Electronic – What does ‘capacitor jumping up and down’ mean and what work does it do

capacitorcapacitor charging

While studying about capacitors, I came across with an explanation talking about "jumping up and down when a capacitor separating two stages".
I understood from several articles here that capacitors block DC when it is fully charged and that the idea of 'charging and discharging' of capacitor.

'This page' explains

  1. If a capacitor has the negative lead connected to the 0v rail, it will charge and discharge
  2. If a capacitor is NOT connected directly to the 0v rail, it will JUMP UP AND DOWN.

and with the following figure, says

the capacitor will 'fall'and the voltage on the negative lead can actually go below the 0V rail

where I totally lost my understanding.

enter image description here
jumping cap
(source: talkingelectronics.com)

(please refer to '4. A capacitor separates two stages' on 'the linked page.')

The pages explains that

By knowing how much a capacitor jumps-up-and-down, you can "see" a circuit working.
and here my questions came.

  1. I can't understand the difference between 'charging/discharging' and 'jump up/down'. I thought even though it's not directly connected to 0V rail, still depending on its reference voltage, it can be charged and discharged. What is the difference in those two expression to comprehend their meaning?
  2. What happens when capacitor jump up and down?
  3. How can I calculate the amount of the 'jumps'?

Best Answer

What the author is describing in that circuit is that if the voltage on the left side of the capacitor suddenly changes level, the voltage on the right side will change by the same amount.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Square-wave passed through a capacitor. (Please excuse arrows as RC discharge curves.)

With the circuit schematic shown above:

  • Initially 'A' is high and 'B' is at 0 V.
  • When Q1 switches on 'A' is pulled ("jumps" in the author's parlance) to 0 V.
  • At the instant of switching the voltage across C1 is V+ so when 'A' is pulled low 'B' is pulled low as well. i.e., Both sides "jump" together as neither side is grounded.

In the case of a filter capacitor one side is usually grounded so this effect is not seen.

I find it useful in circuit analysis to think of the capacitor's action in this fashion. I figure out what the steady-state voltage is across the capacitor and what will happen the right side when the left side suddenly changes voltage.

Simulation waveforms

schematic

simulate this circuit

Figure 2. Test schematic.

enter image description here

Figure 3. 500 Hz, 1 µF, 100 kΩ.

Figure 3 shows what happens when the capacitor is feeding a high resistance load.

  • On the first rising edge of the input the output "jumps" up with it. R1 starts to discharge the right side, however, and at the end of that half-cycle the voltage has drooped a little.
  • On the first falling edge the input drops by 1 V and so does the output. Since the starting point is about +0.9 V the output drops to -0.1 V.
  • This process continues and after a while the waveform settles down centred about the zero-volt line.

enter image description here

Figure 4. 500 Hz, 1 µF, 1 kΩ.

  • Decreasing R1 to 1 kΩ causes the effect to be more pronounced as the capacitor discharges and charges more quickly. Notice how the waveform has settled down after a few cycles.

enter image description here

Figure 5. 500 Hz, 1 µF, 100 Ω.

  • In Figure 5 R1 has been decreased to 100 Ω and we can see that the output waveform has become much more spikey. We can also see that it no longer reaches the +1 V level because the load resistor is so low.

This explanation is deliberately non-mathematical and is intended to give you some mental picture of what's really happening. If you study the maths some more and figure out where the current is flowing you should be able to get a good grasp of how it works.

Simulation

Linear Technology (chip-maker) have their LT Spice simulator available as a free download. I recommend you try this to assist in your learning and understanding.

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