IB: 2P5: Electrical Flashcards

Question

How can you increase the stability of the operating point of a BJT circuit? and what are the trade-offs?

Answer 1

* **Connect R1 to the collector rather than Vcc**. This increases the stability of the biasing as if hғᴇ is above nominal, Vᴄᴇ is lower which reduces Iʙ, giving a degree of self compensation. Similarly if hғᴇ is below nominal, Vᴄᴇ is higher which increases Iʙ * Despite increasing the stability of the biasing, the **small signal input resistance is reduced**. ## Footnote ignore changing values, just an example of a difference circuit

Answer 2

A BJT circuit that provides: * a gain close to 1 * a high input resistance * a low output resistance

Answer 3

* They both have a gain just below 1 * The output impedance of a source follower is not usually as low as that of an emitter follow * The input impedance of a source follower is far higher than that of an emitter follower

Answer 4

They are a circuit that uses an emitter (BJT) or a source (FET) coupled pair of identical transistors to amplify the difference between the voltages at the two separate inputs (v1 and v2)

Answer 5

**The input when v1 = v2.** If v1 and v2 increase together, the currents through R3 and R4 will rise and so will the current through Rs. The coltage at both FET sources rise, reducing the gate-source voltages of Q1 and Q2. As a result, the gain (output/input, which is v3/v1 or v4/v2) will be small.

Answer 6

**The input when v1 = -v2.** Increasing v1 increases the current through R3 and Q1 and the voltage v2 decreases, causing an equal fall in current through R4 and Q2. The current through Rs will reamin constant and the voltage across Rs will remain constant.

Answer 7

The CMRR is a measure of performance of a differential amplifier and is defined as: (Gain for differential mode signals) / (Gain for commmon mode signals)

Answer 8

As v1 = c2, the same changes are occuring on both sides of the circuit, hence we can use a "half circuit approach". **This involves splitting Rs into two parallel resistors, each of value 2Rs**. You can then analyse the SSM to find the common mode gain.

Answer 9

For differential signals the change in current through one transistor is matched by an equal and opposite change through the other one. Hence the current through Rs is constant, so vs is constant also (behaves like a DC source), therefore it is a short circuit straight to ground in the SSM. You can then analyse the SSM to find the differential mode gain.

Answer 10

The Zener diode Z has an essentially constant voltage drop over a range of currents, in this case, flowing through R. The current through R is set to be much greater than the base current of the transistor. If the voltage across the Zener diode is several times the base-emitter voltage (Vʙᴇ), the voltage across the emitter resistor Rᴇ will essentially be constant, even if there are small changes in Vʙᴇ. The emitter current is therefore constant, and as the base current is usually a very small fraction of the emitter current, the collector current Ic is constant also. **Therefore it draws a constant current from the differential amplifier.**

Answer 11

A way of providing a current source in an operational amplifier without using components such as a Zener diode

Answer 12

A current mirror can be used in a differential amplifier to draw a constant tail current. The current mirror can be used as a load for the long tailed pair instead of resistors R3 and R4. As well as providing a single output as is usually desired in operational amplifiers, combining the difference of the outputs from the two transistors increases the differential gain whilst greatly reducing the common mode gain, thereby increasing the CMRR.

Answer 13

* Stabilisation of gain - the gain depeds only on resistor values * Increased input resistance * Reduced output resistance * Increased bandwidth

Answer 14

**Loop gain = AB** * A = open-loop gain of the forward path * B = feedback factor (fraction of output fed back to the input)

Answer 15

* Infinite input resistance, hence I+ = I- = 0 * Zero output resistance * Open loop gain, A, is infinite * v+ = v-

Answer 16

The fractional change in gain is reduced by a factor (1+AB)

Answer 17

* Midband gain is reduced by a factor (1 + A₀B) * Low cut-off frequency is reduced from f₁ to f₁' = f₁ / (1 + A₀B) * High cut-off frequency is increased from f₂ to f₂' = f₂ (1 + A₀B) Feedback is used to sacrifice the gain in order to increase the bandwidth. Gain-bandwidth product stays constant.

