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
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
Put a voltage divider between Vout and Vadjust pins. See page 9 of the datasheet from National's site.
Its 50-0.7=49.3V Using this Formula : PIV Rating = Vout - 0.7V
"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.
It can be answered in two ways : 1. ratio of output & input voltages [Vout / Vin] i.e Drain voltage(Vds)/Source voltage(Vs). 2. multiplication of trans-conductance & drain resistance .
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
de Coinci Gautier has written: 'C'est d'un moine qui vout retolir a une nonne une ymage de Nostre Dame que il li avoit aportee de Jherusalem'