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What will happen if I apply a negative pulse to the capacitor? Is the capacitor voltage increase positive or negative?

The pulse which is shown in the below picture is the output voltage pulse from a photodetector (photo multiplier tube.) The detector is consists of a photoelectric plate (that gives an electron when a light incident on the plate ) and electron multiplier (that multiplies the number of electrons or amplifies the signal.) The multiplied electrons are collected in one end across a resistor as a voltage pulse.

My aim is to measure the amount of charge from the detector. So I use this signal to charge a capacitor then I do some other analyses to measure the charge.

enter image description here

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    \$\begingroup\$ Define negative pulse. Voltage? Current? Energy? \$\endgroup\$
    – winny
    May 19, 2020 at 13:46
  • \$\begingroup\$ ok, the pulse is actually output from a photodetector. and the detector is simply an electron multiplier tube.so the output pulse is made up of a large number of charges(electron).@winny \$\endgroup\$ May 19, 2020 at 13:56
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    \$\begingroup\$ Voltage pulse? Current pulse? What have you measured? \$\endgroup\$
    – winny
    May 19, 2020 at 14:09
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    \$\begingroup\$ Now you have a more tricky situation than stated in your question. Please edit your question to include a schematic with part numbers where appropriate, where your oscillogram is taken from and scales on the axis. \$\endgroup\$
    – winny
    May 19, 2020 at 14:15
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    \$\begingroup\$ One option is to swap the photodetector pins so that they apply a positive charge to the capacitor. But I suspect there is a lot going on here that is not being explained. \$\endgroup\$
    – user57037
    May 19, 2020 at 21:31

4 Answers 4

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Negative voltage means lower potential then reference. So the direction of current flow will be from the reference level towards the negative voltage. So electron flow will be the opposite.

It does not matter if the input voltage is positive or negative for a non-polar capacitor (For example ceramic capacitor). Charges will be stored in it.
enter image description here

But for polar electrolytic capacitor, you have to be careful. They can not be used in bipolar cases. If the negative terminal is at zero volts, then the positive terminal must be zero or greater than zero volts. If you apply a negative voltage, then it will explode.

If the positive terminal is at 0 volts, then the negative terminal must be zero or less than zero volts. If you apply a positive voltage, then it will explode.

I have experienced polar electrolytic capacitor explosions a few times. There's also non-polar electrolytic capacitor available in the market.

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It depends on your capacitor and the corresponding circuit. All capacitors have built-in series resistance (ESR), and the corresponding RC circuit (not to mention the rest of your design) will either smooth that spike into a tiny bump, or it'll look almost identical to the input. Different capacitor sizes, types, and even the physical dimensions will impact their transient response.

The physical design of the capacitor will also react differently to negative voltages. Standard ceramic capacitors won't care about negative voltages as long as the amplitude is within their design parameters. If your capacitor is polarized - like a tantalum or electrolytic capacitor - a negative voltage for an extended period will make it explode (although a small transient like what you have pictured is probably just fine)

You might also look into a Coulomb Counter for your design - they're designed to track the charge of batteries, but some are extremely accurate and will factor in their own consumption.

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Will the capacitor voltage increase is positive or negative?

The capacitor voltage becomes negative. There is nothing surprising or complicated about this. All voltages are measured relative to an arbitrarily chosen reference potential, so negative voltages are entirely common and a normal part of circuit analysis.

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  • \$\begingroup\$ yes, I can understand your answer. But the signal is just a bunch of charges. and the pulse is negative because the charges are electron. then capacitor is a charge storage device right then why its went to negative \$\endgroup\$ May 19, 2020 at 14:57
  • \$\begingroup\$ @krishnamoorthijayakumar, because the capacitor plate with an excess of negative charge is negatively charged relative to the plate with an excess of positive charge. \$\endgroup\$
    – The Photon
    May 19, 2020 at 15:00
  • \$\begingroup\$ but as per my design If the capacitor voltage is increased positively that will be better \$\endgroup\$ May 19, 2020 at 15:01
  • \$\begingroup\$ @krishnamoorthijayakumar, if you want it to end up positively charged (for example, to measure its voltage with a single-supply ADC), then you could send the signal through an inverting amplifier before charging the capacitor. \$\endgroup\$
    – The Photon
    May 19, 2020 at 15:02
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    \$\begingroup\$ @JackDanniels, there's obviously a whole lot left unsaid here because OP has hardly shared any information about their application. I'm trusting them to pick a reasonable buffer circuit, reset circuit, sampling circuit, etc. So I'm going to assume they can also pick an appropriate capacitor for their application. \$\endgroup\$
    – The Photon
    May 19, 2020 at 20:26
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that detector is consists of a photoelectric plate (that gives an electron when a light incident on the plate ) and electron multiplier(that multiply the number of electrons or amplify the signal) the multiplied electron are collected in one end across a resistor as a voltage pulse. My aim is to measure the amount of charge from the detector.

Photomultipliers have multiplication noise due to the electron multiplication process. A single photon might generate on average 1 million electrons, but the variance on that output is also about 1 million electrons (although it varies between devices). For this reason, counting charge is not always the most useful thing to do, since you frequently are just measuring noise in the detector that has no significance.

A common approach, since you have a signal of almost 1 volt into (presumably) 50 ohms, is just to count how many rising edges you have, especially if you don't expect photon pile up.

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