Failed Capacitor/resistors

[Besoeker]

A bit of graphical stuff to add to my previous post.

This is from a spreadsheet that can be used to add waveforms together.

First, with no added harmonics and you get a clean sine wave:

Now add in odd harmonics with an amplitude the inverse of their harmonic number and you get something approaching the square wave I mentioned in another thread.

It would need a greater number of terms beyond the 25th to make it better but, hopefully, it demonstrates the idea.

 

[kwired]

A bit of graphical stuff to add to my previous post.

This is from a spreadsheet that can be used to add waveforms together.

First, with no added harmonics and you get a clean sine wave:

Now add in odd harmonics with an amplitude the inverse of their harmonic number and you get something approaching the square wave I mentioned in another thread.

It would need a greater number of terms beyond the 25th to make it better but, hopefully, it demonstrates the idea.

I think I see what you are saying. In the graph with the harmonics doesn't that sudden 180 degree change at the zero cross over look like a short circuit in a way to a circuit with a capacitor in it? Short circuit might be wrong choice of words but there is significant increase in current within the capacitor at this point isn't there?

I think I remember there is even more complication from the supply transformer, as the current increases there is more saturation of the core and the voltage starts to increase, then the sudden 180 degree change in polarity results in higher voltage across the capacitor and more discharge/recharge current as there is no gradual increase the other direction either like there would be with a true sine wave.

I guess what I am trying to say is there is no gradual change of direction of current with each cycle but rather a nearly instant change of polarity within the cycle resulting in overall higher current levels.

Please correct me if wrong.

 

[Besoeker]

I think I see what you are saying. In the graph with the harmonics doesn't that sudden 180 degree change at the zero cross over look like a short circuit in a way to a circuit with a capacitor in it? Short circuit might be wrong choice of words but there is significant increase in current within the capacitor at this point isn't there?

Nice observation if I may say so.
Yes, if the voltage did do that the capacitor current would be very high.

i=Cdv/dt

But the voltage generally doesn't do that*. I was simply trying to illustrate how a non-sinusoidal waveform can be analysed as a series of sine waves at harmonic frequencies of the fundamental. The choice of a square wave was just to show how it works. You certainly couldn't get by with 46% total voltage distortion.

*One example of where you do get high rates of change of voltage (dv/dt) is the output of a variable frequency. Fitting PFC there is an absolute no no because of that and the peak currents that would flow.

 

[T.M.Haja Sahib]

*One example of where you do get high rates of change of voltage (dv/dt) is the output of a variable frequency. Fitting PFC there is an absolute no no because of that and the peak currents that would flow.

Is not the capacitor also used to smooth out high dv/dt as in rectifiers and so might not be damaged by high dv/dt? Perhaps the reason for not using capacitor in the output of a variable frequency drive is something else.

 

[Besoeker]

Is not the capacitor also used to smooth out high dv/dt as in rectifiers and so might not be damaged by high dv/dt?

In applications using rectifiers or SCRs, a capacitor witha series resistor is frequently used as a "snubber" circuit to limit rate of change of voltage to that within the capability of the device which, for SCRs, is typically in the several hundred volts per us.

Perhaps the reason for not using capacitor in the output of a variable frequency drive is something else.

IGBT switching rate is very fast. This from a report I did a while back:
"Rate of change of voltage (dv/dt)
Rate of change of voltage was very fast. In all cases it was either outside the limits of the IEC TS60034-17 technical specification, or marginal. For the rise times measured the maximum dv/dt should not be greater than approximately 1600V/us. The average dv/dt measured at the pump house was 2500V/us."
Remembering that i=Cdv/dt, this would result in destructively huge currents in the IGBTs which is why you can't do it.

 

[T.M.Haja Sahib]

In applications using rectifiers or SCRs, a capacitor witha series resistor is frequently used as a "snubber" circuit to limit rate of change of voltage to that within the capability of the device which, for SCRs, is typically in the several hundred volts per us.

A capacitor can well do alone the above function of limiting high dv/dt. The reason is a rule from circuit theory: voltage across a capacitor can not change instantaneously.

IGBT switching rate is very fast. This from a report I did a while back:
"Rate of change of voltage (dv/dt)
Rate of change of voltage was very fast. In all cases it was either outside the limits of the IEC TS60034-17 technical specification, or marginal. For the rise times measured the maximum dv/dt should not be greater than approximately 1600V/us. The average dv/dt measured at the pump house was 2500V/us."
Remembering that i=Cdv/dt, this would result in destructively huge currents in the IGBTs which is why you can't do it.

How can it be so? The dv/dt of the circuit with capacitor is much lower than the same circuit without the capacitor due to the smoothing effect of capacitor.

 

[Besoeker]

A capacitor can well do alone the above function of limiting high dv/dt. The reason is a rule from circuit theory: voltage across a capacitor can not change instantaneously.

