Hello, i just didn't understand that why we get negative voltage(-4.7V) from capacitor's negative pole when we apply 0V to capacitor's positive pole.As i figured out, this is used in transistor flip flop.

Power Supply : 5V

The site that i used for simulation is https://www.falstad.com/circuit/circuitjs.html

Circuit Link : Click Here

Please help me, best regards.

  • \$\begingroup\$ Did you know Falstead has an export function that gives you the schematic as text that you can give other people? Even better, you can export it as a link. \$\endgroup\$ – DKNguyen Jun 20 '19 at 18:43
  • \$\begingroup\$ Thank you, i didn't know, here is the URL of circuit : tinyurl.com/y33dm9da \$\endgroup\$ – Androman Jun 20 '19 at 18:45
  • \$\begingroup\$ You can just add that to your orignal post with the "edit" button. \$\endgroup\$ – DKNguyen Jun 20 '19 at 18:47
  • \$\begingroup\$ You can also embed your images in the question so that we don't have to follow links to understand your question. \$\endgroup\$ – Transistor Jun 20 '19 at 20:35
  • \$\begingroup\$ when we apply 0V to capacitor's positive pole ... that is not quite correct .... this is more accurate when you use capacitor's positive pole as ground reference ...... 0V is actually any point where you touch the negative probe of your DVM \$\endgroup\$ – jsotola Jun 20 '19 at 23:31

Since your design has no 0V reference which by definition we call the ground symbol, Falstad has chosen a different reference when the switch to the right is closed.

On a floating circuit the currents are not affected by a single placement of the ground symbol, it only changes your relative voltages.

In my simulation I used >Options> other options< to change the sample time to around 500us with the simulation slider speed to “mid scale” to get closer to a realtime display of 1 second per second as you are using longish time constants. As always all passive parts are ideal and the traces have no inductance, so I add these ESR,DCR,ESL’s for RF work to closer model the real physics.

I also created by inserting a ground symbol on the Cap-V so that Falstad Sim knows to use this for all “absolute” voltage measurements. You can put it anywhere you want as long as if you use more than 1, you are consistent. Parallel Voltages and caps cause infinite circulating currents, so in real physics, there must be some ESR. (mOhm)

Differential V measurements are possible using the right mouse Labels and stuff> Scope Probe and place the + point 1st then drag to any -ve point. Then you can choose V, Vrms, Vpp, Vmax,Vmin, etc. Add line traces to make it less interfere with the components to look good.

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  • \$\begingroup\$ Thank you very much for your explanation and simulation, but when i measure the circuit in reality, i read - voltage between cap's - pole and supply's negative pole, i even lit a led by connecting led's + to supply's - and cap's - pole for a very short time, just didn't understand this part.How supply's - part become + for LED and our cap's - voltage become - according to this situation \$\endgroup\$ – Androman Jun 20 '19 at 19:27
  • \$\begingroup\$ Something is wrong with your test that does not match the simulation. It is impossible here tinyurl.com/y2ezld32 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 20 '19 at 19:35
  • \$\begingroup\$ i meant like this : tinyurl.com/y4glqqmm \$\endgroup\$ – Androman Jun 20 '19 at 19:39
  • \$\begingroup\$ That’s different. Charge current is + for a + charging voltage and discharge current is always -ve for a discharging + voltage. You know this from dV/dt=Ic/V. Thus the voltage is always positive here but Ic changes direction with dV/dt polarity current. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 20 '19 at 19:53

Run your simulation at maximum speed with 75% current speed. Observe the capacitor voltage. Your capacitor is so large and your 10K resistor is so large that after the capacitor charges up, it takes a long time for the capacitor to discharge and while it's discharging the current that if you run at slow speed you are perpetually stuck in a transient state that looks like steady state.

Observe the current direction through the 10K resistor before and after switching and note what voltage drop polarity that represents relative and what it does for the voltage on the negative terminal of the capacitor with respect to ground.

You should see that when if you open the open the rightmost switch and allow the capacitor to charge, then close the switch so the capacitor discharges, the discharge current flows through the 10K resistor in a direction where (using passive sign convention) the resistor terminal for the + end voltage drop is the ground/0V node and the terminal for the negative end of the voltage drop is the capacitor's negative terminal. If you measure with the supply's negative terminal as your reference, then the other end of the resistor (the capacitor's negative terminal) will measure as below ground.

Neat circuit in that capacitor is charged to the same as the supply voltage so it isn't high enough to discharge back into it, but uses the voltage drop across the 10K resistor due to its own discharge current to "boost" the voltage of its positive terminal higher than the power supply so it can discharge back into it.

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  • \$\begingroup\$ Thank you very much for your wide explanation \$\endgroup\$ – Androman Jun 20 '19 at 19:22

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