1 Description of the bandpass

Describe in detail the band model of solids and its significance for distinguishing between conductors, semiconductors and insulators. In particular, address the following aspects:

• Relevant bands and their occupation by charge carriers.

• Significance of the band gap and how it influences the electrical properties of materials.

• Influence of temperature changes on conductivity.

1.1 Lösung:

The band model explains the energy states that electrons can occupy in solids. In a single atom, the electron states are discrete energy levels. See: Section ??

2 The space charge zone in the pn junction

Investigate the properties and formation of the space charge zone in pn junctions of semiconductors. Answer the following questions in your paper:

• What is the space charge zone and how does it form in the pn junction?

• What is the distribution of the electric field in the semiconductor?

• What internal processes affect the free charge carriers?

• How does the space charge zone change when a voltage is applied in the forward and reverse directions?

2.1 Lösung:

The RLZ is a narrow region around the pn junction in which positive ions from the n layer and negative ions from the p layer remain after diffusion is complete. There are no free charge carriers in this zone, as the positive and negative charges neutralise each other. This creates an electric field (\(\vec {E}\)) that generates a drift current and suppresses the diffusion of further charge carriers. The space charge zone acts as a barrier layer and prevents current flow in the reverse direction. See: Section ??

3 Calculation of the voltage drop across a diode

Given is a series connection consisting of a DC voltage source, a resistor and a silicon diode with a barrier voltage of \(U_\mathrm {S}=0.7\,\mathrm {V}\). What is the approximate voltage drop across the resistor \(R\) when the supply voltage is \(U_0=5\,\mathrm {V}\)?

3.1 Lösung:

Since this is a silicon diode, the gate voltage and the resulting voltage drop across the component are approximately \(0.6 \text { to } 0.7\,\mathrm {V}\).

In series connection, the following relationship applies: \begin {equation*} U_0 = U_\mathrm {R} + U_\mathrm {D} \end {equation*} \begin {equation*} U_\mathrm {R} = U_0 - U_\mathrm {D}= 5\,\mathrm {V} - 0,7\,\mathrm {V} = 4,3\,\mathrm {V} \end {equation*} See section: ??

4 Determining the operating point and power of a diode

A diode has the following characteristic curve in the forward direction. It is connected in series with an ideal voltage source \(U_0=1\,\mathrm {V}\) and a resistor with a value of \(R=200\,\Omega \).

• What current \(I\) and what voltage \(U_\mathrm {D}\) are established? Solve the problem graphically.

• What electrical power is converted at the diode?

• How must \(R\) be selected so that a current of \(3\,\mathrm {mA}\) is established?

4.1 Lösung:

• In the first step, the effect of the voltage source on the resistor \(R\) is considered. At \(U_0=1\,\mathrm {V}\), a current of \(I_\mathrm {R}=\frac {U_0}{R}=\frac {1\,\mathrm {V}}{200\,\Omega }=5\,\mathrm {mA}\) flows through the resistor \(R\). The resistance line can be drawn from the origin to the point (\(1\,\mathrm {V}\), \(5\,\mathrm {mA}\)) (left figure).
  According to the circuit design, the resistance line must be mirrored in the next step, resulting in the right-hand figure. The intersection of both lines corresponds to the operating point (\(U_\text {AP}\), \(I_\text {AP}\)), which is: (\(0.75\,\mathrm {V}\), \(1.2\,\mathrm {mA}\)).

• The power is calculated from the previously determined operating point. \begin {equation*} P = U_\text {AP} \cdot I_\text {AP} = 0,75 \, \text {V} \cdot 1,2 \, \text {mA} = 0,9 \, \text {mW} \end {equation*}

• Starting from the point (\(1\,\mathrm {V}, 0\,\mathrm {mA}\)) to the intersection with the diode characteristic curve at \(I=3\,\mathrm {mA}\), we obtain the desired resistance line.

  The intersection point (\(0.86 V, 3 mA\)) can be read off.

  \begin {gather*} U_\mathrm {R} = U_0 - U_\mathrm {D} = 1 \, \mathrm {V} - 0,86 \, \mathrm {V} = 0,14 \, \mathrm {V} \\ I = 3 \, \mathrm {mA} \\ R = \frac {U_{\text {R}}}{I} = \frac {0,14 \, \text {V}}{3 \, \text {mA}} = 46,67 \, \Omega \end {gather*}

See section: ??

5 Dimensioning the series resistor of a Zener diode

The Zener voltage of the diode is \(U_\mathrm {D}=6\,\mathrm {V}\) and the diode must not exceed a maximum power dissipation of \(P_\mathrm {tot}=\mathrm {400\,mW} \). The maximum input voltage is \( U_0 = \mathrm {12\,V} \).
What is the minimum value of resistance \( R \) required to prevent the diode from being destroyed?

5.1 Lösung:

The voltage across the resistor is calculated as follows: \begin {align*} U_\mathrm {R} &= U_0 - U_\mathrm {D} = \mathrm {12\,V} - \mathrm {6\,V} = \mathrm {6\,V} \end {align*}

The maximum power dissipation of the diode is: \begin {align*} P_\mathrm {tot} &= U_\mathrm {D} \cdot I = \mathrm {400\,mW} \end {align*}

This results in the maximum current through the diode: \begin {align*} I &= \frac {\mathrm {400\,mW}}{\mathrm {6\,V}} = \mathrm {66{,}7\,mA} \end {align*}

In order not to exceed this current, the minimum value for the resistance is: \begin {align*} R &= \frac {U_\mathrm {R}}{I} = \frac {\mathrm {6\,V}}{\mathrm {66{,}7\,mA}} \approx \mathrm {90\,\Omega } \end {align*}

See section: ??

