8
\$\begingroup\$

If I have a signal generator I want to inject into mains or DC systems. Does it matter if the capacitors are charged or not, does it make a difference?

If for a example, the reactance at 5MHz at 50nF is 6mohm, if we then charge the capacitor to 230V 50Hz or 60V DC for example, would the 5MHz signal see a different impedance? Or would it be the same?

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
2
  • 3
    \$\begingroup\$ if we then charge the capacitor to 230V 50Hz or 60V 230 V, 50 Hz is AC, a capacitor's charge is DC. In this situation, the capacitor isn't charged to a fixed value, the voltage across and the charge in the capacitor change continuously as you apply AC. The capacitor's value might change a little bit depending on the actual voltage at that point in time but this change is very small, I would not worry about it. So no, the impedance of the capacitor only depends on frequency. \$\endgroup\$ – Bimpelrekkie Feb 23 at 16:07
  • 1
    \$\begingroup\$ You can't charge a capacitor to a frequency. \$\endgroup\$ – user253751 Feb 24 at 11:04
15
\$\begingroup\$

The expression "capacitance changes depending on DC bias" is a bit misleading. It actually comes from the fact it is tested with a DC bias and a tiny AC voltage added to it to measure the capacitance. But in reality, the capacitance of any capacitor always depends on the instantaneous voltage across the capacitor, no matter where this voltage came from. Usually it is DC, but in your case it will be low frequency AC plus high frequency AC.

So you need a cap with low \$ dC/dV \$ or capacitance variation per unit of voltage variation. That includes C0G ceramics and most types of film caps like PPS, PP, etc.

In your case, since mains voltage is involved, be sure to pick a capacitor rated for it. Not just the AC voltage, but it should also be rated X. This is not about movies, rather it means it is safe to use the capacitor across AC mains voltage. Mostly this is about not starting a fire if there is an internal short in the capacitor, so it will have features like self-healing. Note the capacitance value of these tends to decrease over time as they are exposed to voltage spikes which cause internal arcing, which pokes a hole in the dielectric and causes a short. At this point the self-healing feature kicks in and the metal around the hole vaporizes, which means there is no longer a short, and prevents fire. But it also means the capacitance value has decreased a bit. So if you want long term accuracy, maybe pick a higher voltage rated cap, like 600V or 1000V.

Also you should put high value resistors across the caps to discharge them once the device is unplugged, because I guess this device is going to have a male mains plug at the end, and it is never a happy experience to grab one of these when there is a charged capacitor on the other end. The resistors will also balance voltage across the caps so they both have half the mains voltage, so they're both as far as possible from their max voltage rating.

Several other people mentioned Class 2 ceramics are the absolute worst, but due to having strong dC/dV not only will the capacitance be modulated by the 50Hz AC voltage, it will also be modulated by your HF signal which will create all sorts of distortion harmonics. They are also quite lossy, so some of your signal will end up as losses. They are awesome for decoupling power supplies due to high capacitance per volume, low cost, low inductance, but they are terrible for everything else.

\$\endgroup\$
4
  • \$\begingroup\$ Can you use non-X rated capacitors across the line if they are rated and protected by a fuse? Or is this a no no? \$\endgroup\$ – Curious Diode Feb 23 at 17:04
  • \$\begingroup\$ I'm not exactly sure about that... \$\endgroup\$ – bobflux Feb 23 at 17:07
  • \$\begingroup\$ Also the resistance of a fuse varies with current along the AC cycle, that may add inaccuracies \$\endgroup\$ – bobflux Feb 23 at 17:07
  • \$\begingroup\$ In series, the lower-capacitance capacitor gets more of the voltage, so voltage spikes will be more likely to further decrease whichever capacitor is already more "worn out". So that's probably another reason to over-rate the caps (or maybe just use 1? Unless you need both for isolation) \$\endgroup\$ – Peter Cordes Feb 24 at 11:41
10
\$\begingroup\$

Actually it strongly depends on the capacitor itself. E.g. many ceramic capacitors are known for their charge dependency. Just as a starting point: https://www.murata.com/en-global/support/faqs/products/capacitor/ceramiccapacitor/char/0005 .

