Showing posts with label current. Show all posts
Showing posts with label current. Show all posts
Friday, May 17, 2013
High Current Regulated Supply
The High Current Regulated Supply beneath uses an added ambagious or a abstracted agent to accumulation ability for the LM317 regulator so that the canyon transistors can accomplish afterpiece to assimilation and advance efficiency. For acceptable ability the voltage at the collectors of the two alongside 2N3055 canyon transistors should be abutting to the achievement voltage. The LM317 requires a brace added volts on the ascribe side, additional the emitter/base bead of the 3055s, additional whatever is absent beyond the (0.1 ohm) equalizing resistors (1volt at 10 amps), so a abstracted agent and rectifier/filter ambit is acclimated that is a few volts college than the achievement voltage.
The LM317 will accommodate over 1 amp of accepted to drive the bases of the canyon transistors and assumming a accretion of 10 the aggregate should bear 15 amps or more. The LM317 consistently operates with a voltage aberration of 1.2 amid the achievement terminal and acclimation terminal and requires a minimum amount of 10mA, so a 75 ohm resistor was called which will draw (1.2/75 = 16mA). This aforementioned accepted flows through the emitter resistor of the 2N3904 which produces about a 1 volt bead beyond the 62 ohm resistor and 1.7 volts at the base. The achievement voltage is set with the voltage affiliate (1K/560) so that 1.7 volts is activated to the 3904 abject back the achievement is 5 volts. For 13 volt operation, the 1K resistor could be adapted to about 3.6K. The regulator has no achievement abbreviate ambit aegis so the achievement apparently should be fused.
The LM317 will accommodate over 1 amp of accepted to drive the bases of the canyon transistors and assumming a accretion of 10 the aggregate should bear 15 amps or more. The LM317 consistently operates with a voltage aberration of 1.2 amid the achievement terminal and acclimation terminal and requires a minimum amount of 10mA, so a 75 ohm resistor was called which will draw (1.2/75 = 16mA). This aforementioned accepted flows through the emitter resistor of the 2N3904 which produces about a 1 volt bead beyond the 62 ohm resistor and 1.7 volts at the base. The achievement voltage is set with the voltage affiliate (1K/560) so that 1.7 volts is activated to the 3904 abject back the achievement is 5 volts. For 13 volt operation, the 1K resistor could be adapted to about 3.6K. The regulator has no achievement abbreviate ambit aegis so the achievement apparently should be fused.
Tuesday, May 14, 2013
Calculation of Fault Current Using Impedance Diagram
Let us first illustrate the calculation of the fault current using the impedance diagram with the help of the following examples.
Example
Consider the power system of Fig. 6.8 in which a synchronous generator supplies a synchronous motor. The motor is operating at rated voltage and rated MVA while drawing a load current at a power factor of 0.9 (lagging) when a three phase symmetrical short circuit occurs at its terminals. We shall calculate the fault current that flow from both the generator and the motor.
We shall choose a base of 50 MVA, 20 kV in the circuit of the generator. Then the motor synchronous reactance is given by
 per unit |
Also the base impedance in the circuit of the transmission line is
 Ω |
 |
Fig. 1 generator supplying a motor load though a transmission line.
Therefore the impedance of the transmission line is
 per unit |
The impedance diagram for the circuit is shown in Fig. 6.9 in which the switch S indicates the fault.
 |
Fig. 2 Impedance diagram of the circuit of Fig. 6.8.
The motor draws a load current at rated voltage and rated MVA with 0.9 lagging power factor. Therefore
 per unit |
Then the subtransient voltages of the motor and the generator are
 per unit  per unit |
Hence the subtransient fault currents fed by the motor and the generator are
 per unit  per unit |
and the total current flowing to the fault is
 per unit |
Note that the base current in the circuit of the motor is
 A |
Therefore while the load current was 1603.8 A, the fault current is 7124.7 A.
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