6 Use of a Zener diode for voltage stabilisation

A load resistor is to be created with the aid of a Zener diode \( R_\mathrm {L} = \mathrm {160\,\Omega } \) be supplied with a stabilised voltage of \( U_\mathrm {A} = \mathrm {8\,V} \).
The data sheet for the Zener diode shows that at a Zener voltage of \( \mathrm {-8\,V} \) a Zener current of \( \mathrm {-0{,}45\,A} \) flows.

• Calculate the series resistor \( R_\mathrm {V} \) so that, for an input voltage of \( U_\mathrm {E} = \mathrm {10\,V} \), an output voltage of \( U_\mathrm {A} = \mathrm {8\,V} \) is established.

• What are the powers dissipated in \( R_\mathrm {V} \), \( R_\mathrm {L} \), and in the diode \( D_\mathrm {Z} \)?

6.1 Lösung:

• \begin {gather*} I_\mathrm {RL} = \frac {\mathrm {8\,V}}{\mathrm {160\,\Omega }} = \mathrm {0{,}05\,A} \\ I = \mathrm {0{,}05\,A} + \mathrm {0{,}45\,A} = \mathrm {0{,}5\,A} \\ U_\mathrm {RV} = \mathrm {10\,V} - \mathrm {8\,V} = \mathrm {2\,V} \\ R_\mathrm {V} = \frac {\mathrm {2\,V}}{\mathrm {0,5\,A}} = \mathrm {4\,\Omega } \end {gather*}

• \begin {gather*} P_\mathrm {RV} = \mathrm {2\,V} \cdot \mathrm {0{,}5\,A} = \mathrm {1\,W} \\ P_\mathrm {RL} = \mathrm {8\,V} \cdot \mathrm {0{,}05\,A} = \mathrm {0{,}4\,W} \\ P_\mathrm {DZ} = \mathrm {8\,V} \cdot \mathrm {0{,}45\,A} = \mathrm {3{,}6\,W} \end {gather*}

See section: ??

7 Determination of the operating limits of a Zener diode

The following circuit with a Zener diode and two resistors is given. The Zener diode used has a Zener voltage of \( U_\mathrm {Z} = \mathrm {8\,V} \). The maximum power dissipation of the diode is \( P_\mathrm {tot}=\mathrm {8\,W} \). The diode operates in the stabilising range when the current through it is at least \( \mathrm {5\,mA} \). In addition, the following values are given: \(U_\mathrm {E}=20\,\mathrm {V}\pm 10\%\) und \(R_\mathrm {V} = 10\,\Omega \).
What are the minimum and maximum resistance values that the load resistance \( R_\mathrm {L} \) must not fall below or exceed in order for the diode to operate in the stabilising range on the one hand and not be overloaded on the other?

7.1 Lösung:

Given is: \begin {equation*} U_\mathrm {max} = \mathrm {22\,V} \text { und } U_\mathrm {min} = \mathrm {18\,V} \end {equation*}

Fall 1: At \( U_\mathrm {max} = \mathrm {22\,V} \) \begin {gather*} U_\mathrm {RV} = \mathrm {22\,V} - \mathrm {8\,V} = \mathrm {14\,V} \\ I = \frac {\mathrm {14\,V}}{\mathrm {10\,\Omega }} = \mathrm {1{,}4\,A} \\ I = I_\mathrm {D} + I_\mathrm {L} \\ I_\mathrm {D,\ max} = \frac {\mathrm {8\,W}}{\mathrm {8\,V}} = \mathrm {1\,A} \\ I_\mathrm {L} = \mathrm {1{,}4\,A} - \mathrm {1\,A} = \mathrm {0{,}4\,A} \\ R_\mathrm {L} = \frac {\mathrm {8\,V}}{\mathrm {0{,}4\,A}} = \mathrm {20\,\Omega } \Rightarrow R_\mathrm {L,\ max} = \mathrm {20\,\Omega } \end {gather*}

Fall 2: At \( U_\mathrm {min} = \mathrm {18\,V} \) \begin {gather*} U_\mathrm {RV} = \mathrm {18\,V} - \mathrm {8\,V} = \mathrm {10\,V} \\ I = \frac {\mathrm {10\,V}}{\mathrm {10\,\Omega }} = \mathrm {1\,A} \\ I_\mathrm {D,\ min} = \mathrm {5\,mA} \\ I_\mathrm {L} = \mathrm {1\,A} - \mathrm {5\,mA} = \mathrm {0{,}995\,A} \\ R_\mathrm {L} = \frac {\mathrm {8\,V}}{\mathrm {0{,}995\,A}} \approx \mathrm {8\,\Omega } \Rightarrow R_\mathrm {L,\ min} \approx \mathrm {8\,\Omega } \end {gather*}

See section: ??

8 Einsatz einer Diode zur Leistungsreduzierung

In a simple circuit, a diode is used for power reduction. The resistor \(R_\mathrm {L} = 1,\mathrm {k\Omega }\) represents a heating load. The mains voltage is \(230,\mathrm {V}\) RMS at a frequency of \(50,\mathrm {Hz}\).

• What average power is dissipated in \(R_\mathrm {L}\) and in the diode if the diode acts as an ideal rectifier (i.e., only one half-wave is conducted)?

• What would be the average power if the diode were bypassed (full mains voltage applied to \(R_\mathrm {L}\))?

