Experiment # 1 The displayed Vpp on FG is half of the real Vpp because the displayed value represents the value with a load at maximum power transfer.

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Presentation transcript:

Experiment # 1 The displayed Vpp on FG is half of the real Vpp because the displayed value represents the value with a load at maximum power transfer. Experiment # 2 Maximum Power Transfer: Maximum Power Thevenin Equivalent: For RT=100Ω and VT=5V

Experiment # 3: Time constant and transient behavior of RC and LR circuits. For RC circuit: For LR circuit:

Experiment # 3: How to measure the current? The function generator is set for a square wave output with a range [+5v , -5v] at 160kHz. The value of the inductor is L=6.25 mH. In order to observe that the current reaches at least 99.3% (e^(-5)=0.7%) of the static value (fully established value), what is the minimum value of the resistor R we should choose? After 5 time constant, the transient value will reach the 99.3% of the static value (e-5=0.7%). In order to observe this transient behavior, we need one period of the square wave accommodates at least 10 time constants. How much is the time constant?

Experiment # 4: Diode model: If VD>>nVT, If VD<<nVT,

Experiment # 4: Load line:

Experiment # 5: Diode applications: voltage regulation with Zener diode For Zener diode, the load line equation is At Vz=0, Iz=-5mA; At Iz=0, Vz=-5V. We get two points on the line (0, -5) and (-5, 0). Connect these two points we obtain the load line. From the load line, we obtain the interception at Vz=-2V with the Zener diode characteristic curve, therefore, the current must be Therefore, the operation point is (-2V, -3mA)

Ideal op-amp conditions: Experiment # 6: basic properties of operational amplifiers Ideal op-amp conditions: Ip=In=0 No current into the terminals Ri=∞ Infinite input resistance Ro=0 Zero output resistance A ∞ Infinite open loop gain Ip=In=0 and Ri=∞ makes no power demands on the input signal source. Ro=0 makes the output voltage independent of the load . Even though the ideal op-amp model deviates much from the real op-amps, the ideal conditions in the ideal op-amp model are very useful in the design and analysis of circuits.

Inverting Amplifier— Ideal op-amp circuit analysis Non-Inverting Amplifier:

Real Op-Amp Frequency Response Real Op amps have a frequency dependant open loop gain Where

If the open loop bandwidth is so small, how can the op amp be useful? Real Op-Amp Frequency Response If the open loop bandwidth is so small, how can the op amp be useful?

Voltage Transfer Curve In reality A, Ri are finite and Ro>0 The output is limited by power source For a 741 op-amp powered with VCC= +10V and VEE= -10V, Vo will saturate (reach the maximum output voltage range) at about ±10 V. With an A=200,000V/V saturation occurs with an input differential voltage of 10/200,000 = 50μV, a very small voltage.

Experiment # 7: Filters Low-pass filters: At low frequency: At high frequency:

High-pass filters: At low frequency: At high frequency:

Band-pass filter: Center Frequency:

What happens if input square wave? Loading Active filters: Low pass filter Cascade Connections What happens if input square wave? Loading

Experiment # 8 & 9: BJT Common Base Current Gain Common Emitter Current Gain Common emitter circuit configuration

Early Voltage VA Output resistance Q: DC bias point (quiescent point) Common emitter configuration Load Line

Graphic Analysis Common emitter configuration Q: DC bias point (quiescent point) Load Line

Hybrid-∏ Model + - Output Resistance Input Resistance

T-Model Av<1. What’s the purpose?

Experiment #10: AM Transmitted signal: Envelope: Percentage of Modulation: For any message satisfies -1≤ m(t) ≤ +1: Over Modulation:

Experiment #10: AM Envelope Detector: Envelope: How to determine the RC time constant?

Experiment #11 For a coil around a magnetic core, the magnetic field intensity H inside the core is where l is the core mean length The allowed

Coupled Inductors where k is coupling coefficient and 0≤ k ≤1

Coupled Inductors Apply KVL:

Coupled Inductors

Ideal Transformer For k=1, where n is called turns ratio n=N2/N1

Ideal Transformer For an ideal transformer, it does not dissipate power, therefore the input power should equal the output power: Therefore, the input impedance for the transformer:

Non-Ideal Transformer For a non-ideal transformer, we don’t know whether V1 and I1 in phase or not:

Non-Ideal Transformer When ?

Non-Ideal Transformer When we have