A capacitor voltage much higher than the signal generetor output - Page 1
Forum section: electronics / beginners

author: selim
author_rank: Newbie
timestamp: December 30, 2022, 11:32:18 am
content: When i build up the circuit and adjust signal generator to 5Vpp 44.78KHz, The Vc voltage i see at the oscilloscope was 45.7Vpp. So, how is it possible to obtain a capacitor voltage much higher than the signal generator output?

author: jpanhalt
author_rank: Super Contributor
timestamp: December 30, 2022, 11:48:25 am
content: Have you read about "boost" converters?https://en.wikipedia.org/wiki/Boost_converter

author: inse
author_rank: Super Contributor
timestamp: December 30, 2022, 12:25:54 pm
content: Is it getting into resonance?

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: December 30, 2022, 01:56:01 pm
content: Signal generator is designed to put out a signal into a 50 ohm load. Your Load is much higher than that.

author: BillyO
author_rank: Super Contributor
timestamp: December 30, 2022, 03:31:41 pm
content: Or he could simply be using a 1x probe but have the channel set to 10x.Since there is not diode in teh circuit there can be no boost.Without the 50ohm load the voltage will only be a maximum of twice what it would be with the 50 ohm load.

author: fourfathom
author_rank: Super Contributor
timestamp: December 30, 2022, 04:46:39 pm
content: Quote from: BillyO on December 30, 2022, 03:31:41 pmWhile I think the 1x probe is the likely culprit, it is certainly possible to get a boost with just an LC circuit. The values shown will provide a roughly 20dB (10x) voltage gain at the resonant frequency of about 50 KHz, assuming a 1Meg scope input impedance. If the actual coil series resistance is lower than shown then the boost will be greater. We often use LC networks for impedance matching.But if the input stimulus is really a 500Hz square wave, then the observed peak will be much lower, really just a small amount of overshoot.[oops:]I just noticed that the OP was using a 44 MHz squarewave. In that case, with a 10Vpp input the peak voltage into a 10M load is about 60Vpp. Setting the frequency to 50KHz (the resonant point of my simulation) I see an output of 110Vpp. So the boost is certainly happening. The components the OP is using probably resonate at 44MHz.

author: Terry Bites
author_rank: Super Contributor
timestamp: December 30, 2022, 06:57:03 pm
content: If you sim or pen and paper the numbers your circuit it will show a (20bB) peak in its response very close to the measured value.So yes, just series resonance.Frequency fo= 50.33 kHzReactance XL/XC =3.16 kΩQ factor =9.04Impedance at f0 494.97 Ωwww.redcrab-software.com/en/Calculator/Electrics/RCL-Series-Resonance-Circuit

author: wasedadoc
author_rank: Super Contributor
timestamp: December 30, 2022, 07:22:46 pm
content: Quote from: fourfathom on December 30, 2022, 04:46:39 pmThe OP wrote 44.78 kHz. Not MHz.

author: bdunham7
author_rank: Super Contributor
timestamp: December 30, 2022, 07:30:19 pm
content: Quote from: selim on December 30, 2022, 11:32:18 amThat is how series resonant circuits work. At resonance you will see much higher voltages across the inductor and capacitor, but those voltages are out of phase and cancel out over the whole circuit. The lower the resistance, the higher the 'Q' (quality factor) of the circuit and the lower 'D' (dissipation) or damping. In this case you apparently have a Q of about 9 at a resonance of about 45kHz. Using your nominal figures, you would expect a Q of 9.1 at a resonance of 50.4kHz, so that's pretty close. The small discrepancy would be caused by component tolerances and parasitics (including the o-scope probe capacitance). Using a square wave with a DC bias also probably has some effects, but those would be a bit harder to describe and calculate.

author: fourfathom
author_rank: Super Contributor
timestamp: December 30, 2022, 07:31:27 pm
content: Quote from: wasedadoc on December 30, 2022, 07:22:46 pmYes, my typo. My sim was using 44 KHz.