8.1 Lösung:

• In an ideal valve: Once the forward voltage is reached, the diode behaves like a perfect conductor without resistance.
  Since only one half-wave of the alternating voltage is passed through, the power in the load resistor is halved. The diode itself does not convert any power: \begin {align*} P_\mathrm {D} &= 0\,\mathrm {W} \end {align*}

  The average power in the load resistance is: \begin {align*} P_\mathrm {RL} &= \frac {1}{2} \cdot \frac {(U_\mathrm {eff})^2}{R_\mathrm {L}} = \frac {1}{2} \cdot \frac {(230\,\mathrm {V})^2}{1\,\mathrm {k\Omega }} = 26,45\,\mathrm {W} \end {align*}

• When the diode is bridged, the full AC voltage is applied to the load resistor: \begin {align*} P_\mathrm {RL} &= \frac {U_\mathrm {eff}^2}{R_\mathrm {L}} = \frac {230\,\mathrm {V}^2}{1\,\mathrm {k\Omega }} = 52,9\,\mathrm {W} \end {align*}

See section: ??

9 Voltage curve in a bridge rectifier circuit

Sketch the voltage \(U_1\) of the given circuit. Note that \(u_\mathrm {E}\) is an alternating voltage.

9.1 Lösung:

• In a negative half-wave, the current flows via \(D_{\text {2}}\) and \(D_{\text {3}}\).

• In a positive half-wave, the current flows through \(D_{\text {4}}\) and \(D_{\text {1}}\).

See section: ??

10 Calculating the series resistor for an LED

As with all diodes, the current must also be limited for light-emitting diodes (LEDs). For this purpose, a series resistor is connected in series, as shown in the circuit below.
The LED used has a forward voltage of \(U_\mathrm {D} = 2.2,\mathrm {V}\). The ideal brightness is achieved at a current of \(I_\mathrm {D} = 30,\mathrm {mA}\). The input voltage is \(U_\mathrm {0} = 5,\mathrm {V}\).

Which of the following resistor values would you use for \(R\) to operate the LED as close as possible to its optimal operating point (i.e., at optimal brightness)? \(15,\Omega \), \(33,\Omega \), \(68,\Omega \), \(150,\Omega \), \(180,\Omega \)

10.1 Lösung:

Calculation of the voltage across the series resistor: \begin {equation*} U_\mathrm {V} = U_\mathrm {0} - U_\mathrm {D} = 5\,\mathrm {V} - 2,2\,\mathrm {V} = 2,8\,\mathrm {V} \end {equation*}

Calculation of the optimum series resistor for \( I_\mathrm {D} = \mathrm {30\,mA} \): \begin {equation*} R_\mathrm {V} = \frac {U_\mathrm {V}}{I_\mathrm {D}} = \frac {2,8\,\mathrm {V}}{30\,\mathrm {mA}} = 93,3\,\Omega \end {equation*}

Calculation of the current when selecting a specific resistance value: \begin {equation*} I_\mathrm {D} = \frac {U_\mathrm {V}}{R} = \frac {2,8\,\mathrm {V}}{150\,\Omega } = 18,7\,\mathrm {mA} \end {equation*}

Calculation for two resistors in parallel connection with additional series resistance: \begin {gather*} R_\mathrm {ges} = \frac {150\,\Omega \cdot 180\,\Omega }{150\,\Omega + 180\,\Omega } + 15\,\Omega = 96,8\,\Omega \\ I_\mathrm {D} = \frac {2,8\,\mathrm {V}}{96,8\,\Omega } =28,9\,\mathrm {mA} \end {gather*}

Alternative series connection: \begin {equation*} R_\mathrm {ges} = 4 \cdot 15\,\Omega + 33\,\Omega = 93\,\Omega \end {equation*}

Fazit: A combined resistance consisting of \(4 \times 15\,\Omega + 33\,\Omega = 93\,\Omega \) or a parallel connection with \(150\,\Omega \,\|\, 180\,\Omega + 15\,\Omega = 96,8\,\Omega \) leads to a current close to the optimum value of \(30\,\mathrm {mA}\).
Recommended choice: Combination with total resistance close to \(93\,\Omega \).

See section: ??

11 Identification of a component based on its circuit symbol

What type of component is indicated by the following symbol?

11.1 Lösung:

The component shown is an NPN transistor. See: Section: ??

12 Determination of base-emitter voltage and base current

Given is the transistor circuit shown with the following parameters: \(U_0=\mathrm {5\,V}\) , \(U_L=\mathrm {15\,V}\), \(R_\mathrm {V}=\mathrm {100\,k\Omega }\) und \(R_\mathrm {V}=\mathrm {500\,\Omega }\).

• What is the approximate voltage \(U_\mathrm {BE}\) for a silicon transistor?

• How big is the current \(I_\mathrm {B}\)?

12.1 Lösung:

• This voltage is a material-related property of the silicon transistor and remains almost constant, regardless of the base voltage \( U_\text {B} \). An npn transistor has a typical base-emitter voltage in the conductive state of: \begin {equation*} U_\text {BE} \approx 0,6\,\text {V} \text { bis } 0,7\,\text {V} \end {equation*}

  Although \( U_\text {0} = 5\,\text {V} \), the transistor limits the voltage \( U_{BE} \) to approx. \( 0.6\,\text {V} \) as soon as the transistor is conductive. The excess voltage drop is compensated for by other currents and components in the circuit.

• \begin {align*} I_\text {B} = \frac {5\,\text {V} - 0,6\,\text {V}}{100\,\text {k}\Omega } = 44\,\mu \text {A} \end {align*}

See section: ?? und ??