The discharge from the capacitor without the resistor would subject the device it is protecting to excessive di/dt.

How can it be so? The dv/dt of the circuit with capacitor is much lower than the same circuit without the capacitor due to the smoothing effect of capacitor.

Perhaps you don't know the inverter power circuit of a VSD is arranged. At the input to the IGBT bridge is a high value DC link capacitor closely coupled to the IGBTs to make the impedance between capacitor and IGBTs negligible. If you connect a capacitor to the output of the IGBT bridge you then have a circuit comprising two capacitors connected by a switch (IGBT). And negligible impedance.

I'll leave it to you to think about what happens when you close that switch......

 

[kwired]

The capacitor in a rectifier does not see the opposite polarity imposed on it. It is in the DC side of the rectifier, it only sees zero to whatever peak is from the source and is always same polarity.

Voltage across a capacitor can not change instantaneously, that does not mean the source power connected to it can not, which is what is happening in the situation described, but it is happening in an AC circuit not a DC circuit

 

[Besoeker]

The capacitor in a rectifier does not see the opposite polarity imposed on it. It is in the DC side of the rectifier, it only sees zero to whatever peak is from the source and is always same polarity.

I was actually aware of that.....

Toltage across a capacitor can not change instantaneously, that does not mean the source power connected to it can not, which is what is happening in the situation described, but it is happening in an AC circuit not a DC circuit

I think the only way to determine what caused the fuse to fail would be to record the current waveform and do an anysis of it.

 

[kwired]

I was actually aware of that.....

I guess I was addressing TM more so than you with that statement.

 

[T.M.Haja Sahib]

The discharge from the capacitor without the resistor would subject the device it is protecting to excessive di/dt.

Here question is about dv/dt and not about di/dt.

Perhaps you don't know the inverter power circuit of a VSD is arranged. At the input to the IGBT bridge is a high value DC link capacitor closely coupled to the IGBTs to make the impedance between capacitor and IGBTs negligible. If you connect a capacitor to the output of the IGBT bridge you then have a circuit comprising two capacitors connected by a switch (IGBT). And negligible impedance.

I'll leave it to you to think about what happens when you close that switch......

If you explain with a diagram.it would be helpful, because I think the two capacitors would then be connected in parallel.

 

[Besoeker]

Here question is about dv/dt and not about di/dt.

Without the resistor, it would become a di/dt problem.

If you explain with a diagram.it would be helpful, because I think the two capacitors would then be connected in parallel.

You want me to draw a circuit with two capacitors and a switch???

In general:
V1 ≠ V2
C1 ≠ C2

 

[mivey]

You want me to draw a circuit with two capacitors and a switch??

Reminds me of the diagram someone posted a while back when asked to explain wire-pulling. They posted a picture with two guys on either end of the conduit, one labeled "push" and the other labeled "pull'.

 

[Besoeker]

Reminds me of the diagram someone posted a while back when asked to explain wire-pulling. They posted a picture with two guys on either end of the conduit, one labeled "push" and the other labeled "pull'.

A nice parallel.....

 

[T.M.Haja Sahib]

Without the resistor, it would become a di/dt problem.


You want me to draw a circuit with two capacitors and a switch???

In general:
V1 ≠ V2
C1 ≠ C2

It is better to start another thread to discuss further because it is neither above nor high dv/dt that causes failure but it is the high harmonic content in the output of VFD that does not permit use of any capacitor on the output side.

 

[Besoeker]

It is better to start another thread to discuss further because it is neither above nor high dv/dt that causes failure but it is the high harmonic content in the output of VFD that does not permit use of any capacitor on the output side.

If, in the above circuit, you closed SW1 at t=0 and V1 ≠ V2, what do you think the current would do?

 

[T.M.Haja Sahib]

If, in the above circuit, you closed SW1 at t=0 and V1 ≠ V2, what do you think the current would do?

The current would, of course, rise. But if the switching is very fast as in IGBT, the rise in current may not be too high to blow a fuse.

 

[Besoeker]

The current would, of course, rise. But if the switching is very fast as in IGBT, the rise in current may not be too high to blow a fuse.

What would limit the current at t>0 ?

 

[T.M.Haja Sahib]

What would limit the current at t>0 ?

It is a practical circuit we are discussing. So in addition to capacitance, resistance and inductance would also be present and together with them, high speed switching would limit the average voltage to such a value that short circuit would not result.

 

[Besoeker]

It is a practical circuit we are discussing. So in addition to capacitance, resistance and inductance would also be present and together with them, high speed switching would limit the average voltage to such a value that short circuit would not result.

What I have given you IS a practical circuit. The physical design deliberately ensures that impedance between the bucket and the IGBT switching elements is negligible for entirely practical reasons.
So what happens at t>0?