Answer 18

A(f) = (low freq. cut off) x A₀ x (high freq. cut off)

Answer 19

The product of the open loop gain and the bandwidth, it remains constant when negative feedback is added

Answer 20

The input resistance is increased by a factor (1+AB) ## Footnote Note: For a current amplifier the result is opposite, it will decrease the input resistance

Answer 21

The output resistance is decreased by a factor (1+AB) ## Footnote Note: For a current amplifier the result is opposite, it will increase the output resistance

Answer 22

1. Finite but very large (and usually not well-defined) open-loop gain, A > 10⁵ 2. Finite but very large input resistance, Ri = 10⁶ - 10¹² Ω 3. Low output resistance, Ro = 10 - 100 Ω 4. Input bias currents 5. Input offset voltages

Answer 23

An amplifier used the isolate stages. They have close to: * A gain of 1 * Infinite input impedance * Zero output impedance Examples include: * The voltage follower (OpAmp) * Source Follower (FET) * Emitter Follower (BJT)

Answer 24

1. **Input Bias Currents** are required to set the operating points of the input tranistors within the OpAmp. Even if the input signal is zero, the bias currents can generate unwanted voltages, which are then amplified. 2. **Input Offset Voltages**. Applying zero volts to both inputs of the OpAmp does not produce zero output as the input transistors are not perfectly matched. The input offset voltage is the voltage needed at the input that is between the inverting and non-inverting input to drive the output to zero. 3. **Finite open loop gain** 4. **Finite input resistance** 5. **Non-zero output resistance**

Answer 25

The gain of an OpAmp falls with frequency due to stray capacitances (or gain compensation). This can be modelled as a first order low pass filter

Answer 26

In data book

Answer 27

You can use the circuits symmetry to simplify the input stages. Once you have determined the outputs (v1 and v4), you can analyse the following stage as a simple difference amplifier.

Answer 28

An amplifier with an input current and output voltage. The gain is an impedance.

Answer 29

A circuit that simulates an inductor, using an OpAmp, a capacitor, and resistors. It can be particularly useful where an inductor is required by the actual physical size of the component is not practical.

Answer 30

All transistors have (DC) currents flowing through them at all times (in order to establish their operating points). They have a maximum efficiency of 25%. They simply consist of a resistive load in the collector of a bipolar transistor.

Answer 31

In the complementary emitter follower each transistor conducts for half the cycle - this is known as class B operation. Higher efficiency and zero power dissipation are characteristics of this mode of operation. The problem is crossover distortion caused by a large deadzone. They have a maximum efficiency of 78% (π/4)

Answer 32

Operating transistors so that they conduct for more than half of the cycle but less than the whole cycle is called class AB operation. Class AB operation gives a compromise between linearity and power output and efficiency.

Answer 33

This can use an npn and a pnp BJT or an n-channel and a p-channel FET. With no input signal both transistors are non-conducting. As v1 becomes more positive T1 starts to conduct as an emitter follower, T2 remains non-conducting. As v1 becomes more negative T2 starts to conduct and behave as an emitter follower. T1 remains non-conducting. There is a large deadband between -0.7V and +0.7V where both transistors are off. Hence tehre is an absence of an output here, which is known as crossover distortion.

Answer 34

Bias the circuit to the edge of conduction in the state of zero input signal. A circuit which performs this uses the voltage drop across two diodes to set the bias for the two transistors. The voltage drop across the two diodes should provide the correct voltage to bias the two base-emitter junctions into conduction. THIS IS CLASS AB OPERATION.

Answer 35

An oscillator is a circuit that produces a periodic output signal (usually sine or square wave) without any external input signal. Output comes from dc power supplies. The signal is sustained by positive feedback. The frequency and amplitude are set by circuit components

IB: 2P5: Electrical Flashcards

(76 cards)