review for elec 105 midterm exam #1 (fall 2001)
... resistance o test-source method for finding Thévenin (Norton) resistance o treatment of dependent vs. independent sources - concept of a signal; voltage and current signals - concept of a circuit “port” (pair of terminals) - input and output resistances of amplifiers, sources, and loads - distinguis ...
... resistance o test-source method for finding Thévenin (Norton) resistance o treatment of dependent vs. independent sources - concept of a signal; voltage and current signals - concept of a circuit “port” (pair of terminals) - input and output resistances of amplifiers, sources, and loads - distinguis ...
Supercapacitor technical guide
... Discharge method 1. Constant current charging 10mA/F to rated voltage. 2. Constant voltage applied for 5 minutes. 3. Constant current discharge at 10mA/F down to 0.1V C= I*(T2-T1) V1-V2 ...
... Discharge method 1. Constant current charging 10mA/F to rated voltage. 2. Constant voltage applied for 5 minutes. 3. Constant current discharge at 10mA/F down to 0.1V C= I*(T2-T1) V1-V2 ...
Increasing the Output Current from a Signal Generator
... Another option is to add another pair of emitter followers at the input to the emitter-follower stage. This can be used to reduce the distortion in the buffer stage itself. The concept (based on the National Semiconductor LM0002 integrated circuit buffer) is shown in figure 3(b). The input emitter f ...
... Another option is to add another pair of emitter followers at the input to the emitter-follower stage. This can be used to reduce the distortion in the buffer stage itself. The concept (based on the National Semiconductor LM0002 integrated circuit buffer) is shown in figure 3(b). The input emitter f ...
EUP3406 1.5MHz, 600mA Synchronous Step-Down Converter
... Chip Enable pin. Forcing this pin above 1.5V enables the part. Forcing this pin below 0.3V shuts down the device. Do not leave EN floating. ...
... Chip Enable pin. Forcing this pin above 1.5V enables the part. Forcing this pin below 0.3V shuts down the device. Do not leave EN floating. ...
The Field Effect Transistor
... The right value of resistor in the source circuit can lead to a good value of gate-source voltage. Choose a value of Rs to give the following circuit a good operating point. For a good operating point, the drain voltage is between 5 and 10 volts. Note that the AC signal on the input is not relevant ...
... The right value of resistor in the source circuit can lead to a good value of gate-source voltage. Choose a value of Rs to give the following circuit a good operating point. For a good operating point, the drain voltage is between 5 and 10 volts. Note that the AC signal on the input is not relevant ...
Rectifier filter capacitors
... Care is taken to determine that 'Engineering Notes' do not refer to any copyrighted or patented circuit or technique, but ISCE can accept no responsibility in this connection. Users of the information in an 'Engineering Note' must satisfy themselves that they do not infringe any Intellectual Propert ...
... Care is taken to determine that 'Engineering Notes' do not refer to any copyrighted or patented circuit or technique, but ISCE can accept no responsibility in this connection. Users of the information in an 'Engineering Note' must satisfy themselves that they do not infringe any Intellectual Propert ...
How to perform an AC simulation:
... You can use either VAC or VSRC as the input voltage source for this type of simulation. (I prefer to use VSRC). Note the input Q-point, VINQ, is entered as the DC parameter of the Voltage source. The AC parameter is set to 1V; this is a convenient set up for determining gain parameters. At first gla ...
... You can use either VAC or VSRC as the input voltage source for this type of simulation. (I prefer to use VSRC). Note the input Q-point, VINQ, is entered as the DC parameter of the Voltage source. The AC parameter is set to 1V; this is a convenient set up for determining gain parameters. At first gla ...
Lab3Questions
... These capacitors are just to decouple the voltages, and detract noise from the power supply. o Are these values critical or could 0.1 uF, 1,000 pF, 1 uF, etc. capacitors be used? These values could be anything. They are used to hold the voltage at a node to a specific value to detract from sudde ...
... These capacitors are just to decouple the voltages, and detract noise from the power supply. o Are these values critical or could 0.1 uF, 1,000 pF, 1 uF, etc. capacitors be used? These values could be anything. They are used to hold the voltage at a node to a specific value to detract from sudde ...
WRL2089.tmp
... Note that this gain is not equal to the short-circuit current gain Ais. The current gain Ai depends on the source and load resistances, as well as the amplifier parameters. Therefore, we must use a circuit model to determine current gain Ai . Although we can use either model, we will find it easier ...
... Note that this gain is not equal to the short-circuit current gain Ais. The current gain Ai depends on the source and load resistances, as well as the amplifier parameters. Therefore, we must use a circuit model to determine current gain Ai . Although we can use either model, we will find it easier ...
VOLTAGE TO CURRENT CONVERTER USING OP AMP
... With Grounded Load In this circuit,one terminal of the load is grounded and load current is controlled by an input voltage.The analysis of the circuit is accomplished by first determining the voltage V1 at the non inverting input terminal and then establishing the relationship between V1 and the ...
... With Grounded Load In this circuit,one terminal of the load is grounded and load current is controlled by an input voltage.The analysis of the circuit is accomplished by first determining the voltage V1 at the non inverting input terminal and then establishing the relationship between V1 and the ...
Differential Amplifier Model: Basic
... Inverting Amplifier Input and Output Resistances Rout is found by applying a test current (or voltage) source to the amplifier output and determining the voltage (or current) after turning off all independent sources. Hence, vs = 0 vx i R i R ...
... Inverting Amplifier Input and Output Resistances Rout is found by applying a test current (or voltage) source to the amplifier output and determining the voltage (or current) after turning off all independent sources. Hence, vs = 0 vx i R i R ...
Downlaod File
... charge flows from one end to the other. 2- what is Ohm’s Law? States that the current in a circuit varies in direct proportion to the potential difference, or voltage, and inversely with the resistance. 3- what is Electric Power? Rate at which electric energy is converted into another form. 4- What ...
... charge flows from one end to the other. 2- what is Ohm’s Law? States that the current in a circuit varies in direct proportion to the potential difference, or voltage, and inversely with the resistance. 3- what is Electric Power? Rate at which electric energy is converted into another form. 4- What ...
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.