Q of LC circuit

I connected the function generator (50 ohm) C =3D 80nF, L =3D 50uF, Resonant Freq. was found to be in the range of 70KHz to 72KHz. Sine wave with amplitude 10 Volts peak to peak was used. voltage across

50ohm resistor =3D 350mV peak to peak.

o----------o-------o-----o measure here | | C C C | | _____ | | C | C =3D=3D=3D C | | | '---o---o-----o and here | .-. | | 50 ohms | | '-' | --- GND

I connected the function generator (50 ohm), Sine wave with amplitude

10 Volts peak to peak was used. C =3D 80nF, L =3D 50uF, L=3D 50uF Resonant Freq. was found to be in the range of 52KHz to 53KHz. Voltage across 50 ohm resistor =3D 440mV peak to peak.

Than tried to get the same result using driving H bridge circuit but MOSFETS get hot again. No secondary coil was present.

=3D

Driven with H bridge. I used 1 Ohm resistor values of 10 Watts rating. The voltage acorss the resistors changes with time. It starts from 0.7 Volts and gets to 2 volts after 20 to 25 minutes. So, the maximum current is about 2A. But the MOSFETS do not get hot as they used to.

Above experiment showed that the 100KHz resonant frequency was the resonant frequency. But 53.5 KHz frequency showed the same results when used to drive the circuit using H bridge. I am thinking that the resonant circuit should be driven with a sine wave not a square wave generated by H bridge. Any comments.

jess

Sorry, I meant to say that the 100KHz is not the resonant frequency but 53.3 KHz did not show any signs of improvements either when used with H bridge driver. MOSFETS still get hot. jess

Sorry, I meant to say that the 100KHz is not a resonance frequency but

53.3KHz did not show the right results when used to drive H bridge.

jess

You seem to be ignoring a lot of what people are trying to tell you.

#1. Remove the COIL at the load of bridge and replace it with a 50 ohm R

#2. Perform a basic test on the bridge output to make sure you have the

+&- 12 volt alternating signal.

#3.

Leave the the 50 ohm R connected across the bridge at your output section and now couple the COIL to it. Take some more readings!

#4.

Do not make this circuit resonant!~!!!!!!!!!!!!!!!!!!!!!!!!!!!

#5. Operate the frequency well out side the self resonation point.

P.S.

With a 50 Ohm R only load, your MOSFETS should not be getting hot, not even warm!. If they are, fix your bridge before you go any farther.

Jamie

Been reading this thread and thinking. (Bad habit, i know.) But since you are using Helmholz coils you should resonate both coil halves with a single capacitor. The issue is that to have the renowned even field you have to have the same circulating currents in both coil halves. Then to reduce the Q add a small resistor between the coil halves and the drive point.

?-)

I suspect that it is used to produce low standby power, especially when the toothbrush is not in the cradle.

?-)

How did you measure the inductance? The coils are magnetically coupled.

Connect the two coils finish-to-start, in their normal position. Measure the total inductance of the series combination. Without altering their position, connect them start-to-start. Measure the total inductance of the series combination again. The difference in readings is four times the mutual inductance of the two coils, from which the coupling coefficient can be calculated.

Two magnetically coupled 50uH coils in series are not 100uH.

Hi,

Did you mean I should calculate the total inductance twice. First, when coils are just lying on top of each other or next to each other not mounted on the box and second time when the coils are mounted on the box approximately 7 inches apart. jess

NO NO NO...

Coils in position as mounted. NOT connected to anything else.

Connect them in series, with the start of one to the finish of the other (series-aiding).

*Measure* the total inductance of the pair. Don't try to use a resonance method, unless you have access to accurately known capacitors (1%).

Connect them, still in series, but with the start of one to the start of the other (series opposing).

Measure the total inductance again, it shouldn't be the same, if they couple magnetically.

In the first case, you have L1+L2+2M In the second case, you have L1+L2-2M Subtracting gives 4M M/sqrt(L1*L2) gives the coupling coefficient K.