So always keep in mind that the given capacitance is often only valid for small signals and can strongly deviate.

\$\endgroup\$
0
8
\$\begingroup\$

Yes, it matters, for real world capacitors the effective capacitance is a function of the voltage applied.

The strength of this effect is dependent on capacitor technology, it is most pronounced in ceramic capacitors with class 2 dielectrics, aka MLCC , the effect can be as strong as -70% effective capacitance when at 100% of rated voltage for smaller sized ceramic capacitors and low grade dielectrics .

The mechanism of this effect is piezoelectric (electrostatic) in nature, the ceramic expands under applied electric fields.

\$\endgroup\$
3
  • \$\begingroup\$ Ah! "... the ceramic expands under applied electric fields ..." I never considered the piezoelectric effect. Thanks. \$\endgroup\$ – Transistor Feb 23 at 17:12
  • \$\begingroup\$ @Transistor it further goes to show that the quality or type of the bulk material is the primary differentiator in capacitor technology even within a common application (ceramic chip). Class 1 dielectrics are of a composition to minimize the piezoelectric effect and, as a corollary, the temperature dependency. \$\endgroup\$ – crasic Feb 23 at 19:31
  • \$\begingroup\$ See also microphonics; for piezoelectric materials there is a voltage <--> audio coupling effect. \$\endgroup\$ – Technophile Feb 24 at 21:13
3
\$\begingroup\$

It doesn't matters if capacitor is charged or not, but only in case it is an ideal capacitor. However ideal elements don't exist in reality. Real capacitors have voltage dependend capacities.

New contributor
Mandalorian is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
\$\endgroup\$
1
  • \$\begingroup\$ thanks ariser for telling and editing my answer \$\endgroup\$ – Mandalorian Feb 24 at 2:54
2
\$\begingroup\$

For nearly all applications, it doesn't matter as long as the TOTAL voltage across the capacitor doesn't exceed its rated voltage.

There are some voltage-dependent effects that are more pronounced in some types and designs of capacitor but for the application you are proposing, they will be insignificant effects.

\$\endgroup\$
2
  • \$\begingroup\$ See crasic's answer. He explains the mechanism of how capacitance does change with voltage. \$\endgroup\$ – Transistor Feb 23 at 17:13
  • \$\begingroup\$ Strictly speaking he's correct. Does that matter in this case? Likely not. \$\endgroup\$ – jwh20 Feb 23 at 17:25
1
\$\begingroup\$

Regarding the stated purpose of this circuit:
Technically, you put a capacitive "transformer" before an inductive one. This means, with 50nF at 230VAC you will get a AC current of about 3,5mA, which goes through your inductive transformer and thus will appear on the other side. Your signal generator probably won't like this. Also, depending on the size of your pulse transformer, you might get core saturation effects, making the transformer non-linear too.
Additionally, without a resistor in series you can get a surprisingly high inrush current if you connect AC at a bad time.

I'd recommend to use far less than 100nF - for 5MHz, 1nF should be more than enough - and a current limiting resistor. And just in case a suppressor diode across the signal generator input.

But I think you could get better results if you inject your signal inductively, rather than capacitively, into the system you want to test.

\$\endgroup\$
1
  • 2
    \$\begingroup\$ It doesn't seem like you really answered the question. Instead you suggest changes or modifications to the circuit in the question, so this should probably be a comment rather than an answer. \$\endgroup\$ – Elliot Alderson Feb 26 at 13:00
1
\$\begingroup\$

it may matter, depending on your application, let's say you use a ceramic capacitor with a class 2 dielectric, then the capacity will change with the bias dc voltage, therefore you may have to account for that especially if you are close to the rated voltage.

Also you can't "charge" a capacitor with an AC voltage as per your example, it will be perpetually charged and discharged with an average charge and voltage of 0. But when you cut the power, there might be considerable charge left depending on the position in the cycle.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.