V when it levels off to become a horizontal line. As base current increases, this graph moves up according to Ic=βIb.
There is also a diagonal line down from Vcc/RL value for Ic to Vcc value on x axis. The intersection between this line and curved bit of other line is “on”, the point on this line meeting the x axis (Vcc) is “off”.

37
Q

Power loss for a BJT acting as a switch for its on and off states

A

On: power loss is VccIc=VccxV/RL. Vcc is small so low power loss.
Off: power loss is Vccx0 so no power loss.
Never operate between states because this leads to large power and the transistor vaporises.

38
Q

What is considered when transistors as switches are designed?

A

Design for largest collector current that will be encountered.
Design for smallest β possible.
Always have a resistor for the base current because a large base current will destroy the transistor.

39
Q

What would a simple inductive load switch look like?

A

Has voltage (Vcc) going into inductor (having resistance) which goes into collector part of transistor (no base drawn) and the emitter goes down to earth.

40
Q

Current or Vce against time graph for simple inductive load switch and why this model doesn’t work

A

When on (starting period) the current and Vce are both horizontal lines (Vce very low). When enter off state, current is vertical line down. Vce goes up very high in very short period of time and comes back down very quickly to a higher horizontal line than before. This large voltage spike kills the transistor.

41
Q

Freewheeling diode solution to inductive load switch

A

Has loop with a diode from between resistor and transistor back to between Vcc and inductor. The current in this loop is ID. For the Vce against time graph, the on section is the same but off section goes up to horizontal line (much lower than previous spike) and then lowers a bit to same horizontal line height as simple model. Area under higher horizontal line to this line is same as that for the large spike.

42
Q

Problem with freewheeling diode solution to inductive load switch

A

At the off stage, the current in loop (ID) decays exponentially to 0 with a time constant L/R (R is resistance of inductor). This means a large turn-off time. It is when ID=0 that the Vce drops a bit.

43
Q

What is the best solution for inductive load switches?

A

Same circuit as freewheeling diode but with a resistor (RD) in the extra loop with the diode. Means Vce initially goes higher than without RD but decays back to original horizontal line. The turn off time is shorter because time constant is now L/(R+RD). This makes a faster switch. The height difference between peak and what Vce decays to is 0.7+RDID.

44
Q

What does operational amplifier look like?

A

Triangle pointing right. Top corner has plus (to non-inverting input). Bottom corner has minus (to inverting input). Right corner has Av (voltage gain) and connects to output. In reality has a positive and negative supply going into near right corner but neither drawn.

45
Q

Formula for vout from an operational amplifier

A

vout=Av(v+ - v-)

Where the + and - correspond to the voltages at those terminals.

46
Q

Principles of operational amplifiers

A

Output does not depend on supply voltage (but is limited by it).
Gain is very large and unpredictable (Av typically 10^6-10^9 and changes with temperature, frequency of input, batch, per device).
The inputs draw no current (high impedance) and the output has low impedance.
It is normal to assume that Av is infinite for ideal op-amps.

47
Q

Where does the feedback for an operational amplifier go?

A

To the inverting input

48
Q

Review last page on op-amps

A

Examples from implications

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