13 Calculation of the resistances of a transistor circuit

The transistor circuit shown has the following parameters: \(U_0=\mathrm {5\,V}\) , \(U_\mathrm {R1}=\mathrm {0,6\,V}\), \(U_\mathrm {L}=\mathrm {15\,V}\), \(R_\mathrm {1}=\mathrm {300\,\Omega }\), \(R_\mathrm {2}=\mathrm {2\,k\Omega }\) und \(R_\mathrm {V}=\mathrm {500\,\Omega }\). In order for the silicon transistor shown to switch, it needs a base current of\(50 \, \mathrm {\mu A}\) to switch through.

• How large must the resistance \(R_\mathrm {V}\) be selected so that the transistor switches through?

• Assuming that the voltage drop across the transistor \(U_\mathrm {CE}\) is \(7,5 \, \mathrm {V}\), how large is the current through \(I_\mathrm {E}\)?

13.1 Lösung:

• \[ \frac {5\,\text {V}}{2\,\text {k}\Omega + 0,3\,\text {k}\Omega } = \frac {U_1}{300\,\Omega } \Rightarrow U_\text {1} = 300\,\Omega \cdot \frac {5\,\text {V}}{2\,\text {k}\Omega + 0,3\,\text {k}\Omega } = 0,65\,\text {V} \] \[ 5\,\text {V} = 0,65\, \text {V} + U_\text {RV} + 0,6\,\text {V} \] \[ U_\text {RV} = 5\,\text {V} - 0,65\,\text {V} - 0,6\,\text {V} = 3,75\,\text {V} \] \[ R_\text {V} = \frac {3,75\,\text {V}}{50\,\mu \text {A}} = 75\,\text {k}\Omega \]

• \[ I_\text {C} = \frac {15\,\text {V} - 7,5\,\text {V}}{500\,\Omega } = 15\,\text {mA} \] \[ I_\text {E} = I_\text {C} + I_\text {B} = 15\,\text {mA} + 50\,\mu \text {A} = 15,05\,\text {mA} \] \[ \frac {I_\text {C}}{I_\text {B}} = \beta = 300 \]

See section: ?? und ??

14 Switching behaviour of a transistor

Given is the transistor circuit shown with the following parameters: \(U_0=\mathrm {5\,V}\) und \(U_L=\mathrm {15\,V}\). In order for the silicon transistor shown to switch, it requires a base current of \( I_\mathrm {B}=\mathrm {50\,\mu A} \), to reach the conductive state.

• How large must the series resistor \( R_\mathrm {V} \) be selected so that the transistor switches on?

• The transistor has a current gain factor \( \beta = 180 \) and a collector resistance \( R_\mathrm {C} = \mathrm {200\,\Omega } \). How large are the collector current \( I_\mathrm {C} \) and the voltage drop between the collector and emitter? \( U_\mathrm {CE} \)?

14.1 Lösung:

• Calculation of the series resistance for the base current: \begin {equation*} R_\mathrm {V} = \frac {U}{I} = \frac {\mathrm {5\,V} - \mathrm {0{,}6\,V}}{\mathrm {50\,\mu A}} = \frac {\mathrm {4{,}4\,V}}{\mathrm {50\,\mu A}} = \mathrm {88\,k\Omega } \end {equation*}

• Calculation of the collector current with current amplification factor: \begin {gather*} \beta = \frac {I_\mathrm {C}}{I_\mathrm {B}} \\ I_\mathrm {C} = \beta \cdot I_\mathrm {B} = 180 \cdot \mathrm {50\,\mu A} = \mathrm {9\,mA} \end {gather*}

  Calculation of the voltage drop across the collector resistor: \begin {equation*} U_\mathrm {RC} = I_\mathrm {C} \cdot R_\mathrm {C} = \mathrm {9\,mA} \cdot \mathrm {200\,\Omega } = \mathrm {1{,}8\,V} \end {equation*}

  Calculation of the voltage between collector and emitter: \begin {equation*} U_\mathrm {CE} = \mathrm {15\,V} - \mathrm {1{,}8\,V} = \mathrm {13{,}2\,V} \end {equation*}

See section: ?? und ??

15 Transistor circuit with LED

Using a bipolar transistor, an LED is to be operated. For the LED to reach its optimal luminous intensity, a current of \(I_\mathrm {D}=30,\mathrm {mA}\) must flow through it.  The transistor used has a current gain of \(\beta = 50\) and, in the conducting state, a collector–emitter voltage of \(U_\mathrm {CE} = 0.3,\mathrm {V}\). In addition, the following parameters are given: \(R_1=3.3,\mathrm {k\Omega }\), \(R_2=270,\mathrm {\Omega }\), and \(U_\mathrm {L}=9,\mathrm {V}\).

What voltage \(U\) must be applied to the base for the transistor to conduct?

15.1 Lösung:

Calculation of the collector current with known voltage and load resistance: \begin {align*} I_\mathrm {C} & = \frac {U}{R}\\ I_\mathrm {C} & = \frac {9\,\mathrm {V} - 0,3\,\mathrm {V}}{270\,\Omega } = 32,2\,\mathrm {mA} \end {align*}

Calculation of the required base current using the current amplification factor \(\beta = 50\): \begin {align*} I_\mathrm {B} & = \frac {I_\mathrm {C}}{\beta }\\ I_\mathrm {B} & = \frac {32,2\,\mathrm {mA}}{50} = 0,644\,\mathrm {mA} \end {align*}

Voltage drop across the series resistor: \begin {align*} U_\mathrm {V} = 0,644\,\mathrm {mA} \cdot 3,3\,\mathrm {k\Omega } = 2,12\,\mathrm {V} \end {align*}

Total base voltage(incl. \( U_\mathrm {BE} \)): \begin {align*} U = U_\mathrm {V} + U_\mathrm {BE} = 2,12\,\mathrm {V} + 0,6\,\mathrm {V} = 2,72\,\mathrm {V} \end {align*}

Antwort: The base voltage must be \( \boxed {2{,}72\,\mathrm {V}} \) for the transistor to conduct.