Alternatively, it's possible to calculate the circuit constants from

*accurate* dimensions, number of turns, wire size, spacing, etc. using finite element analysis.

Once you know what the load looks like electrically, *then* it's time to think about how to drive it.

Please read that for a while.

Look for the subject on Mutual Inductance in transformers.

Also the whole guide as a whole maybe a good read for you, since it covers a lot of what you're doing.

In any case, putting 2 coils together in the same magnetic(B) flux stream is called mutual inductance and simply put, if one coil has moving currents it will be induced into the other and the effects will not be what you think as far as the value of induction.

For example, if you where to have a 2 winding coil on the same form tightly wound together to get maximum coupling efficiency you'll either get the effects of canceling out each other if connected together in parallel in opposite directions, there by producing what looks like 0 self induction, which leads to 0 ohms in the reaction or, if you connect them so they don't cancel, you'll get the self induction of one of the coils but not both together with half the value in mind, like a pair of R's would. This is because they are in the same magnetic B field and thus balanced out between the two of them, more or less. You can think of it as one coil inducing motor forces into the other coil and visversa. Each helping each other. They sort of equalize.

Another analogy to it is, take 2 DC generators, connect both shafts to the same driving force. Connect both outputs of each generator in parallel. Connect a load to the generator. Now spin the generators. The driving force that is spinning the generators will feel lets say 1 foot pound. while spinning the generators, disconnect one generator electrically going to the load and let it remain mechanically connected to the driving force.

Now the generator that is still connected to the load will now have to work twice as hard how ever, the load on the drive shaft for both generators is still the same and thus, it would appear that nothing has changed.

Now, connect these generators backwards in parallel. And now have their shafts driven. With this configuration, one is canceling out the other, the end results is, excessive drive load on the drive shafts, and in the world of induction. 0 henry effects.

: Back to Coils! :)

This only means is if you had 2,3,4 etc. coils all of the same value of induction when measured individually, put together in parallel, you still will only see the value of 1 and not any reduction of. This of course is in a ideal inductor..

Now, having two coils of the same source of signal and you start to pull them apart, you'll start getting different effects. More closely to what you were looking for. Also there is a point there where it may seem like they almost have a null notch when playing with the spacing and verifying L on the coils.

These little tricks have been used to make things like linear inductive potentiometers.

Please read the file I posted. It may take a little to get it sunk in but you don't need to understand all of it, just some of it. :)

Jamie

That's a Q of 105, at 100KHz, assuming R at 100KHz = R at DC (which it probably won't due to skin and other effects).

I assume you made an allowance for the resistance of your meter leads. 0.3 ohms really needs four-wire measurement.

On Wed, 21 Dec 2011 14:19:16 -0800, Fred Abse wrote:

Without knowing the coupling between coils, I brewed a FEMM model of two 2 inch, 25 turns of AWG22 coils, spaced 4 inches apart, and came up with the LTSpice suggestion below, driven by your FET bridge, minus the drain-to-source networks, (which don't seem to do much), using a series resonant arrangement.

About 8 amps RMS in the coils, and about half a watt dissipation per FET. No nasty voltages at the FETs.

Not knowing what the IC is, I modeled the drive as plain pulse voltage sources.

I noticed a "mic input" on your schematic. If you're going to modulate it, you might need to lower the Q a bit.

Since I don't know the dimensions of your Helmholz pair, this is all a complete guess.

