Q of LC circuit

Not before time.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

Oh, I hadn't thought about that. Is it like coupled (mechanical) oscillators? Where you get an 'avoided' crossing. Strong coupling, and a lot of loading is then going to 'play havoc' with the OP=92s high Q system.

My boss is playing around with a mechanical oscillator that's got a bit of an anharmonic term, (like a pendulum), so the resonance frequency depends of the amplitude too.

George H.

=A0 =A0 ...Jim Thompson

=A0 =A0| =A0 =A0mens =A0 =A0 |

=A0 | =A0 =A0 et =A0 =A0 =A0|

=A0|

=A0 =A0 =A0 |

ide quoted text -

this project makes no sense at all, but with any coupling between resonant circuits, the higher the Q the more fussy everything becomes about how close the tuning is.

Try this: 4.1KHz to 21.5KHZ bandwidth, entirely dependent on coupling.

Version 4 SHEET 1 880 680 WIRE -32 96 -64 96 WIRE 96 96 48 96 WIRE 144 96 96 96 WIRE 336 96 272 96 WIRE 448 96 336 96 WIRE 144 128 144 96 WIRE 272 128 272 96 WIRE 336 128 336 96 WIRE 96 144 96 96 WIRE -64 160 -64 96 WIRE 448 160 448 96 WIRE 96 224 96 208 WIRE 144 224 144 208 WIRE 144 224 96 224 WIRE -64 320 -64 240 WIRE 64 320 -64 320 WIRE 144 320 144 224 WIRE 144 320 64 320 WIRE 272 320 272 208 WIRE 272 320 144 320 WIRE 336 320 336 192 WIRE 336 320 272 320 WIRE 448 320 448 240 WIRE 448 320 336 320 WIRE 64 368 64 320 FLAG 64 368 0 FLAG 272 96 V1 SYMBOL ind2 128 112 R0 SYMATTR InstName L1 SYMATTR Value 1m SYMATTR SpiceLine Rser=0 SYMATTR Type ind SYMBOL cap 112 144 M0 SYMATTR InstName C1 SYMATTR Value 2.53303e-9 SYMBOL voltage -64 144 R0 SYMATTR InstName V1 SYMATTR Value AC 1 SYMBOL res 64 80 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName Rg SYMATTR Value 10k SYMBOL ind2 256 112 R0 SYMATTR InstName L2 SYMATTR Value 1m SYMATTR SpiceLine Rser=0 SYMATTR Type ind SYMBOL cap 320 128 R0 SYMATTR InstName C2 SYMATTR Value 2.53303e-9 SYMBOL res 432 144 R0 SYMATTR InstName R1 SYMATTR Value 10k TEXT 96 376 Left 2 !.ac dec 1000 50k 150k TEXT -80 408 Left 2 !.measure tmp max mag(V(V1))\n.measure BW trig mag(V(V1))=tmp/sqrt(2) rise=1 targ mag(V(V1))=tmp/sqrt(2) fall=last TEXT 176 112 Left 2 !K1 l1 l2 {K1} TEXT 80 392 Left 2 !.step param K1 0.01 0.2 0.02 TEXT 312 392 Left 2 !.probe V(V1)

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

Fred wrote: Try this: 4.1KHz to 21.5KHZ bandwidth, entirely dependent on coupling.

I did not understand the concept. Are you suggesting to use two LC circuits which will coupled each other via magetic field. or you made a change to my present helmholtz coil by adding an extra capacitor. Because one coil was out of tune in my schematic. please explain.

I have following questions

1. I am using following capacitors, any comment!

2) How can I lower the Q of the system?

3) I am tryning to get my system to resonate properly. That might cool down the transistors. In case if it did not work out than I will try using following options

1. use series LC circuit for driving. or try to get both helmholtz coils resonante with their separtae resonant capacitor. I am using parallel LC circuit as secondary or receiving coil with rectifying diodes.

1. Put big heat sinks on the transsitors and a Fan to cool it off. I am using the following transistors

My questions are

a. How big should be the heat sinks (need four of them) ? How can I calculate the size of the heat sink. I measured that the transistors get hot upto 45 C in half an hour. This transistor gets derated 20C /W over temparatures of 25C according to data sheet. And how big should be the size of the Fan . How to calculate the size of fan and CFM to keep the transistors between 20 and 25 C?

I am trying to measure the current in the primary.

thanks

jess

It was an illustration of the concept of coupling coefficient affecting bandwidth of two identical coupled tuned circuits, in response to George Herold's comments. Purely academic.

Add resistance in series or parallel with the inductors.

