Electronics 16-1a Semiconductors They collect a positive electric - - PowerPoint PPT Presentation

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16-1a Electronics

Semiconductors

• They “collect” a positive electric charge on a small minority of the

atoms.

• If a voltage is applied, the electron goes to the positive terminal.

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16-1b Electronics

Semiconductors

• If the semiconductor is doped with atoms that have three valence

electrons, each dope atom forms three covalent bonds with its neighboring Si or Ge atoms, resulting in one neighbor atom in the lattice with no atom to bond with.

• If a semiconductor is doped with atoms that have five valence

electrons, each dope atom forms four covalent bonds with its neighbors, resulting in one unshared electron in the dope atom, causing the dope atom to donate a free electron.

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16-1c Electronics

Semiconductors

• p-n Junction – when p-type and n-type doping occur next to each
• ther in the same crystal
• Diffusion Current – free electrons from the n-type material combine

with the holes in the p-type material near the junction

• Depletion Region – area near the junction
• Drift current – the potential difference creates an electric field that

pushes electrons back toward the n-type material from the p-type material

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16-1d Electronics

Semiconductors

Example 1 (FEIM): Which of the following is NOT true for intrinsic semiconductors? (A) There are holes in intrinsic semiconductors. (B) There are free electrons in intrinsic semiconductors. (C) They make good insulators. (D) Increasing thermal energy increases their conductivity. Intrinsic semiconductors will carry current, so answer (C) is not true. Therefore, (C) is the answer.

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16-1e Electronics

Semiconductors

Example 2 (FEIM): In the depletion region of a semiconductor p-n junction, there (A) is an electric field. (B) are more holes than outside the depletion region. (C) are more free electrons than outside the depletion region. (D) is current perpendicular to the current outside the depletion region. Answers (B) and (C) are wrong because the depletion region has fewer holes and free electrons than outside the depletion region. Answer (D) is

• nonsense. However, there is an electric field.

Therefore, (A) is correct.

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16-1f Electronics

Semiconductors

Diode Symbol

• P-type – anode
• N-type – cathode

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16-2 Electronics

P-N Junction Biasing

• Forward biased
• Reverse biased

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16-3 Electronics

Diode Characteristics

• Static forward resistance
• Breakdown voltage

For an ideal diode in series with a voltage: For an ideal diode with zero resistance in the forward bias direction and infinite resistance in the reverse bias direction:

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16-4a Electronics

Special Diodes

Zener Diodes

• They have a high doping concentration.
• Avalanche – the effect of the e– in the depletion region accelerating

and colliding.

• For an ideal Zener diode, Vo = 0, rf = 0, and ra = 0.

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16-4b Electronics

Special Diodes

Example (FEIM): When a Zener diode suffers breakdown, it (A) is immediately destroyed. (B) behaves as a reversed biased ideal diode. (C) becomes an open circuit. (D) behaves as a voltage source. Since the Zener diode is at the Zener voltage in the reverse bias direction when it suffers breakdown, (D) is correct. Note that answers (B) and (C) are the same (just worded differently), so they both must be wrong. Therefore, the answer is (D).

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16-5a Electronics

Diode Applications

Half-Wave Rectifiers

• Half of a symmetric AC signal gets through
• Used in AC-to-DC converters

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16-5b Electronics

Diode Applications

Full-Wave (Bridge) Rectifiers

• Current is always going in the

same direction

• Used in AC-to-DC converters,

and are more efficient than half-wave rectifiers

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16-5c Electronics

Diode Applications

Clamping Circuits Output Voltage: Vout = Vin+ Vp - Vm where Vin = the input voltage Vp = the clamping voltage Vm = the maximum voltage of the input For a clamping circuit output with a sinusoidal input:

• Average Voltage: Vave = Vp – Vm
• RMS Voltage:

Vrms = 1 2 Vm +Vp Vm

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16-5d Electronics

Diode Applications

Base Clipper

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16-5e Electronics

Diode Applications

Peak Clipper Valley Clipper

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16-5f1 Electronics

Diode Applications

Combined Clipper

• 1. Valley clipper + peak clipper

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16-5f2 Electronics

Diode Applications

Combined Clipper (cont.)

