Trouble understanding calculating capacitor ESR from dissipation factor

Thread Starter

SiCEngineer

Joined May 22, 2019
207
Hello,

I am trying to calculate the ESR of some high voltage capacitors so I can model them in LTSpice.
Taking an 8kV CG0 capacitor with 6.8nF of capacitance, and a dissipation factor of maximum 0.1% @ 25 degrees celcius, the calculation yields:

ESR = DF/Xc = 0.1/(2*pi*1MHz*6.8^10-9)

= 2.341Ohms.

Now, I am no expert, but that seems quite large of an ESR for a NPO/COG capacitor, despite even the high voltage.
I know that the ESR of the capacitor is directly related to transient response. Therefore this high value of ESR will probably cause a large deviation in output voltage, which I definitely am not able to tolerate with my converter.

I used 1MHz because the data sheet tends to specify their data at this frequency. Maybe this is where I am going wrong.
The reason I looked into C0G was because it was supposed to have much lower ESR than other high voltage capacitors and is stable over the operating temperature, freq. etc... but the ESR seems to be no better, leading me to believe I'm probably misunderstanding something.
 

MrAl

Joined Jun 17, 2014
7,592
The dissipation factor is the ratio of energy lost in the ESR and magnitude of the reactive energy of the capacitance. That means if you multiply the dissipation factor by the magnitude of the reactive energy you get the ESR:
ESR=DF*magnitude(Xc)
which it looks like you have done.

So if you get 3 Ohms then you get 3 Ohms.
I think the dissipation factor goes down with frequency but you should check the data sheet of the capacitors you intend to use. The dissipation factor probably goes down with increasing temperature up to a point and then it probably goes up again, but again check the data sheet.

When the ESR of a single cap is too large for converter application the typical thing to do is place several capacitors in parallel. Each cap only has to be 1/N times the original value where N is th number in parallel. The ESR is then also 1/N if all the caps have the same capacitance value.

So for your example if you used 10 caps in parallel each with a value of 6.8e-10 farads and DF=0.1 the total ESR would be 0.3 Ohms and the capacitance would be 6.8nf.
The key is to find lower value capacitors that have the right DF, or just go with a larger total value capacitance for your application.
Another key point is to wire them in parallel with as little lead length as possible to avoid adding more ESR to the mix.
 
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Thread Starter

SiCEngineer

Joined May 22, 2019
207
The dissipation factor is the ratio of energy lost in the ESR and magnitude of the reactive energy of the capacitance. That means if you multiply the dissipation factor by the magnitude of the reactive energy you get the ESR:
ESR=DF*magnitude(Xc)
which it looks like you have done.

So if you get 3 Ohms then you get 3 Ohms.
I think the dissipation factor goes down with frequency but you should check the data sheet of the capacitors you intend to use. The dissipation factor probably goes down with increasing temperature up to a point and then it probably goes up again, but again check the data sheet.

When the ESR of a single cap is too large for converter application the typical thing to do is place several capacitors in parallel. Each cap only has to be 1/N times the original value where N is th number in parallel. The ESR is then also 1/N if all the caps have the same capacitance value.

So for your example if you used 10 caps in parallel each with a value of 6.8e-10 farads and DF=0.1 the total ESR would be 0.3 Ohms and the capacitance would be 6.8nf.
The key is to find lower value capacitors that have the right DF, or just go with a larger total value capacitance for your application.
Another key point is to wire them in parallel with as little lead length as possible to avoid adding more ESR to the mix.
okay, thanks for that brilliant explanation it really helped. One thing though. If I split say a 0.06uF capacitor into three 0.02uF as you describe, assuming the same DF, the ESR of the capacitor will increase by 3. Because the capacitor reactance will reduce by 3. So paralleling three of them will give me the total capacitance, but how does it give a third of the ESR if the ESRof the new smaller capacitors is 3 times larger than the single capacitor alone? ... god, that makes sense in my head - but let me know if I am getting confused! Cheers.
 

MrAl

Joined Jun 17, 2014
7,592
okay, thanks for that brilliant explanation it really helped. One thing though. If I split say a 0.06uF capacitor into three 0.02uF as you describe, assuming the same DF, the ESR of the capacitor will increase by 3. Because the capacitor reactance will reduce by 3. So paralleling three of them will give me the total capacitance, but how does it give a third of the ESR if the ESRof the new smaller capacitors is 3 times larger than the single capacitor alone? ... god, that makes sense in my head - but let me know if I am getting confused! Cheers.

Hello again,

I think you are right in questioning this because i may not have explained it correctly the first time.

What i meant to say was that if you have one cap that was 3uf and ESR=3 Ohms, if you find three 1uf caps that each have ESR=3 Ohms then you connect them in parallel, the capacitance is again 3uf but the ESR is 3/N which means it becomes 1 Ohm.

So really you have to find lower value capacitors that have the same ESR or just parallel the larger value caps if you can use the higher value uf.

As you can now see it is easier to think of in terms of the ESR rather than the DF. The capacitance value itself is in the denominator of the reactance Xc so as the capacitance goes down with constant DF we get higher ESR as i think you noted, so that would mean the DF would have to be 3 times SMALLER in our 3 cap examples.
But to be sure, just calculate the ESR's and go from there.

Oh yeah BTW because power dissipation is then distributed over 3 caps instead of just 1, if the body size is similar the temperature rise should be much less for each cap so the longevity goes up. The temperature rise should go down to roughly 40 percent. With 10 caps down to roughly 20 percent, very roughly.
 
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Thread Starter

SiCEngineer

Joined May 22, 2019
207
Hello again,

I think you are right in questioning this because i may not have explained it correctly the first time.

What i meant to say was that if you have one cap that was 3uf and ESR=3 Ohms, if you find three 1uf caps that each have ESR=3 Ohms then you connect them in parallel, the capacitance is again 3uf but the ESR is 3/N which means it becomes 1 Ohm.

So really you have to find lower value capacitors that have the same ESR or just parallel the larger value caps if you can use the higher value uf.
Perfect, thanks for your help!
 

Thread Starter

SiCEngineer

Joined May 22, 2019
207
0.1% expressed in decimal is 0.001.
another question if you don’t mind.. the large majority of data sheets I’ve seen for capacitors specify the frequency of test for the caps as either 1kHz or 1MHz depending on the size of the capacitance. My capacitance values often fall into the 1kHz values - but my power supply switches > 100kHz. Should I use 100kHz for the f value within the capacitor reactance? It makes sense to me to use the value at which I am actually switching the capacitors, but what about for DC output and input capacitors?
 
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