Why Does Inductive Kickback Voltage Vary in a Solenoid?

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SUMMARY

The discussion focuses on the calculation of inductive kickback voltage in a solenoid with a supply voltage of 12V, inductance of 8mH, and a current of 1A. The theoretical formula for inductive electromotive force (emf) is given as φ = -L (dI/dt), which indicates that the voltage surge is dependent on the rate of change of current. The participant observed voltage surges ranging from 80V to 200V, which exceed the calculated -8V due to an underestimation of the instantaneous rate of change of current. The discussion concludes that measurement instruments may capture peak signals, leading to higher observed voltages.

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  • Understanding of inductive circuits and solenoid behavior
  • Familiarity with the formula for inductive emf: φ = -L (dI/dt)
  • Knowledge of current measurement techniques and instrumentation
  • Basic principles of electrical engineering related to inductance
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  • Research methods to accurately measure instantaneous current changes in inductive circuits
  • Learn about peak voltage measurement techniques and their applications
  • Explore advanced solenoid design considerations for minimizing inductive kickback
  • Study the effects of different inductance values on kickback voltage in various circuits
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Electrical engineers, hobbyists working with solenoids, and anyone involved in circuit design and analysis of inductive components will benefit from this discussion.

arul005
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please help me with the theoretical calculation of inductive kick back voltage for a solenoid(supply voltage is 12V,L=8mH and current passing through is 1A)

i tried experimenting i got surges around 80-200V. why is it varying from time to time. please tell me a method to predict the exact surge without experimentation.
 
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The formula for the inductive emf is:

<br /> \mathcal{E} = -L \, \frac{d I}{d t}<br />

As you can see, it depend on the rate of change of the current in the circuit. Your data is insufficient to solve the problem.
 
current changes from 1A to 0A(in 1 ms) when i switch off the supply of 12 V


so according to the formula:


e=-(8*.001)*(1/.001)
=-8V

but the actual surge voltage is greater than 80V every time.
 
That's cause you estimated the rate with the average rate. The instantaneous rate of change might be 10 times higher at the instant when you read the voltage. Your instrument might be in a mode where it measures the peak signal.
 

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