*helmholz.asc Version 4 SHEET 1 1352 680 WIRE 320 -160 256 -160 WIRE 320 -128 320 -160 WIRE 256 16 256 -160 WIRE 256 16 -32 16 WIRE 592 16 256 16 WIRE -288 32 -352 32 WIRE 928 32 880 32 WIRE -32 48 -32 16 WIRE 592 48 592 16 WIRE 928 48 928 32 WIRE -352 64 -352 32 WIRE -224 64 -256 64 WIRE -128 64 -160 64 WIRE 736 64 704 64 WIRE 848 64 800 64 WIRE 304 112 176 112 WIRE 464 112 384 112 WIRE -288 128 -288 32 WIRE -256 128 -256 64 WIRE -256 128 -288 128 WIRE -240 128 -256 128 WIRE -128 128 -128 64 WIRE -128 128 -160 128 WIRE -80 128 -128 128 WIRE 304 128 304 112 WIRE 384 128 384 112 WIRE 704 128 704 64 WIRE 704 128 640 128 WIRE 736 128 704 128 WIRE 848 128 848 64 WIRE 848 128 816 128 WIRE 880 128 880 32 WIRE 880 128 848 128 WIRE -352 160 -352 144 WIRE -32 160 -32 144 WIRE -32 160 -352 160 WIRE 112 160 -32 160 WIRE 176 160 176 112 WIRE 464 160 464 112 WIRE 592 160 592 144 WIRE 592 160 464 160 WIRE 928 160 928 128 WIRE 928 160 592 160 WIRE -32 176 -32 160 WIRE 592 176 592 160 WIRE -32 208 -32 176 WIRE 592 208 592 176 WIRE -224 224 -256 224 WIRE -128 224 -160 224 WIRE 736 224 704 224 WIRE 848 224 800 224 WIRE 304 240 304 208 WIRE 384 240 384 208 WIRE 384 240 304 240 WIRE -256 288 -256 224 WIRE -256 288 -352 288 WIRE -240 288 -256 288 WIRE -128 288 -128 224 WIRE -128 288 -160 288 WIRE -80 288 -128 288 WIRE 704 288 704 224 WIRE 704 288 640 288 WIRE 736 288 704 288 WIRE 848 288 848 224 WIRE 848 288 816 288 WIRE 928 288 848 288 WIRE -352 304 -352 288 WIRE 928 304 928 288 WIRE -32 336 -32 304 WIRE 256 336 -32 336 WIRE 592 336 592 304 WIRE 592 336 256 336 WIRE 256 368 256 336 FLAG 256 368 0 FLAG 320 -48 0 FLAG -32 176 L IOPIN -32 176 BiDir FLAG 592 176 R IOPIN 592 176 BiDir FLAG 928 384 0 FLAG -352 384 0 SYMBOL subckt_nmos -80 48 R0 SYMATTR InstName U1 SYMATTR Value irfb4115pbf SYMBOL subckt_nmos -80 208 R0 SYMATTR InstName U2 SYMATTR Value irfb4115pbf SYMBOL subckt_nmos 640 48 M0 SYMATTR InstName U3 SYMATTR Value irfb4115pbf SYMBOL subckt_nmos 640 208 M0 SYMATTR InstName U4 SYMATTR Value irfb4115pbf SYMBOL voltage 320 -144 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V1 SYMATTR Value 12 SYMBOL voltage -352 48 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V2 SYMATTR Value PULSE(0 12 0 100n 100n 4u 10u) SYMBOL voltage 928 32 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 WINDOW 3 -103 -25 Left 2 SYMATTR InstName V3 SYMATTR Value PULSE(0 12 5u 100n 100n 4u 10u) SYMBOL res -144 112 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R1 SYMATTR Value 1.5 SYMBOL diode -160 48 R90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName D1 SYMATTR Value 1N4148 SYMBOL res -144 272 R90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R2 SYMATTR Value 1.5 SYMBOL diode -160 208 R90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName D2 SYMATTR Value 1N4148 SYMBOL res 720 112 M90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R3 SYMATTR Value 1.5 SYMBOL diode 736 48 M90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName D3 SYMATTR Value 1N4148 SYMBOL res 720 272 M90 WINDOW 0 0 56 VBottom 2 WINDOW 3 32 56 VTop 2 SYMATTR InstName R4 SYMATTR Value 1.5 SYMBOL diode 736 208 M90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName D4 SYMATTR Value 1N4148 SYMBOL ind2 288 112 R0 SYMATTR InstName L1 SYMATTR Value 53.62e-6 SYMATTR Type ind SYMATTR SpiceLine Rser=0.560812 SYMBOL ind2 368 224 M180 WINDOW 0 36 80 Left 2 WINDOW 3 36 40 Left 2 SYMATTR InstName L2 SYMATTR Value 53.62e-6 SYMATTR Type ind SYMATTR SpiceLine Rser=0.560812 SYMBOL cap 176 144 R90 WINDOW 0 0 32 VBottom 2 WINDOW 3 32 32 VTop 2 SYMATTR InstName C1 SYMATTR Value 23.0405e-9 SYMBOL voltage 928 288 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 WINDOW 3 -96 130 Left 2 SYMATTR InstName V4 SYMATTR Value PULSE(0 12 0 100n 100n 4u 10u) SYMBOL voltage -352 288 R0 WINDOW 123 0 0 Left 2 WINDOW 39 0 0 Left 2 SYMATTR InstName V5 SYMATTR Value PULSE(0 12 5u 100n 100n 4u 10u) TEXT 616 344 Left 2 !.lib irfb4115pbf.spi TEXT 616 368 Left 2 !.tran 0 800u 300u TEXT 280 264 Left 2 !K1 L1 L2 25.161320e-3 TEXT 168 48 Left 2 ;Derived from FEMM model of Helmholz pair of two\n2" dia. coils of 25 turns 22AWG wire, spaced\n4" apart. TEXT 64 216 Left 2 ;To resonate at 100kHz

