DC to 100Hz Power Amp - Coil/Loop Driver

What is your ultimate goal? If it is "to generate the strongest possible radiated signal", perhaps you should use a coil with 'way fewer turns - that you could resonate.

Even if you're stuck with this coil, you need to give some hint of what constitutes a "best design".

HTH...

I assume you want a bridge drive, to get 48V peak across the coil.

Any dual power opamp would work. A couple were mentioned in a nearby thread. The max current will only be 38 mA, limited by the resistance.

If you reconfigure the winding into two coils of 600 turns each, in parallel, you get 4x the current and twice the field from your available voltage.

Or four windings of 300t each, 4x the field, 600 mA total.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

** No - it is very inductive, about 3H.

ail voltage to generate the strongest

** Contradicts your 5Vp-p OR is that the signal you have to amplify ? ** Really?

Try using the AC supply ( 120 or 240) for a test.

.... Phil

Interesting. How did you calculate that?

5Vpp is the input signal. I wanted to maximize the voltage out to increase the coil's EMF given the fixed coil specs.

That was straight Ohm's law. 0.4mA. Why do you believe not?

Kevin Foster

As I understand, bridge drive means current reversal. This means the polarity of the field alternates. Not sure if this is suitable, but I will check.

I had thought of that, but according to my source, the coil is supposed to resonate within the applied frequency range of DC to 100Hz.

As Phil pointed out, the inductance is actually quite high.

BTW how would an expert such as yourself go about winding this? 4000'

30ga. 1m dia.

Kevin Foster

Resonating a coil at DC will be *interesting* :)

piglet

** Google " inductance calculator".

** You must know a different Georg Ohm to me.

..... Phil

Kevin Foster prodded the keyboard with:

P= V X I (24 X 0.0004 = 0.0096W)

--
Best Regards: 
                      Baron.

I did it different, 48Vpp equals 17VRMS. P=V^2/r (17x17)/1,274 = 0.227 Watts. But! that doesn't add any inductive component, so power will me lower as frequency goes up. Unless the L is resonated. Mikek Mikek

It only reverses if you want it to. The point is that if you apply +24 to one end of the coil and -24 to the other, you get 48 volts.

If you don't need bipolar field, use one opamp but connect the other end onto one of the rails.

Your source is not making sense.

If you want higher field using the available +-24 supplies, I would have used a shorter length of heavier wire, fewer turns. It's ohmic limited now. Do the math before winding that giant coil!

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

!

! Yup, I generally match the coil to the power supply.

30V, 3A supply means about a 10 ohm coil. Maximum B field comes from current and number of turns... and then turn the design crank. (you often have to go back and change something...)

George H.

I think that the dominant parameter is how many pounds of copper and how much power supply you're willing to buy. After that, it's a matter of matching the coil resistance to the available power supply.

We don't know what the application is, so there's no point in speculating (or calculating) a lot.

--

John Larkin         Highland Technology, Inc 

lunatic fringe electronics

Generally, winding different sized wire into the same space results in a coil that's scaled in impedance, but has all of the same power properties. There's probably some magic with inter-winding capacitance, though.

At some point you stop worrying about whether your power supply is herky enough, and start worrying about whether your coil is going to melt.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com 

I'm looking for work -- see my website!

Whether or not there's an advantage to a constant-current drive depends on what you're trying to do. If you want the magnetic field to match the drive command, and if inter-winding capacitance is negligible, then a constant-current drive will achieve that for you, at the cost of needing more drive voltage at higher frequencies, and needing careful design so the thing remains stable.

--

Tim Wescott 
Wescott Design Services 
http://www.wescottdesign.com 

I'm looking for work -- see my website!

Auto front wheel axle and brake rotor, affixed to a plywood disk of 1m diameter. Vinyl gutter (with heat-gun assist) wrapped around the disk rim to form the spool. Drive with a crank pin on the disk, and foot pedal (just look up 'spinning wheel' on images.google.com to see the mechanism).

Turns counting is traditional, but if you want to use 4000' of wire, you'll need a wire-feed drum of 1' diameter on its own counter, to count out feet.

Correct, it'll appear inductive above 25 Hz, about j 2k at 100Hz. So with voltage drive his complex waveform will lose some magnitude above 50Hz or so. For applications like this I've grabbed a 50W Radio-Shack P.A. amplifier from the local store, and used the 70V tap on its output transformer. Hey, this provides a high enough output to let one linearize the coil with a small ballast resistor.

--
 Thanks, 
    - Win

According to the website below, 48V (see OP "+/-24VDC" rails) into 1.2K is 0.04A at 1.92W.

Kevin Foster

That's a great solution. But I hope you can answer a couple of questions.

Given the output transformer, would the output be linear (and 70V) across the low end, say between 1 and 30Hz.

Is it necessary to short any on-board caps to pass these frequencies?

How would I determine the correct value for the ballast resistor you suggest? If I decided to resonate the output coil with a parallel cap for higher output, does this still apply.

Kevin Foster

Which brings up a question re: heatgunning PVC vs. easier not to.

At ELF is there really much difference in terms of internal field topology between a circular loop and an octagonal one?

Kevin Foster