Q of LC circuit

I believe so. I only skimmed the article, but it looks like everything is there. The exception is that you don't call out an adjustable capacitance; if this is so then you need to adjust your signal generator to the resonance frequency of the tank circuit, rather than adjusting the cap in the tank circuit to resonance.

(This will also inform you of the resonant frequency, which is a good thing)

Keep in mind that what you're doing is measuring the response of a filter, and determining the effective Q from that.

Find some other way to excite the coils, and measure amplitudes like they do.

For your purposes, it may be better to run the coils from the generator with a 50 ohm resistor to ground, and look at the current in the resistor as you vary the frequency. This will show you resonance (where the current hits a minimum) and it will show you what the current _is_ at resonance. Then you can go design things.

That depends on what's making them hot. Find out that, and fix it or deal with it.

I would start by measuring the current at the bottom of the half bridges, preferably on a two-channel scope. Measuring the current and the voltage on one half bridge will also be informative.

Either your current is just generally too high because you're loading it too much, or you have some typical switching amplifier problem like shoot- through or slow turn-off, or you just under-designed your amplifier for the load that's being presented to it -- these are vastly different problems with vastly different solutions, but they all show up as hot transistors.

That depends on what L1 is doing to the circuit. If it is dramatically lowering the Q then yes, it could be the problem. If it is changing the resonance away from your design value, then yes, that could be the problem. If not -- not.

Keep in mind that if you're going to shove something into those Heimholtz coils to test its response to a magnetic field, you're going to see a change -- perhaps a marked one -- in the coil set's resonant frequency. If this is going to mess you up, you need to consider how to fix the problem.

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Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

On a sunny day (Thu, 15 Dec 2011 09:39:48 -0800 (PST)) it happened Jessica Shaw wrote in :

I was under the impression that Helmholz coils needed DC. Those are used to cancels magnetic fields. There is no omega in those equations on the wikipedia page. Or did I mis something?>

What are you trying to do with it?

Did you mean 50 Ohm resistor in parallel to LC circuit? Can you show me in the schematic where to place the resistor.?

ge

Did you mean accross the drain and source of the NMOSFETS by putting a resistor? I do not have a current probe to measure the current. I think the internal diodes of the MOSFETS condct lots of current and thats why MOSFETS get hot. thanks

jess

signal in o----------o-------o-----o measure here | | C | C === C | | | '---o---o-----o and here | .-. | | 50 ohms | | '-' | --- GND

| | ||-+ ||-+ ||

L1 and L2 are each 1/2 of the Helmholtz pair? Is the capacitance added or is it the distributed capacitance of the coils?

You can hit the circuit with a step or a pulse and look at the ringing in the response to get a measure of the Q and resonant frequency. What are you trying to do? Do you want a high Q for lots of B field or low Q for quick changes?

For low Q I'd be happy to share my favorite coil 'trick'.

George H.

a

Helmholtz coils give you a uniform B field in the center. The inhomogeneity goes something like the (sample radius/ coil radius) to the fourth power. (In theory) They are not very efficient at turning current into magnetic field.

George H.

Capacitor is in parallel with the inductor. Its a capacitor added to the circuit.

I tired to calculate the Q of the circuit. At resonant frequency of

99.5 KHz , the amplitude across the Capacitor is 304 volts peak to peak t. I increased the frequency to 99.95 Khz and volts decreased to 210 volts peak to peak that is 0.707 x 99.5 KHz. Than I decreased the frequency to 99.15 KHz and the voltage drops to 210 volts peak t peak again which is equal to 0.707 x 210.

Q = 99.5 / ( 99.95 - 99.15) = 124.37 . I am little confused. Is this right?

jess

i think you have it how ever, for some reason i have .77 in mind instead of .707. I need to look this up for a better check.. Q = Fc / BW.. Bw being the factor of .77 below and above the Center frequency of peak resonance. Fc = Frequency of center/resonance.

If you're looking for Q between R and X then those are different maths.

P.S. Like I said, I may have the .77 factor incorrect. It could be the .707 and 1.414, like you're using..

Jamie

Yup. A Q over 100 is certainly possible. You can measure a few more points and then fit the curve, to get a better number... but that's hardly ever necessary.

Did you measure the DC resistance of the coil?

You can see how well that fit's with the Q=3DZ(inductor)/R... and then try and understand the difference.

George H.

s

s.

Jamie, the sqrt of 2 is 0.707..., the sine of 45 degree's. It's how the first Boeing 707 got it's number. (At least the myth I heard)

George H.

Actually the square root of 2 is 1.414 or so -- 0.707 is the approximate square root of 1/2.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

And since the OP was looking at the 3 dB points, 20*log(1.414) is ~~ 3 dB.

tm

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ttdesign.com- Hide quoted text -

Opps... Thanks Tim.

George H.

The OP should be looking for the 3dB _down_ points; 10 * log(1/2) ~= 20 * log(0.707) ~= -3dB.

Which is all mostly argument over terms -- as long as the OP knows what to do.

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

Hi,

I do not know what this number tells me about my system. Is it a good Q value or bad. Q is about how much energy stored divided by how much energy dissipiated in one cycle. How can this number tells me that how much energy my system has lost and stored.

jess

The R term is the real part of the reactance which will be DC resistance + loss due to proximity and skin effects (AC resistance).

I meant the number that I calculated. Q = 124.37

jess

If you don't know -- why are you measuring it?

The Q of your coil will _indirectly_ tell you how your impedance will change as you fall off from resonance. But it might be easier to just use the expression for the impedance of an RLC tank circuit.

What are you going to do with these Heimholtz coils? You realize that as soon as you put your cat or the neighbor's noisy kid or whatever it is in there that you want to subject to a 100kHz magnetic field, that the resonant frequency, the Q, and everything else is going to change -- yes?

--
My liberal friends think I'm a conservative kook.
My conservative friends think I'm a liberal kook.
Why am I not happy that they have found common ground?

Tim Wescott, Communications, Control, Circuits & Software
http://www.wescottdesign.com

Firstly, you are not measuring the Q of the coil, you are measuring the Q of the coil, damped by whatever resistance happens to be in parallel with it.

Classical methods of measuring Q seldom, if ever, (apart from an ancient Marconi dielectric test set), use a parallel circuit. The reason for this is, theoretically, you need an exciting source with infinite impedance, the impedance of the source affects the measured Q, and needs to be known accurately inorder to apply corrections.

Better to use a series circuit:

Internal Gen R ___ V1 -----------|___|--------- | | | C| | C|L | C| /+\ | ( )Generator | \-/ V2| | | | | | | | --- | --- C | | | | | | -------------------------

Pick a suitable value for C to resonate the inductor at the frequency of interest, as near as you can.

Adjust the generator frequency for maximum voltage across C.

The Q of the LC combination is then V2/V1

Thge value of R has no effect on V2/V1, but it might make V1 too low to measure. Any reactive component in the generator impedance will shift the resonant frerquency from the true one, but won't affect the indicated Q enough to worry.

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