See section: ?? und ??

16 Analysis of the operating range of a transistor

The current amplification factor of the silicon transistor used is \( \beta = 150\) and the supply voltage is \(U_\mathrm {V} = 15\,\mathrm {V}\). The circuit should be dimensioned so that the collector current is \(I_\mathrm {C} = 2\,\mathrm {mA}\).

• Dimension the resistor \(R_3\) so that, at an output voltage of \(U_\mathrm {A}=3\,\mathrm {V}\), the current through \(R_3\) is exactly \(75\%\) of the emitter current.

• Determine the resistances \(R_1\) and \(R_2\), assuming that the collector-base voltage \(U_\mathrm {BE} = 0.6\,\mathrm {V}\) and that approximately 9 times the base current \(I_\mathrm {B}\) flows through \(R_2\).

16.1 Lösung:

• Calculation of the base current using the current amplification factor: \begin {gather*} I_\mathrm {B} = \frac {I_\mathrm {C}}{\beta } = \frac {\mathrm {2\,mA}}{150} = \mathrm {13,3\,\mu A} \\ I_\mathrm {E} = I_\mathrm {C} + I_\mathrm {B} = \mathrm {2\,mA} + \mathrm {13,3\,\mu A} = \mathrm {2,013\,mA} \\ I_\mathrm {R_3} = 0,75 \cdot I_\mathrm {E} = 0,75 \cdot \mathrm {2{,}013\,mA} = \mathrm {1,51\,mA} \\ R_3 = \frac {\mathrm {3\,V}}{\mathrm {1,51\,mA}} = \mathrm {1,99\,k\Omega } \end {gather*}

• Calculation of the current through \( R_2 \): \begin {gather*} I_{R_2} = 9 \cdot I_\mathrm {B} = 9 \cdot \mathrm {13,3\,\mu A} = \mathrm {0,12\,mA} \\ R_2 = \frac {U_\mathrm {a} + U_\mathrm {BE}}{I_\mathrm {R_2}} = \frac {\mathrm {3\,V} + \mathrm {0,6\,V}}{\mathrm {0,12\,mA}} = \mathrm {30\,k\Omega } \end {gather*}

  Calculation of the current through \(R_1\), which is equal to \(I_\mathrm {R_2}\): \begin {gather*} R_1 = \frac {U_\mathrm {BE}}{I_\mathrm {R_1}} = \frac {0,6\,\mathrm {V}}{0,12\,\mathrm {mA}} = 5,6\,\mathrm {k\Omega } \end {gather*}

See section: ??

17 Calculations in transistor circuits

The following circuit with the output characteristic curve of a transistor is given.

• Wie groß ist der maximale Kollektorstrom des idealen Transistors (bei \(R_\mathrm {CE} = 0\,\Omega \)), wenn der Kollektorwiderstand \(R_\mathrm {C} = 7,5\,\Omega \) beträgt und die Versorgungsspannung \(U_\mathrm {V} = 3\,\mathrm {V}\) ist?

• What is the maximum collector current of the ideal transistor (for \(R_\mathrm {CE} = 0,\Omega \)) if the collector resistance is \(R_\mathrm {C} = 7.5,\Omega \) and the supply voltage is \(U_\mathrm {V} = 3,\mathrm {V}\)?

• For a new operating point, only \(1,7,\mathrm {V}\) now drops across the collector resistor \(R_\mathrm {C}\), while the base current is still \(2,\mathrm {mA}\). Determine the new resistance \(R_\mathrm {C}\).

17.1 Lösung:

• Calculation of the maximum collector current for an ideal transistor (\( R_\mathrm {CE} = 0 \)): \begin {equation*} I_\mathrm {C} = \frac {U_\mathrm {V}}{R_\mathrm {C}} = \frac {3\,\mathrm {V}}{7,5\,\Omega } = 400\,\mathrm {mA} \end {equation*}

• Determination of voltages and collector current according to characteristic curve field:

  \begin {gather*} U_\mathrm {CE} = 1,2\,\mathrm {V} \\ U_\mathrm {RC} = 1,8\,\mathrm {V} \\ I_\mathrm {C} = \frac {U_\mathrm {RC}}{R_\mathrm {C}} = \frac {1,8\,\mathrm {V}}{7,5\,\Omega } = 240\,\mathrm {mA} \end {gather*}

• Calculation of the new resistance value for the same \(I_\mathrm {C}\) and new voltage \(U_\mathrm {RC} = 1,7\,\mathrm {V}\): \begin {equation*} R_\mathrm {C} = \frac {U_\mathrm {RC}}{I_\mathrm {C}} = \frac {1,7\,\mathrm {V}}{240\,\mathrm {mA}} = 7,08\,\Omega \end {equation*}

See section: ??

18 Calculation of the current gain of a transistor circuit

The following circuit with the characteristic curve of the transistor is given. The supply voltage is \( U_\mathrm {V} = \mathrm {6\,V} \). The operating point is at half the operating voltage.

