Lecture-3: Transistors - Dr. Imtiaz Hussain
... – Cutoff region (for digital circuit) – Saturation region (for digital circuit) – Linear (active) region (to be an amplifier) – Breakdown region (always be a disaster) ...
... – Cutoff region (for digital circuit) – Saturation region (for digital circuit) – Linear (active) region (to be an amplifier) – Breakdown region (always be a disaster) ...
KIRCHOFF`S VOLTAGE LAW: EXAMPLE 1
... The voltage drops across both resistors were equal even though the currents were different. The voltage drop is ALWAYS the same across two resistors in parallel. Notice that IR1 + IR2 = I. This means that current is conserved. We will learn later that this is an application of Kirchhoff’s Current ...
... The voltage drops across both resistors were equal even though the currents were different. The voltage drop is ALWAYS the same across two resistors in parallel. Notice that IR1 + IR2 = I. This means that current is conserved. We will learn later that this is an application of Kirchhoff’s Current ...
0 - 30 v Adjustable voltage, current stabilized voltage supply
... (inverse Clockwise rotation To the minimum).Adjust the RV1 to make output voltage as 0V(may appear negative voltage and the value is very small,please use Digital multimeter do this).The maximum output voltage no need to adjust,it is about 33V when the input voltage is AC 24V. 2.Current Calibration ...
... (inverse Clockwise rotation To the minimum).Adjust the RV1 to make output voltage as 0V(may appear negative voltage and the value is very small,please use Digital multimeter do this).The maximum output voltage no need to adjust,it is about 33V when the input voltage is AC 24V. 2.Current Calibration ...
CIRCUIT IDEAS FOR DESIGNERS Current Source/Current Sink
... to the positive supply the circuit will act as a current source with node B as the output. When node B is connected to ground the circuit will act as a current sink with node A as the input. The current is set by the resistor R. If the current has a tendency to rise, the drop across R increases whic ...
... to the positive supply the circuit will act as a current source with node B as the output. When node B is connected to ground the circuit will act as a current sink with node A as the input. The current is set by the resistor R. If the current has a tendency to rise, the drop across R increases whic ...
FET Current Mirrors
... and are almost impossible to fabricate on an integrated circuit. • Instead, current mirrors are fabricated. ▫ These are circuits that contain two or more FETs, where the drain of one of the FETs is connected to the rest of the circuit. ▫ This FET is operating in the saturation/pinch-off mode. Thus, ...
... and are almost impossible to fabricate on an integrated circuit. • Instead, current mirrors are fabricated. ▫ These are circuits that contain two or more FETs, where the drain of one of the FETs is connected to the rest of the circuit. ▫ This FET is operating in the saturation/pinch-off mode. Thus, ...
KIRCHOFF`S VOLTAGE LAW: EXAMPLE 2
... (a) First, we identify the loops in the circuit. As shown below, we can choose any two of the three loops. ...
... (a) First, we identify the loops in the circuit. As shown below, we can choose any two of the three loops. ...
The Input Bias Current
... Example: The Input Bias Current Q: How do input bias currents I B 1 and IB 2 affect amplifier operation? A: Consider both inverting and non-inverting configurations. Inverting Configuration ...
... Example: The Input Bias Current Q: How do input bias currents I B 1 and IB 2 affect amplifier operation? A: Consider both inverting and non-inverting configurations. Inverting Configuration ...
Reference Directions in Voltage and Current Division
... the current measured by the meter, i m . (a) Suppose v s = 15 V . Determine the value of the resistance R that causes the value of the current measured by the ammeter to be i m = 5 A. (b) Suppose v s = 15 V and R = 24 Ω. Determine the value of the current measured by the ammeter. (c) Suppose R = 24 ...
... the current measured by the meter, i m . (a) Suppose v s = 15 V . Determine the value of the resistance R that causes the value of the current measured by the ammeter to be i m = 5 A. (b) Suppose v s = 15 V and R = 24 Ω. Determine the value of the current measured by the ammeter. (c) Suppose R = 24 ...
AC circuits ch 23 S2017
... fields and store magnetic energy, just like capacitors with electric fields. 2. Inductors & capacitors are used in tuning circuits in selecting signals. 3. Inductive-loops are used to detect vehicles at traffic lights. ...
... fields and store magnetic energy, just like capacitors with electric fields. 2. Inductors & capacitors are used in tuning circuits in selecting signals. 3. Inductive-loops are used to detect vehicles at traffic lights. ...
Lab Guide - inst.eecs.berkeley.edu - University of California, Berkeley
... 1. First, calculate the value of the series resistor RS to control current through the LED diodes. RS is chosen such that the voltage (VCC-VDIODE) across RS results in a limited but sufficient current IDIODE through itself and the diodes. The design formula is RS = (VCC-VDIODE)/IDIODE. Use IDIODE=10 ...
... 1. First, calculate the value of the series resistor RS to control current through the LED diodes. RS is chosen such that the voltage (VCC-VDIODE) across RS results in a limited but sufficient current IDIODE through itself and the diodes. The design formula is RS = (VCC-VDIODE)/IDIODE. Use IDIODE=10 ...
OSCILLATOR, VERY LOW FREQUENCY - 0.1Hz
... The input is 12V.AC only and the output is 15Volt peak to peak at 5mA max. load. The starting of the oscillation is controlled by a small press button and the output is electrically protected against short circuit. Using small lamps or LEDs and resistors, the oscillator can be used to demonstrate ha ...
... The input is 12V.AC only and the output is 15Volt peak to peak at 5mA max. load. The starting of the oscillation is controlled by a small press button and the output is electrically protected against short circuit. Using small lamps or LEDs and resistors, the oscillator can be used to demonstrate ha ...
1080AF-AIFO-2.0
... 1. The slope of the RF output power vs. the input signal voltage curve shall be non-zero and positive at all points between 0 and 1.0 Volts input, inclusive. 2. Output power factory set to 2 W at 1 Volt input. Power stability less than 5% over the heat sink's ambient temperature range of 0-40° C, af ...
... 1. The slope of the RF output power vs. the input signal voltage curve shall be non-zero and positive at all points between 0 and 1.0 Volts input, inclusive. 2. Output power factory set to 2 W at 1 Volt input. Power stability less than 5% over the heat sink's ambient temperature range of 0-40° C, af ...
Wilson current mirror
A Wilson current mirror is a three-terminal circuit (Fig. 1) that accepts an input current at the input terminal and provides a ""mirrored"" current source or sink output at the output terminal. The mirrored current is a precise copy of the input current. It may be used as a Wilson current source by applying a constant bias current to the input branch as in Fig. 2. The circuit is named after George R. Wilson, an integrated circuit design engineer who worked for Tektronix. Wilson devised this configuration in 1967 when he and Barrie Gilbert challenged each other to find an improved current mirror overnight that would use only three transistors. Wilson won the challenge.