.SUBCKT irfb4115pbf 1 2 3

• SPICE3 MODEL WITH THERMAL RC NETWORK

**************************************

• Model Generated by MODPEX *

*Copyright(c) Symmetry Design Systems*

• All Rights Reserved *
• UNPUBLISHED LICENSED SOFTWARE *
• Contains Proprietary Information *
• Which is The Property of *
• SYMMETRY OR ITS LICENSORS *

*Commercial Use or Resale Restricted *

• by Symmetry License Agreement *

**************************************

• Model generated on Apr 30, 10
• MODEL FORMAT: SPICE3
• Symmetry POWER MOS Model (Version 1.0)
• External Node Designations
• Node 1 -> Drain
• Node 2 -> Gate
• Node 3 -> Source M1 9 7 8 8 MM L=100u W=100u .MODEL MM NMOS LEVEL=1 IS=1e-32

+VTO=5.44978 LAMBDA=10 KP=1000 +CGSO=4.63102e-05 CGDO=6.88156e-07 RS 8 3 0.0101692 D1 3 1 MD .MODEL MD D IS=7.26144e-08 RS=0.00468718 N=1.30545 BV=150 +IBV=0.00025 EG=1 XTI=1 TT=1e-07 +CJO=5.9001e-09 VJ=0.5 M=0.597926 FC=0.5 RDS 3 1 1e+07 RD 9 1 0.0001 RG 2 7 6.86555 D2 4 5 MD1

• Default values used in MD1:
• RS=0 EG=1.11 XTI=3.0 TT=0
• BV=infinite IBV=1mA .MODEL MD1 D IS=1e-32 N=50

+CJO=1.95946e-09 VJ=0.500143 M=0.899735 FC=1e-08 D3 0 5 MD2

• Default values used in MD2:
• EG=1.11 XTI=3.0 TT=0 CJO=0
• BV=infinite IBV=1mA .MODEL MD2 D IS=1e-10 N=0.4006 RS=3e-06 RL 5 10 1 FI2 7 9 VFI2 -1 VFI2 4 0 0 EV16 10 0 9 7 1 CAP 11 10 1.97021e-09 FI1 7 9 VFI1 -1 VFI1 11 6 0 RCAP 6 10 1 D4 0 6 MD3
• Default values used in MD3:
• EG=1.11 XTI=3.0 TT=0 CJO=0
• RS=0 BV=infinite IBV=1mA .MODEL MD3 D IS=1e-10 N=0.4006 .ENDS irfb4115pbf

