Electronics Formulas
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Enter:
Enter Diameter of Coil = 2.378 inches
Enter:
Enter Wire Size = .0201 inches (See Table = 24 gage)
Enter:
Enter Inductance = Micro Henries Wanted = 603
Answer:
T = Number of Turns = 122.87
Answer:
Rejected Answer
Answer:
Number Under the Square Root Bracket
Formula used here:
Let b1 = (d*d)*(T*T)
Let b2 = (18*d + 40*W*T)
=> L = b1/b2 = Micro-Henries = Inductance
d = Coil Diameter (Inches)
W = Wire Diameter (Inches)(See Table Below)
This is a Quadradic Equation:
Solving for T = Number of Turns:
A = d*d
B = 40*W*L
C = 18*d*L
=> A*T*T - B*T - C = 0
=> Quadradic Formula:
T = (-B + SqRt(B*B - 4*A*C))/(2*A)
T = (-B - SqRt(B*B - 4*A*C))/(2*A)
The length of the coil used in the inductor
should be equal to 0.4 times the diameter of
the coil.
As shown in the equation, inductance of the
air-core inductor varies as the square of the
number of turns. Thus the value ‘W*T’ is
multiplied four times if the value of ‘T’ is
doubled. The value of ‘W*T’ is multiplied by
two if the value of ‘T’ is increased up to 40%.
W*T = Length of the Coil = (Wire-Diameter)*(Number-of-Turns)
LC Frequency Formula:
Let b1 = 2*Pi
Let b2 = SqRt(L*C)
=> F = 1/(b1*b2)
L = Henries
C = Farads
F = Hertz
Simplified Formula:
Let b1 = 159
Let b2 = SqRt(L*C)
=> F = b1/b2
L = Micro-Henries
C = Pico-Farads
F = Mega-Hertz
Enter:
L = Micro-Henries = 2 = Inductance of Coil
Enter:
C = 50 Pico-Farads = Capacitor
Answer:
F = 15.9 Mega-Hertz
Ohm's Laws:
E = I*R
P = I*E
P = (E*E)/R
P = (I*I)*R
P = Power = Watts
E = Volts
I = Amps = Current
R = Ohms = Resistance = Load
Enter:
Enter E = 9 Volts
Enter:
Enter R = 10 Ohms
Answer:
I = Current = .9 Amps
Answer:
P = Power = Watts = 8.1 Watts