Determine the following values from the characteristic curve field:

• the pairs \( I_\mathrm {B} / U_\mathrm {BE} \) and \( I_\mathrm {C} / U_\mathrm {CE} \) at the operating point

• The current amplification \( \beta \)

18.1 Lösung:

The following values are given at the operating point: \begin {gather*} U_\mathrm {CC} = \mathrm {6\,V} \\ U_\mathrm {CE} = \mathrm {3\,V} \\ I_\mathrm {C} = \mathrm {6\,mA} \\ U_\mathrm {BE} = \mathrm {0{,}62\,V} \\ I_\mathrm {B} = \mathrm {20\,\mu A} \end {gather*}

Calculation of current amplification \( \beta \): \begin {equation*} \beta = \frac {I_\mathrm {C}}{I_\mathrm {B}} = \frac {\mathrm {6\,mA}}{\mathrm {20\,\mu A}} = \mathrm {300} \end {equation*}

See section: ??

19 Behaviour of a transistor under alternating voltage

What happens when a pure alternating voltage is applied to a transistor?

19.1 Lösung:

When a pure alternating voltage is applied to a transistor, continuous amplification does not occur.

This leads to the following effects:

• Distortions in the output signal, as only the positive half-waves are amplified.

• The negative half-wave is cut off, which is similar to rectification of the signal.

• There is no linear amplification, as the transistor switches between the blocking and saturation states.

See section: ??

20 Creation of a small-signal equivalent circuit diagram

Draw the small-signal equivalent circuit diagram of the given circuit.

20.1 Lösung:

See section: ?? und ??.

21 Creation of a small-signal equivalent circuit diagram

Draw the small-signal equivalent circuit diagram of the given circuit.

21.1 Lösung:

See section: ?? und ??.

22 Determination of the gate-source voltage of a field-effect transistor

At what source-gate voltage does a junction transistor conduct the most current?

22.1 Lösung:

As a rule, the amount is \(U_\mathrm {BE}=0,7\, \mathrm {V}\).

See section: ??

23 Investigation of current flow in the field-effect transistor (FET)

Where does most of the current flow in the transistor shown?

23.1 Lösung:

Between gate and source.

See section: ??

24 Interpretation of the characteristic curve of a junction transistor

Into which two areas can the output characteristic curve of the junction transistor be divided?

24.1 Lösung:

In saturation and resistance range.

See Section: ??

25 Detection of doping patterns in transistors

Without applying a voltage, the following distributions of p- and n-regions result. What type of transistor is this?

b) b)

25.1 Lösung:

• n-channel self-locking (Depletion type) See section: ??

• p-channel self-locking (Depletion type) See section: ??

• n-channel self-conducting (Enrichment type) See section: ??

• p-channel self-conducting (Enrichment type) See section: ??

26 Explanation of how CMOS technology works

What does the C in CMOS stand for?

• Distinction between enrichment and depletion types

• The use of JFETs and MOSFETs in a circuit

• Reflects the transfer characteristic curve

• Indicates the voltage sensitivity of MOSFETs

• Both p-channel and n-channel transistors are used in a circuit

26.1 Lösung:

• Both p-channel and n-channel transistors are used in a circuit.

See Section: ??

27 Transistor composite amplifier

Two npn bipolar transistors together form a low-noise amplifier for signals in the range below \(1,\mathrm {\mu V}\). The supply voltage is \(U_1 = 10,\mathrm {V}\). At the collector of transistor \(\mathrm {T_2}\), a voltage of \(4,\mathrm {V}\) with respect to ground is present. At the operating point of \(T_1\) and \(T_2\), a base–emitter voltage of \(U_\mathrm {BE} = 600,\mathrm {mV}\) can be assumed for each transistor. The collector currents \(\mathrm {I_C}\) are assumed to be much larger than the base currents \(I_\mathrm {B}\); therefore, to a good approximation, \(I_\mathrm {C} = I_\mathrm {E}\). In addition, the following parameters are given:

\(R_1 = 10\, \mathrm {k\Omega }\), \(R_2 = 10\, \mathrm {k\Omega }\), \(R_3 = 2\, \mathrm {k\Omega }\), \(R_4 = 300\, \mathrm {k\Omega }\), \(R_5 = 10\, \mathrm {M\Omega }\)
\(C_1 = 1\, \mathrm {\mu F}\), \(C_2 = 1\, \mathrm {\mu F}\)
\(U_1 = 10\, \mathrm {V}\), \(\hat {u}_\mathrm {E} = 0,1\, \mathrm {mV}\)

• What is the maximum current through \(R_2\) at the operating point (\(I_\mathrm {R2} = I_\mathrm {C,,T2}\))?

• What is the approximate current through \(R_3\) if \(R_4 \gg R_3\)?

• What is the voltage drop across \(R_3\)?

• What is the DC voltage at the base of \(T_2\) (\(U_\mathrm {B,,T2} = U_\mathrm {CE,,T1}\))?

• What is the current through \(R_1\) (\(I_\mathrm {R1} = I_\mathrm {C,,T1}\))?

• Explain the operating point stabilization of the entire circuit due to the negative feedback via \(R_4\).

• What is the transconductance \(S\) of \(T_1\) if additionally \(U_\mathrm {T} = 26,\mathrm {mV}\) is given?

• What is the voltage gain \(A_\mathrm {V} = \frac {\Delta U_\mathrm {CE}}{\Delta U_\mathrm {BE}}\) of the first transistor stage?

• What is the voltage gain \(A_\mathrm {V}\) of the second transistor stage, using the definition \(A_\mathrm {V,T2} = \frac {\Delta U_\mathrm {C}}{\Delta U_\mathrm {B}}\)? To solve this, the resistors \(R_2\) and \(R_3\) must also be taken into account.

• What is the voltage gain of the two stages \(T_1\) and \(T_2\) together?

• Draw the small-signal equivalent circuit of the entire circuit. In doing so, replace both transistors at the operating point by a controlled current source \(S = \beta \cdot i_\mathrm {B}\) and a differential input resistance \(r_\mathrm {BE}\).