*SPICE Thermal Model Subcircuit .SUBCKT irfb4115pbft 4 1

R_RTHERM1 4 3 0.050025 R_RTHERM2 3 2 0.146073 R_RTHERM3 2 1 0.204102 C_CTHERM1 4 1 0.001039480 C_CTHERM2 3 1 0.003203877 C_CTHERM3 2 1 0.023037495

.ENDS irfb4115pbft

*helmholz.plt [Transient Analysis] { Npanes: 3 Active Pane: 2 { traces: 1 {34603010,0,"I(L2)"} X: ('u',0,0,5e-005,0.0005) Y[0]: (' ',0,-12,2,12) Y[1]: ('K',1,1e+308,100,-1e+308) Amps: (' ',0,0,0,-12,2,12) Log: 0 0 0 GridStyle: 1 }, { traces: 2 {524291,0,"V(N001)*Ix(U1:D)+V(N003)*Ix(U1:G)+V(L)*Ix(U1:S)"} {524292,0,"V(L)*Ix(U2:D)+V(N007)*Ix(U2:G)"} X: ('u',0,0,5e-005,0.0005) Y[0]: ('K',1,-100,100,1500) Y[1]: ('K',1,1e+308,100,-1e+308) Units: "W" ('K',0,0,1,-100,100,1500) Log: 0 0 0 GridStyle: 1 }, { traces: 2 {524293,0,"V(N001)*Ix(U3:D)+V(N004)*Ix(U3:G)+V(R)*Ix(U3:S)"} {524292,0,"V(L)*Ix(U2:D)+V(N007)*Ix(U2:G)"} X: ('u',0,0,5e-005,0.0005) Y[0]: ('K',1,-100,100,1500) Y[1]: ('K',1,1e+308,100,-1e+308) Units: "W" ('K',0,0,1,-100,100,1500) Log: 0 0 0 GridStyle: 1 } }

I don't see how that's possible, Fred, unless the source voltage gets pumped up, and some physical laws makes me think that's not happening.

Any stored energy is returned to the source via the MOSFET body diodes. She doesn't need the snubbers because everything is clamped to the rails. Her voltage measurement is at some weird point in her load.

She's trying to drive about an ohm of loss in the coils and capacitor with 12Vp-p. Its gonna take some current.

I don't see, either, but that's what she wrote.

The IC is HIP4081A.

jess

Hi,

I can not open your file in LTspice. LTSpice is givingme errors that it does not have the models for the MOSFETS.

jess

No model available.

Take a look at the circuit I posted.

If I knew the dimensions of the coil assembly, I could home in on a more accurate model.

The model was included in the posting: .SUBCKT irfb4115pbf 1 2 3 et seq.

You'll need this symbol file as well as the model LTSpice's native NMOS symbol only works with models, not subcircuits.

Put it where your LTSpice keeps its symbols (/lib/sym probably)

*subckt_nmos.asy

Version 4 SymbolType CELL LINE Normal 48 48 48 96 LINE Normal 16 80 48 80 LINE Normal 40 48 48 48 LINE Normal 16 48 40 44 LINE Normal 16 48 40 52 LINE Normal 40 44 40 52 LINE Normal 16 8 16 24 LINE Normal 16 40 16 56 LINE Normal 16 72 16 88 LINE Normal 0 80 8 80 LINE Normal 8 16 8 80 LINE Normal 48 16 16 16 LINE Normal 48 0 48 16 WINDOW 0 56 32 Left 0 WINDOW 3 56 72 Left 0 SYMATTR Value NMOS SYMATTR Prefix XM PIN 48 0 NONE 0 PINATTR PinName D PINATTR SpiceOrder 1 PIN 0 80 NONE 0 PINATTR PinName G PINATTR SpiceOrder 2 PIN 48 96 NONE 0 PINATTR PinName S PINATTR SpiceOrder 3