Why is the voltage negative in this circuit

[FreshmanNewbie]

I was trying some simulations and tried the below circuit.



My question is, how is the below marked voltage negative 3V?

I thought like, since there is a diode, there will be approximately 4.3V across the capacitor and 0.7V across the diode. Why is this not the case in the above circuit? What am I missing to understand.

 

[barry]

-3.09V relative to what? Where’s your ground?

 

[BradtheRad]

That looks like Falstad's simulator. I tried a similar circuit, sending DC through a diode to a cap without any series resistance.

The cap charged higher than supply voltage. Obviously an error. It happens even if a node is assigned to 0V ground.

I found I need to install a resistor in the current path. Without it the algorithm has a problem calculating a time constant for the capacitor. Current flow doesn't behave sensibly, nor volt levels.

Some resistance needs to be present in most cases inline with a capacitor, or else some resistive component.

 

[KlausST]

Hi,

again a simulation problem ... that adds to my list agains the usefulness of simulation tools.
And: No, I´m not against simulation /-tools at all.

Often the problem is caused by the user. Mainly not to use realistic parts.
(But I agree that the GND symbol is missing)

How to debug the OP´s simulation?
* I guess the capacitor becomes charged by a current (non realistic current)
* thus there needs to be a source for this current.
* the characterisitic of a load (dissipating power) is that current is positive when voltage is positive
* but here we are looking for a source. Thus I expect a device where the voltage is positive while the current is negative..

Checking voltage and current part by part, maybe one can find the faulty device..

Klaus

 

[danadakk]

SIM should look like :

Regards, Dana.

 

[KlausST]

Hi,

seems O.K. from the voltages and currents.

But a critical view shows that a 1N4148 is not suitable. It´s datasheet specifies max. 2A for 1us.
I^2t (heating) is 4uA^2s
This is violated to almost 30uA^2s

Klaus

 

[FvM]

Think sole purpose of the SPICE simulation is to show that this circuit topology doesn't generate negative voltage. Diode selection is beyond scope, notice that a real circuit also doesn't raise input voltage in no time.

 

[danadakk]

Hi,

seems O.K. from the voltages and currents.

But a critical view shows that a 1N4148 is not suitable. It´s datasheet specifies max. 2A for 1us.
I^2t (heating) is 4uA^2s
This is violated to almost 30uA^2s

Klaus

Was not paying attention to diode chosen, fixed. Also changed timestep to 1 nS :

Regards, Dana.

 

[KlausST]

Hi,

Apologies. My post was not intended to accuse anyone personally.

Rather, it should show to OP what information can be obtained from a simulation such as the one shown.
It was a "critical look" at the values of the chart (I and t)... and how to read and validate it.

It was meant to not just see a the current goes up and down .... but also focus on the values.
Like: Wow, 4 Amperes! I did not expect that.

Now one can validate the values .. and check if the parts can stand them. (like I did)
Or one can ask oneself whether the values (and the simulation) really is realistic (like FvM did).

And me re-thinking about it ... maybe the first question should be whether the simulation is realistic ...and after that whether the parts can stand it.

Klaus

 

[Ratch]

I was trying some simulations and tried the below circuit.



My question is, how is the below marked voltage negative 3V?

I thought like, since there is a diode, there will be approximately 4.3V across the capacitor and 0.7V across the diode. Why is this not the case in the above circuit? What am I missing to understand.

The mathematical (convential)l current direction is clockwise (CW). Since the charge carriers (electrons) in the wire are negative, we have to reverse the direction to get the true charge flow, which will be CCW. That means electrons will be piling up on the bottom of the capacitor and depleting at the top until the cap is fully energized. Therefore, a 5 volt positive voltage will develop at the top of cap with respect to the bottom. I don't know where the 2 volts went in the simulation or why the polarity is reversed. In the real world, the diode stops conducting significantly at 0.6-0.8 volts, but if you wait long enough, the minuscule current leakage below 0.7 volts will eventually top off the capacitor at 5 volts.

 

[dick_freebird]

Your first question should be, "Are any of the netlist nodes, '0'? ".

If no then you can forget sanity elsewhere.