Calculate delay in long line

I found this calculator on the web. It seems to use the correct formula and that is: -

Force exerted = \$(NI)^2 \times \mu_0 \times\dfrac{A}{2g^2}\$

Where

• N is number of turns
• I is amps
• A is cross sectional area of coil
• \$\mu_0\$ is permeability = \$4\pi\times 10^{-7}\$
• g is the gap between the solenoid and the thing it attracts to

Note that the permeability of the material you wind it on does produce more flux density but the permability of the gap is the totally dominant part of the equation and material permeability is not included because it is so very minor.

An an electrical analogy, having something with high conductance (ferrite) in series with something with low conductance (air) still means something with low conductivity i.e. air is dominant except when the gap becomes small as in gapped ferrite transformers or inductors (which you haven't got)

Doubling the turns or doubling the current produces 4 times the force but doubling the gap quarters the force. Doubling the diameter of the solenoid produces four times the area and hence 4 times the force too.

Increasing the wire diameter allows more current to flow (for a given power supply voltage) BUT only if the supply can give the extra current. Ideally, wind with silver wire (costly) thus keeping the resistance low and feed as much current in as you can. Make the coil diameter as big as you can too. As many turns as you can manage also.

Bifilar winding won't help - this will be a dc solenoid with maybe a low frequency excitation and skin effect will be almost negligible - better to use two wires in series because N doubles rather than split current into two parallel paths (in effect this is just one turn still).

JohnD is correct: the LDR and R1 need to be interchanged for the circuit to work as you want. Assuming you interchange the LDR and R1 (the LDR should be connected from the transistor base to emitter), here's a way to calculate the required value of R1:

Let's say the the LED should turn off for a condition somewhere in between Case3 (LDR= 8.6kΩ) and Case4 (LDR= 2.5kΩ). Let's choose LDR=5kΩ as a suitable condition. For this condition, the base voltage of the transistor should be approximately at the turn-on Vbe (about .6V from the 2N3904 datasheet available here http://www.jameco.com/Jameco/Products/ProdDS/783421.pdf). So Vout= .6 = 5* LDR/(LDR+R1) = 5* 5k/(5k+R1). Solving for R1 gives R1~36kΩ.

Next we should check how much current can be supplied to the LED in full darkness: In full darkness, the LDR resistance is very high, and approximately an open circuit. R1 will therefore supply the entire base current, calculated as Ib = (5-.6)/R1 = 122 µA. With a transistor gain of ~70, this would result in a maximum collector current through the LED of app 8.5 mA. This may be enough for a nightlight LED. If you find the LED is not bright enough, then using two transistors in a Darlington arrangement would provide a much higher current gain and thereby allow more LED current.