Practical implementation of an RLC circuit

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8dm7bz

Joined Jul 21, 2020
133

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BobaMosfet

Joined Jul 1, 2009
1,208
Hello,
I want to implement an RLC circuit with a 1.49mH coil (https://www.grupopremo.com/3dcc28-3d-coil-cube-emitter-for-vr-applications-395x395x386mm-/1982-3dcc28-a-0150j.html I used the x-winding). I used a 47nF cap and a 20 ohms resistor. I added 3 ohms to the 20 ohms because that is the DC resistance of the coil.
Now, the green sine wave in the simulation is the voltage across the capacitor. But as you can see, in my measurement the voltage is much lower.
Why is that ?

thanks,
8
When you actually grasp how an inductor and a capacitor behave in conjunction with the resistor, it will begin to make sense.
 

Thread Starter

8dm7bz

Joined Jul 21, 2020
133
Could you elaborate a little more please.

Maybe my question was a little wiffy formulated. What I'm asking is why isn't my implementation at least near the simulation I made ? I used the same values for the components
 

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8dm7bz

Joined Jul 21, 2020
133
The circuit I built is in the attachments. I used the MCP1402 to switch the current. Measuring the output of the two mosfet driver gives me indeed 10Vpp. So there is no problem there.

any ideas ?
 

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Papabravo

Joined Feb 24, 2006
14,410
Could you elaborate a little more please.

Maybe my question was a little wiffy formulated. What I'm asking is why isn't my implementation at least near the simulation I made ? I used the same values for the components
There could be lots of actual reasons, but the main one that comes to mind is that the simulation is probably using ideal components with no parasitic properties. If you sweep an inductor or a capacitor on a VNA you will see them tun into each other. That is a mind bending concept to get your head around. You try to build a circuit with components you think you understand and suddenly they are turning into each other. Your two drivers do not appear to be related to a voltage source and GND.
 
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Thread Starter

8dm7bz

Joined Jul 21, 2020
133
thank you @Papabravo for explaining that. I indeed didn't change to much with the simulation values of the inductor and capacitor. I kind of hoped that it wouldn't be that bad ^^.
But now I know, thanks
 

Papabravo

Joined Feb 24, 2006
14,410
Something was bothering me about this thread and my original answer. Taking a closer look at the simulation as opposed to the Eagle(?) schematic, I'm wondering if the scope measurement is not being taken where you think it is. You should be able to estimate the AC current flowing in the series circuit. The voltage across the capacitor will be the Capacitative Reactance times the series current. The difference between 35 Volts and 5 Volts is huge. In your LTspice simulation it looks like one squarewave trace is going between 0 and 5 volts while tho other one is going between -5 volts and +5 volts. Can you clarify what is happening.

Some other points of interest:
Resonant Frequency ≈ 19018 Hz
XL = XC ≈ 178 Ω
Q = 178 / 23 = 7.74, respectable but not overly large
BW ≈ 19018 / 7.74 = 2457
So your 20 kHz excitation frequency is well within the bandwidth of the circuit
If we assume that the 5 Volts appears across the resistor, that current would be 217 mA. If that current appears across a reactance of 178 Ω, you would get about 38.7 volts.

So I think your original query had some substance to it that cannot be easily dismissed. I apologize for not looking at the whole post, and responding to just the portion that I saw,
 

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8dm7bz

Joined Jul 21, 2020
133
Thanks for looking further into this. First of, yes it's eagle. And what's happening is: I get a 20kHz square wave from 0V to 5V from the AD9833. With that signal I drive one MCP1402. Then I take the same signal and put it through an inverter (74HC04) and drive the second MCP1402 with that. This way I get -5V and +5V (10V in total). In the simulation I only displayed one of the 0V to 5V waves (blue simulation wave) and then the resulting wave (red one, which is behind the blue one: -5V to +5V).

Yes I did the same calculations, that's why I am so confused.

Maybe it's of interest that I take the 5V rail (for the MCP's and the power for every component) from an arduino due which is capable of outputing up to 800mA according to their website.

