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clamped inductive load circuit

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Junior Member level 1
Jun 20, 2011
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i)what is the criteria needed for inductor, diode and resistor?
ii)what is the variation of input supply and threshold voltage affect the switching performance of the device

This circuit uses the bipolar transistor as a switch, and the transistor needs to be turned either fully "OFF" (cut-off) or fully "ON" (saturated).
In order for the base current to flow, the base input terminal must be made more positive than the emitter by increasing it above the 0.7 volts needed for a silicon device. The value of the base resistor determines how much input voltage is required and corresponding base current to switch the transistor fully "ON". When maximum collector current flows the transistor is saturated. With inductive loads a flywheel diode is placed across the load to dissipate the back EMF generated by the inductive load when the transistor switches "OFF" to protect the transistor from damage.
Voltage at that time is equal to inductance times the rate of change of the current (U=L*di/dt).

The diode needs to have at least as high a current rating as the load current, as when the circuit is switched off, the diode will initially handle the load current due to the inductance. Voltage needs to be at least as high as the operating voltage. Also, if the on-off duty cycle is high, the thermal conditions may require a large diode and heat sinking.
The input current of the base should be typically about 1/10-1/20 of the output current for operation as a switch to saturate the transistor and get a low Vce voltage drop.

The base current can be calculated using (Vin-0.7v)/Ibase so for example with 5v input voltage if you want 10mA base current (5v-0.7v)/0.01A = 430 ohm

i would like to simulate the circuit above in pspice in order to get the turn-off time... how to simulate it if the input voltage is 15V , the +12V is changed to +500V where current flowing though the inductor is 8A with base resistor 22 ohm in order to get the turn off time delay??

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