author: fourfathom
author_rank: Super Contributor
timestamp: December 30, 2022, 07:40:22 pm
content: Quote from: bdunham7 on December 30, 2022, 07:30:19 pmOr, if it's easier, consider it as a parallel-resonant circuit, being shunt-fed from a low-impedance source. It's all the same.

author: BillyO
author_rank: Super Contributor
timestamp: December 30, 2022, 10:58:44 pm
content: Quote from: fourfathom on December 30, 2022, 04:46:39 pmI see what you mean. I would call that resonance, not boost.

author: nigelwright7557
author_rank: Frequent Contributor
timestamp: December 31, 2022, 01:50:35 am
content: Because it uses inductor and capacitor they react together to generate a much higher voltage.This why a class d amp should always be loaded.On class d amp I always use an output cap rated 3 times voltage of amp voltage on output filter in case user disconnects speaker.

author: EPAIII
author_rank: Super Contributor
timestamp: December 31, 2022, 02:22:08 am
content: A signal generator with a 50 Ohm internal resistance and a 50 Ohm load would need only a 10 Volt internal generator to provide a 5 Volt signal at the output. The internal 50 Ohm resistance and the external 50 Ohm load form a Voltage divider with a factor of 0.5.With the higher impedance load, I would expect to see a Voltage near 8.6 Volts at the output terminal of the generator.That is no way near the Voltage observed by the OP.Quote from: Wallace Gasiewicz on December 30, 2022, 01:56:01 pm

author: CaptDon
author_rank: Super Contributor
timestamp: December 31, 2022, 07:02:23 pm
content: In a class D amplifier the output filter is never designed to be resonant anywhere in the audio bandpass of interest or in the range of the switching clock frequency. All overshoots and undershoots are clamped back to the power supply rails with snubber diodes and body diodes within the mosfets. All properly designed class D amps should function with reasonable linearity from open circuit all the way down to minimum designed load impedance with no need for 3X rated components. You shouldn't be seeing voltages higher than the + or - rail and no more than the combination of the pair. Class D isn't 'flyback' and shouldn't show any flyback overshoot tendencies.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: January 02, 2023, 11:18:22 am
content: Do this in a simulator:1. Remove the resistors, so yo just have a generator and an LC combination.2. Change the generator to a single step from 0 to 5V.3. Run simulation and observe the output.In the beginning you will see only a slow change in current through the inductor. Inductors "resist" change in current.Some while later, the capacitor is charged to the same 5V as the source. Up to that time the current through the inductor has been increasing.Because there is still current flowing through the inductor, it will keep on pushing that into the capacitor.Because the voltage over the capacitor is now higher then the voltage of the source, the current through the inductor will get less again.By the time the current through the inductor is zero, the capacitor will be charged to twice the voltage of the source.If the source can also sink current, then the current through the inductor will reverse, and current will flow from the capacitor through the inductor back to the source.When there is a voltage over a capacitor has energy stored. Likewise, if there is current flowing through an inductor, it has also stored energy.Without resistors, the energy will keep on exchanging between the capacitor and the inductor, and you will see sinewaves on your simulation, even though the initial input is just a single step.With a series resistor, there is energy loss, and the sinewave will dampen out.When you change the single step of the source to a square wave of the right frequency, you will pump a bit of energy into the LC combination during each period, and the output voltage will increase. In and ideal LC combination, the output voltage will be infinite, but in reality there is always some resistive part (such as wire resistance) and the output voltge will stop increasing when the energy dissipated is the same as the energy pumped into the system.

author: nigelwright7557
author_rank: Frequent Contributor
timestamp: January 05, 2023, 01:30:41 am
content: Quote from: CaptDon on December 31, 2022, 07:02:23 pmYour right on the left side of the LC but on the right side you dont have any clamping and you can get very high voltages due to ringing.Have a play with LTSPICE with a square wave and a series LC circuit. You can get many times the supply voltage across the C.Its basic AC theory that voltages across components can be much more that the supply voltage.