coilinductancemeasurementohms-lawresistance
I have a hand-wound coil I'm using in an LC Tank circuit. Using my multi-meter, I found out that the coil is drawing .6mA at 9v. Using Ohm's Law…
$$R = V/I$$
$$R = 9/.0006$$
$$R = 15kΩ$$
Is there a way to calculate the inductance of my coil from the DC resistance?
There is a web-calculator for this. And here, you can find an excellent paper about methods of calculation for different shapes of coils.
In your circuit you are using the tank as the load on the op-amp buffer circuit. The op-amp is trying to produce 4.5 V, but its output is short-circuited by the inductor.
To generate oscillations in a linear circuit, you need to meet the Barkhausen criterion:
The usual way to use a tank circuit to set the frequency of an oscillator is to put it in the feedback loop where it can cause the Barkhausen criterion to be met at (or very near) its resonance frequency.
In your circuit, the tank is not involved in feedback, so it can have no influence on oscillation.
A relatively similar circuit that can oscillate is shown here:
(source)
Notes:
R1 prevents the op-amp output being short-circuited at DC.
The tank forms a divider with R1 so that positive feedback is very low except when the tank is at resonance.
Best Answer
An ideal coil has an inductance \$L = mew \times turn\text_density^2 \times Area \$.
The resistance of your coil is \$R = \rho \times \frac{length\text_wire}{cross\text_section\text_area\text_of\text_wire}\$. This can be simplified to \$R = constant\times length\text_wire\$
Naturally the if the coil has a greater turn_density or area, then the length of wire would be longer, so they do go up together, but the inductance is highly dependent on the geometry of the coil, so you are probably better off trying to calculate the inductance with the equation.
For instance if you increased the length of the coil while keeping the same density and area, the resistance would go up, but the inductance would not. So they cannot be related