If any other information is required please let me know
 

Papabravo

Joined Feb 24, 2006
14,410
Thanks for looking further into this. First of, yes it's eagle. And what's happening is: I get a 20kHz square wave from 0V to 5V from the AD9833. With that signal I drive one MCP1402. Then I take the same signal and put it through an inverter (74HC04) and drive the second MCP1402 with that. This way I get -5V and +5V (10V in total). In the simulation I only displayed one of the 0V to 5V waves (blue simulation wave) and then the resulting wave (red one, which is behind the blue one: -5V to +5V).

Yes I did the same calculations, that's why I am so confused.

Maybe it's of interest that I take the 5V rail (for the MCP's and the power for every component) from an arduino due which is capable of outputing up to 800mA according to their website.

If any other information is required please let me know
What did you get for the current in the RLC loop?
 

Thread Starter

8dm7bz

Joined Jul 21, 2020
133
Would I measure that with a 1 ohms resistor ?

I did another measurement with only one MCP1402 and an external power supply set to 10V on the input of the mosfet. The results are better (attachment). I wonder why. Also the wave has more positive voltage, why is that ?
 

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Thread Starter

8dm7bz

Joined Jul 21, 2020
133
I also experimented a bit more and found that I could get to about 33Vpp across the capacitor with one MCP1402 and the external power supply set to 10V with a frequency of 17kHz.
I guess that's the expected output. But not yet with bipolar drive :/
 
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Papabravo

Joined Feb 24, 2006
14,410
Would I measure that with a 1 ohms resistor ?

I did another measurement with only one MCP1402 and an external power supply set to 10V on the input of the mosfet. The results are better (attachment). I wonder why. Also the wave has more positive voltage, why is that ?
We don't want to introduce new components to a system that is already producing results we cannot account for. You measure that AC current, with an AC ammeter by breaking the loop and putting the ammeter in series with the 23 Ω resistor. The only thing to be aware of is what the ammeter is displaying for AC current. It could be peak or it could be RMS or even something else. To tell you the truth I've never had much experience with measuring AC current.
 

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8dm7bz

Joined Jul 21, 2020
133
I just thought about something. Could it be that the resistor value behaves weird because it may not be rated for high wattage ?
 

Thread Starter

8dm7bz

Joined Jul 21, 2020
133
Another thing I found which I find rather strange is that when I connect a small 100ohms resistor between the AD9833 output and ground, I get a 4Vpp square wave. Is that maybe related ?
 

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Papabravo

Joined Feb 24, 2006
14,410
I attached the waveform of the voltage across the resistor. I think that would result to around 30mA ?
I don't quite come up with the same result. 1.72 Vpp divided by 2 gives 860 mV peak. 860 mV divided by 23 gives 37.4 ma. 37.4 ma times 178 Ω gives 6.65 Volts
That number is in the ballpark of your original scope measurement and way far away from the value recorded in you simulation.
So we are no closer to understanding the discrepancy.

When you measure something on your scope, are you using the ground clip and the probe or are you measuring differentially WITH NO GROUND CLIPS?
Connecting the scope ground clip to your RLC circuit is going to really disturb things.

One more thing occurs to me that your driver chips are not very good approximations to a voltage source. You are not getting anything like the current I was expecting. Go back to your LTspice simulation and measure the current through the 23 Ohm resistor. It should be way more than 37 ma Peak.
 
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Thread Starter

8dm7bz

Joined Jul 21, 2020
133
Yes I use one single ended probe with a ground (the arduino I use as a power supply gets it's power from a power bank and is not connected to earth ground). I just read up on why that's not right. But can I do two single ended probe measurements and do math on those 2 signals to get it right ? If so, how would I connect it ? A differential probe is quite expensive for me.

And for the current, I get about 200mA thorugh the resistor in the simulation.
 

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Papabravo

Joined Feb 24, 2006
14,410
200 mA is consistent with the voltage waveform you are seeing in the simulation (35 Vpeak) with the RLC circuit being driven by two low-impedance voltage sources.
To make a differential measurement, you use two probes and have the scope tell you the difference between the voltages on the two probes. If you like you can connect the two ground clips together and leave them float.
 
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