Won't do any good, it'd just look like a resistor at its resonant frequency.

or try to get both helmholtz coils

Better

I am using parallel LC

The secondary will affect the resonant frequency of the whole system. Take it away, and it will detune the primary coils.

I get the impression that you need to take a crash course in basic resonant circuit theory.

I have no idea how you are driving the Helmholz, or with what waveform. I guess I missed that bit.

Measured how? 45 C at the die itself won't hurt anything. 45 C at the heatsink surface will be much higher at the die. To be safe, the die should be temparatures of 25C according to data sheet. And how big should be the

Heatsinks are rated in degrees per watt, ie. its rise in temperature per watt sunk.

You need to know the thermal resistance from device die to its mounting base, the thermal resistance from mounting base to heatsink, including any insulting washer, and the thermal resistance of the heatsink to ambient. From that, you can calculate the rise in temperature of the die itself per watt dissipated. Thermal resistances add, same as electrical resistances in series.

A copy of the Aavid Thermalloy heatsink catalog might help you.

An ohm, or less, depending on current, in series with the primary, with a differential oscilloscope or AC meter known to be good at the frequency of interest across it.

This is all basic stuff. Spend some time with a few good textbooks.

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

This is how I am driving them

jess

You're exceeding the BVdss of the FETs.

What is the IC?

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

--
On the schematic, I noticed that you have 400VPP on the source of M3,
which is grounded.

How is that possible?

I didn't quite hit critical coupling on the steps. See if you can find it ;-)

I lent my Terman to a "friend". Never saw it again, the friend either. There are a few paperback on Amazon for silly money. Mine was hardback ;-(

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

Scriblo? (That's a handwritten typo)

--
"For a successful technology, reality must take precedence 
over public relations, for nature cannot be fooled."
                                       (Richard Feynman)

Sorry, it was a wrong schematic.

The correct schematic is as follows

400 volts peak to peak is acorss the capcitor measured by oscilloscope.

jess

If so, then you're capacitor has 7 amps through it and its rating is 3 amps (all according to you).

It's not hot?

Get rid of the RC snubbers. They do nothing for you other than waste power.

John S

It's not and you know it. She(He?) is very difficult to help, perhaps due to language or education or...

1. There are RC snubbers around the FETs which do nothing (other than, perhaps, moderate dv/dt).
2. The OP's conditions change constantly. Shown is a .08uF cap around one of the 50uH coils and the statement is that it is resonant at

100kHz. Not so. It takes about .1uF to do so (series resonant).

1. At series resonance, the only limiting factor is the so-called Helmholtz coil series resistance and the losses in the capacitor. So, why is the OP trying to dump many watts of power into the coil resistances?

2. You got me.

If you want an LTSpice simulation of what I think the OP has, let me know.

John S

Looking at that, it appears that you maybe driving the coil with square wave? Not a good idea with what you're doing. Especially at generating harmonics.

In your circuit it would appear, that cross over on the bridge is going to give you some side effects with out load on the coil, why? Because as the bridge switches, the out coming voltage from the coil will be working against your circuit due to the polarity of it and your circuit is going to not symmetrically quenching it.

To avoid these problems from what I can see, I would suggest that you place a load R across bridge/coil connections. This will give you a none inductive load and there for, wheeling voltages will not be elevating your bridge circuit and you should be getting a more solid transition. Otherwise, if you view that with the scope, I'll be you're getting distorted wave forms?

I don't know what your L is for the coil, but may I suggest you start with what ever XL is for the load R.

Also, Do not resonant this circuit, the R Load across the bridge should remove the parasitic problems you're having. You should have a rather nice looking sine wave on the other coupled coil. Make sure you are not saturating the coil to the DC R point before switching the bridge. If you are, you won't see nice looking waves ;)

Hope that helps.

Jamie

Something is very wrong there, You should not be getting no where near that voltage!!!!!!!!!!!

Place a 50 ohm 3 Watt R across the bridge. You are getting wheeling voltages due to the square wave and slow cross over you have. This voltage is not benefiting you at all.

When you probe at A and B, which is your bridge output, you should not see much more than 12 volts however, current will be in the coil.

Remove the coils and have only the R load there to make sure you do have a nice looking wave between A and B of your bridge. If you don't have a scope to check this at the moment, you can use your DMM meter on AC. You should be getting an RMS value from the 12VDC.

Also, I don't understand your use of two coils, one with a cap across it? I must be missing something.

Jamie

--
Can you elaborate on that for the moment and leave the "how to" for
later, please?

Some recent cell-phone designs use inductive coupling for charging, so you don't have to fiddle with plugging them in. ...Jim Thompson

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