27.1 Lösung:

• \begin {align*} I_\mathrm {C} & = \frac {6\,\mathrm {V}}{10\,\mathrm {k\Omega }} = 0,6\,\mathrm {mA} \end {align*}

• \begin {align*} I_\mathrm {C,T2} & = 0,6\,\mathrm {mA} \end {align*}

• \begin {align*} U_{\mathrm {R3}} & = 0,6\,\mathrm {mA} \cdot 2\,\mathrm {k\Omega } = 1,2\, \mathrm {V} \end {align*}

• \begin {align*} 1,2\, \mathrm {V} + 0,6\,\mathrm {V} = 1,8\,\mathrm {V} \end {align*}

• \begin {align*} I_\mathrm {C,T1} & = \frac {8,2\,\mathrm {V}}{10\,\mathrm {k\Omega }} = 0,82\,\mathrm {mA} \end {align*}

• An increase in \(I_\mathrm {C}\) leads to a rise in the voltage across \(R_4\), which in turn reduces the base voltage of \(T_\mathrm {1}\), thereby reducing \(I_\mathrm {C}\) again. The negative feedback stabilizes the operating point.

• \begin {align*} S & = \frac {I_\mathrm {C,T1}}{U_\mathrm {T}} = \frac {0,82\,\mathrm {mA}}{26\,\mathrm {mA}} = 31,54\, \frac {\mathrm {mA}}{\mathrm {V}} \end {align*}

• \begin {align*} A_\mathrm {V} & = S \cdot (-R_\mathrm {C}) = -315 \end {align*}

• \begin {align*} A_\mathrm {V} & = -\frac {R_\mathrm {C}}{R_\mathrm {E}} = \frac {10\,\mathrm {k\Omega }}{2\,\mathrm {k\Omega }} = -5 \end {align*}

• \begin {align*} -5 \cdot -315 = 1575 \approx 1600 \end {align*}

• Equivalent circuit diagram
  

See section: ??

28 Rectifier circuit

Here is a rectifier circuit with pn silicon diodes. The voltage source \(V_1\) supplies a sinusoidal \(50,\mathrm {Hz}\) AC voltage with an amplitude of \(5,\mathrm {V}\).

• What DC voltages occur across \(R_1\), \(R_2\) and \(R_3\)?

• What are the functions of capacitors \(C_1\) and \(C_2\)?

• What is the particular advantage of this rectifier circuit, e.g. for applications with operational amplifiers?

• Sketch another voltage doubler circuit in which the maximum positive DC voltage (output voltage) is available with respect to the ground potential of the supplying AC voltage \(V_1\).

28.1 Lösung:

• • \(U_\mathrm {R1} = 4.3 , \mathrm {V} \Rightarrow \) since \(0.7 ,\mathrm {V}\) drop across \(D_1\)
    

  • \(U_\mathrm {R2} = -4.3 , \mathrm {V} \Rightarrow \) since the same applies for the negative half-wave
    

  • \(U_\mathrm {R3} = 8.6 ,\mathrm {V} \Rightarrow \) since \(R_3\) takes up both voltage values
    

• The capacitors \(C_1\) and \(C_2\) have the function of smoothing the signals, both from the negative and positive half-wave.

• The advantage of this rectification is that both positive and negative voltages can be tapped.

• Circuit

See section: ??

29 Investigation of a simple rectifier

Analyse the following rectifier. The following parameters are also given:

\(\hat {u}_\mathrm {E} = 5\,\mathrm {V}\), \(R_\mathrm {L} = 1\,\mathrm {k\Omega }\) and a diode forward voltage of \(U_\mathrm {f} = 0,7\,\mathrm {V}\).

• Calculate the output voltage \(U_\mathrm {A}\).

• Determine the current \(I_\mathrm {L}\) flowing through the load resistor \(R_\mathrm {L}\).

29.1 Lösung:

• Calculation of the output voltage \(U_\mathrm {A}\)
  The given circuit consists of a diode in the forward direction and a load resistor. The output voltage \(U_\mathrm {A}\) corresponds to the input voltage minus the diode forward voltage: \[ U_\mathrm {A} = U_\mathrm {E} - U_\mathrm {f} \] Applying the values: \[ U_\mathrm {A} = 5\,\mathrm {V} - 0{,}7\,\mathrm {V} = 4{,}3\,\mathrm {V} \]

• Determination of the load current \(I_\mathrm {L}\)
  The current flowing through the load resistance \(R_\mathrm {L}\) is calculated according to Ohm’s law: \[ I_\mathrm {L} = \frac {U_\mathrm {A}}{R_\mathrm {L}} \] Applying the values: \[ I_\mathrm {L} = \frac {4{,}3\,\mathrm {V}}{1\,\mathrm {k\Omega }} = 4{,}3\,\mathrm {mA} \]

See section: ??

30 Analysis of a voltage doubler circuit

Analyse the function of the Delon circuit shown for voltage doubling and calculate the output voltage for different input voltages. The following parameters are also given:

\(U_\mathrm {E1,eff}=110\,\mathrm {V}\), \(U_\mathrm {E2,eff}=230\,\mathrm {V}\) and the diode forward voltage of \(U_\mathrm {f} = 0,7\,\mathrm {V}\).

• Explain the operating principle of the Delon circuit and how it doubles the input voltage.

• Calculate the theoretical no-load output voltage \(U_\mathrm {A}\) for the given input voltages.

• Discuss how the output voltage \(U_\mathrm {A}\) behaves under load conditions and which factors influence voltage stability.