• 2. Two Zener diodes in series in the opposite direction

The ideal model for the Zener diode:

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16-5g Electronics

Diode Applications

Example (FEIM): What is the average current through the resistor in the rectifier shown? Assume ideal diodes. (A) 0 A (B) 0.76 A (C) 3.06 A (D) 4.80 A This is a full-wave rectifier, so Therefore, (C) is correct. Vave = 2Vpeak

• = (2)(120 V)
• Iave = Vave

R = 240 V

• 1

25

• = 3.06 A

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16-6a1 Electronics

Operational Amplifiers

• An electronic device used to perform mathematical operations on

analog signals.

• Two inputs, one output, small current, and large gain

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16-6a2 Electronics

Operational Amplifiers

EIT8 Table 51.1

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16-6b Electronics

Operational Amplifiers

Example (EIT8):

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16-7a1 Electronics

Input Impedance

Example 1 (FEIM): What is the input impedance as seen by the source va of the following circuit? (A) 5 kΩ (B) 7.5 kΩ (C) 10 kΩ (D) 12.5 kΩ

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16-7a2 Electronics

Input Impedance

The input voltage is 15 V. The input current is The input impedance is the absolute value of the input voltage over the input current. So the input impedance is Therefore, (B) is correct. iin = 15 V 5 V 5 k = 2 mA Zin = Vin iin = 15 V 2 mA = 7.5 k

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16-8a Electronics

Amplifiers

Therefore, I1 = va vb R1 I2 = vo vb R2 va vb R1 = vo vb R2 vo = R2 R1 va + 1+ R2 R1

• vb

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16-8b Electronics

Amplifiers

Noninverting Amplifiers

• va = 0
• v1 = v2

Since v2 is a voltage divider circuit of the operational amplifier output, v2 = R1 R2 +R1

• vout

vout = 1+ R2 R1

• vb

Since vb = v1 = v2, we can substitute and solve for vout:

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16-8c Electronics

Amplifiers

Inverting Amplifiers

• vb = 0
• v1 = v2 = 0

iin = va R1 Since v1 = v2 = 0, input current: current through the feedback resistor: if = vout R2 Since iin = –if, va R1 = vout R2 vout = R2 R1 va

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16-8d Electronics

Amplifiers

Summing Amplifiers

• Superposition theorem – currents

can be treated as independent forces trying to push electrons into (or out

• f) node A.

For the noninverting amplifier: if = i1 + i2 + i3 +L vout Rf = v1 R1 + v2 R2 + v3 R3 +L vout = Rf v1 R1 + v2 R2 + v3 R3 +L

• vout = 1+ R2

R1

• v1

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16-8e Electronics

Amplifiers

Integrating Amplifiers

• Similar to inverting amplifier, feedback current has to be equal and
• pposite to the input current.
• The output voltage is the voltage across the capacitor.

Assume the initial voltage on the capacitor = 0; the voltage

• n the capacitor is:

Applying Ohm's law: vout = 1 C i dt

• vout = 1

RC vin dt

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16-8f Electronics

Amplifiers

Differentiating Amplifiers

• Feedback current has to be equal and opposite to the input current.
• The output voltage is the voltage across the resistor.

