A: I could tell you but it is better if you learn how to do it on your own look up datasheetlocator.com LM78XX . For these numbers there are SOT-223 and TO-252 and TO-220. These numbers specify packaging You probably have TO-220 in that case looking from the top left is adj vout vin. The metal can be Vout again it depends on the packaging use a VOM to ascertain the metal tab connection.
4.5 Volts
Put a voltage divider between Vout and Vadjust pins. See page 9 of the datasheet from National's site.
It should be ~180 degrees out of phase, because a CE amplifier is an inverter. A BJT CE amplifier is a good example to look at. The output is across CE, and at a minimum total output voltage is split across CE and some resistor R. As a higher voltage is applied to the base, the current flow through CE increases as a result of the resistance of CE decreasing. This boils down to a simple voltage divider at the output, Vout = CE / (R + CE). As CE decreases as a result of the input increasing, Vout will decrease.
Self biasind is called self because in this biasing, the variation due to change in temperature increases the the collector current, which hence decreases the output voltage i.e Vout=VCC-ICR and maintains the stability
dee-vow-t
I am assuming this configuration: -----------|Capacitor 2|----------|--------|------- | | Vin [R] [C1] Vout | | | | ------------------------------------------------------- 1. Using the Laplace transformer, C1 impedance equals 1/ sC1 (similar for C2). 2. Then treat C1 as you would a resistor to reduce the network: For two parallel resistors, the combined resistance = R1*R2 / (R1 + R2) Zparallel R and C1: R*1/sC1 / (R + 1 / sC1) = R / (RsC1 + 1) 3. Then to determine how the input varies from the output, Vout / Vin, using voltage divider Voltage divider for two resistors: Vout = Vin * R1 / (R1+R2) For the above: Vout = Vin * [R/RsC1 + 1] / ([R/RsC1 + 1] + 1/sC2), If you do a little manipulation, you will get: Vout = Vin * RsC1 / (RsC2 + RsC1 + 1)
The amplification factor Vout/Vin determines the voltage gain.
Most linear regulators absorb the amount of energy equal to (Vin-Vout)*I where Vin is the voltage into the regulator, Vout is the regulated output voltage, and I is the current flowing through the regulator. This energy is dissipated as heat. As to much heat will burn the device out, a heatsink is needed where the die of the device cannot adequately dissipate the waste heat into the atmosphere.
A: I could tell you but it is better if you learn how to do it on your own look up datasheetlocator.com LM78XX . For these numbers there are SOT-223 and TO-252 and TO-220. These numbers specify packaging You probably have TO-220 in that case looking from the top left is adj vout vin. The metal can be Vout again it depends on the packaging use a VOM to ascertain the metal tab connection.
4.5 Volts
The Acoelco vout GND 5V in the 2008 Chevrolet Colorado is part of the vehicle's electrical system, providing a regulated 5-volt output for various sensors and modules. It serves as a reference voltage for components such as the engine control unit (ECU) and other electronic devices that require a stable power supply for proper operation. This ensures accurate readings and reliable performance of the vehicle's systems.
Its 50-0.7=49.3V Using this Formula : PIV Rating = Vout - 0.7V
Put a voltage divider between Vout and Vadjust pins. See page 9 of the datasheet from National's site.
A voltage divider is an electrical circuit that divides the input voltage into smaller output voltages. It consists of two resistors connected in series. The output voltage is determined by the ratio of the two resistors. The formula for calculating the output voltage is Vout Vin (R2 / (R1 R2)), where Vin is the input voltage, R1 is the resistance of the first resistor, R2 is the resistance of the second resistor, and Vout is the output voltage.
"Vout" typically refers to the output voltage of a circuit or device. It is the voltage level that is generated or present at the output terminal or node of the circuit. Voltage output is a common measurement parameter in electronics and electrical engineering.
It should be ~180 degrees out of phase, because a CE amplifier is an inverter. A BJT CE amplifier is a good example to look at. The output is across CE, and at a minimum total output voltage is split across CE and some resistor R. As a higher voltage is applied to the base, the current flow through CE increases as a result of the resistance of CE decreasing. This boils down to a simple voltage divider at the output, Vout = CE / (R + CE). As CE decreases as a result of the input increasing, Vout will decrease.