30.1 Lösung:

• Functional principle of the Delon circuit
  The circuit operates as a cascade rectifier. During the positive half-wave of the alternating voltage, \(C_1\) charges via \(D_1\) to the input peak voltage \(\hat {U}_\mathrm {E}\). During the negative half-wave, \(C_2\) charges via \(D_2\). This adds the voltages of \(C_1\) and \(C_2\), resulting in a theoretical DC voltage of approximately \(2 \cdot \hat {U}_\mathrm {E}\) at the output.

• Calculation of the no-load output voltage \(U_\mathrm {A}\) The input voltage is given as the effective value \(U_\mathrm {E}\). The peak voltage is: \[ \hat {U}_\mathrm {E} = \sqrt {2} \cdot U_\mathrm {E} \] Since the circuit doubles the voltage, the open-circuit output voltage is: \[ U_\mathrm {A} = 2 \cdot (\hat {U}_\mathrm {E} - U_\mathrm {D}) = 2 \cdot (\sqrt {2} \cdot U_E - 0{,}7\,\mathrm {V}) \] For \( U_\mathrm {E} = 110\, \mathrm {V} \): \[ U_\mathrm {A} = 2 \cdot (\sqrt {2} \cdot 110\, \mathrm {V} - 0{,}7\, \mathrm {V}) \approx 308\, \mathrm {V} \] For \( U_\mathrm {E} = 230\,V \): \[ U_\mathrm {A} = 2 \cdot (\sqrt {2} \cdot 230\,\mathrm {V} - 0{,}7\, \mathrm {V}) \approx 644\, \mathrm {V} \]

• Behaviour under load conditions The output voltage remains high when idle. Under load, the voltage drops depending on the internal resistance of the capacitors, the load \(R_\text {load}\) and the diode losses. The voltage drop depends on the residual ripple, which is influenced by the charging and discharging times of the capacitors. A high load current reduces the voltage because the capacitors discharge more quickly.

  Factors affecting voltage:

  • Internal resistance of the diodes

  • Capacitance of the capacitors \(C\)

  • Load due to \(R_\text {Load}\)

  • Mains frequency (\(50, \text {Hz}\) or \(60, \text {Hz}\))

31 Basic behaviour of an NPN bipolar transistor

Analyse the basic behaviour of a bipolar transistor in a basic circuit. The npn transistor has a current gain factor of \(\beta = 100\). A base current of \(I_\mathrm {B} = 20\,\mathrm {\mu A}\) is assumed.

• Calculate the collector current \(I_\mathrm {C}\).

• Determine the collector-emitter voltage \(U_\mathrm {CE}\) when the collector resistance is \(R_\mathrm {4} = 1\,\mathrm {k\Omega }\) and the supply voltage \(U_\mathrm {2} = 12\,\mathrm {V}\) amounts to.

31.1 Lösung:

• Calculation of the collector current \(I_\mathrm {C}\) The collector current of a bipolar transistor is calculated using the following equation: \[ I_\mathrm {C} = \beta \cdot I_\mathrm {B} \] with the given values: \begin {equation*} I_\mathrm {C} = 100 \cdot 20\,\mathrm {\mu A} = 2\,\mathrm {mA} \end {equation*}

• Determination of the collector-emitter voltage \(U_\mathrm {CE}\) The collector-emitter voltage is calculated using the following equation: \[ U_\mathrm {CE} = U_\mathrm {CC} - I_\mathrm {C} \cdot R_\mathrm {4} \] Applying the values: \begin {equation*} U_\mathrm {CE} = 12\,\mathrm {V} - (2\,\mathrm {mA} \cdot 1\,\mathrm {k\Omega }) = 12\,\mathrm {V} - 2\,\mathrm {V} = 10\,\mathrm {V} \end {equation*}

See section: ??

32 Drawing and calculation of a collector circuit

• Draw the following circuit: Connect an NPN transistor with a base series resistor of \(R_\mathrm {B} = 4,7\,\mathrm {k\Omega }\), an emitter resistance of \(R_\mathrm {E} = 1\,\mathrm {k\Omega }\) und einer Spannungsversorgung von \(U_\mathrm {0} = 15\,\mathrm {V}\). Switch on a \(10 \,\mathrm {V}\)-Z-diode in reverse direction from the base to ground.
  Only use a power supply that supplies both the base and the collector.

• Calculate the emitter voltage \(U_\mathrm {E}\) and the emitter current. \(I_\mathrm {E}\).

• Calculate the power at the emitter resistor. \(P_\mathrm {RE}\).

32.1 Lösung:

• Circuit:

• Calculation of the emitter voltage \(U_\mathrm {E}\)
  The Zener diode fixes the base voltage of the transistor at \(U_\mathrm {Z} = 10\,\mathrm {V}\). The base-emitter voltage is \(U_{BE} = 0,7\,\mathrm {V}\). \begin {gather*} U_\mathrm {E} = U_\mathrm {Z} - U_\mathrm {BE} = 10\,\mathrm {V} - 0,7\,\mathrm {V} = 9,3\,\mathrm {V} \\ I_\mathrm {E} = \frac {U_\mathrm {E}}{R_\mathrm {E}} = \frac {9,3\,\mathrm {V}}{1\,\mathrm {k\Omega }} = 9,3\,\mathrm {mA} \end {gather*}

• Calculation of power at the emitter resistor \(P_{RE}\)
  \[ P_\mathrm {RE} = I_\mathrm {E}^2 \cdot R_\mathrm {E} = (9,3\,\mathrm {mA})^2 \cdot 1\,\mathrm {k\Omega } = 86,49\,\mathrm {mW} \]

See section: ??