Since v1 = v2 = 0: i = C dvin dt Substituting into vout = iR gives:vout = RC dvin dt

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16-8g Electronics

Amplifiers

Low-Pass Filters The output voltage divided by the feedback impedance is equal and

• pposite to the input voltage divided by the input impedance.

vout vin = Zf Zin = 1 ZinYf = 1 Ri 1 Rf + jC

• vout

vi = Rf Ri 1+ jRfC

( )

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16-8h Electronics

Amplifiers

Example 1 (FEIM): What is the input impedance of the following ideal amplifier? Therefore, (B) is correct. (A) R1 (B) R3 (C) R2 R1 +R3 (D) R1R3 R1 +R2 iin = ia + i3 ia = 0, because this is an ideal op amp. Rin = Vin iin = Vin i3 = R3

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16-8i Electronics

Amplifiers

Example 2 (FEIM): What is the input impedance of the ideal amplifier shown? Therefore, (A) is correct. (A) R1 (B) R2 (C) R2 R1 (D) R1 R1 +R2 Rin = Vin iin iin = Vin Va R1 Va = 0 because this is an ideal op amp. Substituting yields: Rin =Vin R1 Vin

• = R1

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16-8j1 Electronics

Amplifiers

Example 3 (FEIM): The 700 Hz signal shown is applied to the ideal amplifier circuit shown. What will the output signal be?

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16-8j2 Electronics

Amplifiers

Both the DC and AC part are multiplied by –3. Therefore, (B) is correct. The input current and feedback currents must be equal and opposite, so: iin = Vin 3 k = ifeedback = Vout 9 k Vout Vin = 9 k 3 k = 3

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16-8k1 Electronics

Amplifiers

Example 4 (FEIM): Two AC signals V1 and V2 are to be combined such that The following subtracting amplifier circuit is used. What must be the values of R1, R2, R3, and R4? Vout = 3 2V2 5 2V

1.

(A) R1 = 2 k, R2 = 2 k, R3 = 5 k, R4 = 3 k (B) R1 = 2 k, R2 = 4 k, R3 = 5 k, R4 = 3 k (C) R1 = 4 k, R2 = 8 k, R3 = 10 k, R4 = 2 k (D) R1 = 5 k, R2 = 3 k, R3 = 4 k, R4 = 2 k

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16-8k2 Electronics

Amplifiers

Solving for Vout yields: Va =V2 R4 R2 +R4

• So the possibilities narrow to (A),

(B), and (C). Trying R1 = 2 kΩ in the other relationship yields 4R4 = 3R2 Answer (B) fits because R4 = 3 kΩ, and R2 = 4 kΩ. Plugging in to confirm, Therefore, (B) is correct. i1 = V

1 Va

R1 = iout = Vout Va R3

• Vout = R1 +R3

R1

• R4

R2 +R4

• V2 R3

R1 V

1 = 3

2V2 5 2V

1

R3 R1 = 5 2 R1 +R3 R1

• R4

R2 +R4

• = 2+5

2

• R4

R2 +R4

• = 3

2 Vout = 2+5 2

• 2

4+3

• V2 5

2V

1 = 3

2V2 5 2V

1

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16-8l1 Electronics

Amplifiers

Example 5 (FEIM): A 30 mV sinusoidal signal must be inverted, amplified to 6 V (peak), and chopped at 4 V. If the following circuit is used, what are the values of R1 and R2, and the avalanche voltage of the Zener diodes Z? Assume –0.7 V forward bias voltage drop and negligible diode resistance. (A) R1 = 1 kΩ, R2 = 20 kΩ, Z = 4 V (B) R1 = 1 kΩ, R2 = 200 kΩ, Z = 4 V (C) R1 = 2 kΩ, R2 = 400 kΩ, Z = 3.3 V (D) R1 = 2 kΩ, R2 = 800 kΩ, Z = 3.3 V

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16-8l2 Electronics

Amplifiers

The amplification without clipping is This narrows the choices to (B) and (C). When |Vin| > 2 mV, the diodes will be forward biased and reversed biased respectively, so the voltage across the two diodes in series will be the Zener voltage plus the forward bias voltage. Thus, the Zener voltage is 3.3 V. Therefore, (C) is correct. Vout Vin = R2 R1 = 6 V 3010

3 V

R2 = 200 R1