How to crappify a cap

[kfitch42]

Ran across this http://mightyohm.com/blog/2013/02/simple-fix-for-bus-pirate-power-supply-oscillations/

The end result of that is he solved a problem by adding a high(er) ESR cap in parallel with a low ESR cap.

That got me thinking (Yeah dangerous I know :) ... just for the sake of argument, can I effectively increase the ESR of cap by adding a resistor?

Would adding R1 (see attached) be the right way to add 1 Ohm of ESR?

[c4757p]

Yes.

Except "in" and "out" should be the same node.

[kfitch42]

Ahhh, think I have it now.

Would it matter if I put the resistor on the other side of the cap?

[c4757p]

It is not uncommon to see discrete analog circuits stabilize capacitors like this with separate resistors from a couple ohms to a couple tens of ohms.

Would it matter if I put the resistor on the other side of the cap?

Not one bit.

[PA0PBZ]

Interesting that on the blog he paralleled the already existing cap with another one.
Not sure what the net effect is but it will not increase the ESR.

[c4757p]

Good point, I misunderstood the blog entry. I thought he was saying that the cap he replaced was parallel to another one, like how large electrolytics are often bypassed with small ceramics, and that he changed the larger one and kept it parallel to the smaller one. Not sure what's going on here.

[mariush]

That's what they get for not respecting what datasheets say and for trying to be cheap
. 
They could have used a 22-33uF 10v electrolytic from the start and still be fine, just the initial output voltage will ramp up a bit slower. And such electrolytic would have high enough esr to not be an issue.

[jimmc]

As well as causing stability problems with the voltage regulator, a low ESR ceramic decoupling capacitor can form a high Q resonant circuit with the inductance of the power supply tracks and make matters worse at certain frequencies.
Connecting a suitable low value resistor across the ceramic cap will damp out this resonance but will also short out the supply!
Adding a capacitor in series with the resistor will prevent the supply short.
Using a capacitor with a poor ESR is equivalent to a good capacitor with a series resistor so no additional resistor is required.

Jim

[DavidDLC]

I just checked my Bus Pirate v4.0 and it has the same problem on the 3.3 V power supply rail.

I need to make the fix.

David.

[gotwood]

can I asked what you used to draw your schematic?  freeware?

[kfitch42]

can I asked what you used to draw your schematic?  freeware?

LTSpice ("free as in beer" ware)... windows only, unfortunately. One of these days I will have to see how well it runs under Wine.

I drew it in LTSpice, hit "Print Screen", pasted into MS Paint, cropped, saved as .png.

[tld]

LTSpice ("free as in beer" ware)... windows only, unfortunately. One of these days I will have to see how well it runs under Wine.

I'm using LTSpice under Wine, on OS X.

tld

[c4757p]

Running perfectly under Wine on Debian for me.

[amspire]

Interesting that on the blog he paralleled the already existing cap with another one.
Not sure what the net effect is but it will not increase the ESR.

Yes it will.

Sounds wrong - how can a resistor in parallel with zero ohms increase the resistance?

The reason is the ESRs of the two caps are not directly in parallel.

If you had a perfect 1uF cap with zero ESR and put a 1uF electro with 1 ohm ESR in parallel, you will end up with a net 2uF capacitance with ESR. The point is at lower frequencies, half the current is flowing through the second cap, and that current is seeing ESR. Some resistive power is being dissipated and hence you have ESR.

Obviously if the frequency goes well above the  f =  1/(2*pi*R_esr*C) frequency, more of the current will start to go through the perfect zero ESR cap and so the net  ESR will disappear.

You can actually guess the approximate value of the ESR of the 1uF electro used to snub the oscillations: it would have to add a zero at 15Khz or a bit under. A zero at 15KHz would be 10 ohms ESR using the equation above, so the 1uF electro he picked probably had an ESR roughly in the 5 to 10 ohms range. Above that or below that, it would probably not have stopped the oscillation.

[PA4TIM]

Adding a capacitor in series with the resistor will prevent the supply short.
Using a capacitor with a poor ESR is equivalent to a good capacitor with a series resistor so no additional resistor is required.

Jim

Yes and no, the effect is the same but it is ( in some cases) better to dissipate the power in a external resistor as in the ESR of the capacitor because dissipation is heating, and heat is the one thing caps do not like.

[The Electrician]

Not sure what the net effect is but it will not increase the ESR.

This seems to be a common misunderstanding.  If a couple of caps are parelleled, the ESR of the parallel combination will be greater than the ESR of the cap with the smallest ESR, if the frequency is less than:
Sqrt[(C2*R2 - R1*(2*C1 + C2))/(4*C1^2*C2*Pi^2*(R1^3 + R1^2*R2))]

which is around 12 MHz or so.  At two orders of magnitude or so below this frequency, the ESR of the parallel combination will be fairly well approximated by:

ESR~(C1^2*R1 + C2^2*R2)/(C1 + C2)^2

which is the case here.

I took a high quality 1 uF film cap in place of the ceramic cap mentioned in the first post and measured its ESR versus frequency.  I did the same for a 10 uF electrolytic.  The ESR of the electrolytic is much higher than that of the film cap.

Without a full analysis, one might think that the ESR of paralleled caps must behave like the resistance of a pair of resistors--the paralleled result should be less than either resistance before paralleling.  But, the mysteries of complex arithmetic don't give this result for the paralleled caps.

The first image gives a mathematical analysis:

Here are the sweeps of the caps on the impedance analyzer.  The second image shows the ESR of the 1 uF film cap which I'm using in place of a 1 uF ceramic cap.  The third image shows the ESR of the 10 uF electrolytic.  The fourth image shows the individual cap ESRs plus the ESR of the parallel connection of the two caps.  The middle curve is the ESR of the combination.

Note that the ESR of the combination is much greater than the ESR of the 1 uF cap by itself, and the value is essentially what is predicted by the mathematical analysis.

This is not unexpected if we think about the law of conservation of energy.  Putting the cap with the high ESR in parallel must dissipate some additional power in its ESR, and that additional loss will show up as an increase in the ESR of the combination.  Edit:  This sounds good, but it isn't always true.  See my later post.

[PA0PBZ]

I sort of see what you mean, but I'm still confused. Using 2 equal caps gives me half the ESR, but the frequency formula turns negative so I'm not sure what to do in that case. (Both caps 1uF, ESR 0.01 gives ESRparr 0,005 but a maximum frequency of Sqrt[-2.5E14])
I probably have to draw the complex vectors to see what happens here...

[AlfBaz]

Sqrt[(C2*R2 - R1*(2*C1 + C2))/(4*C1^2*C2*Pi^2*(R1^3 + R1^2*R2))]

Just testing latex...



Look about right?

[AlfBaz]

I'm curious Leco, in that first image you say
let z1=r1 + Xc1. Is it because you are using the complex reactance that we dont need to do the sum of squares... ie 1/j = -j, which takes the 90deg phase shift and direction into account?

[PA4TIM]

This is not unexpected if we think about the law of conservation of energy.  Putting the cap with the high ESR in parallel must dissipate some additional power in its ESR, and that additional loss will show up as an increase in the ESR of the combination.

Can you explain that ? It think you are forgetting something. If the Current is 1 A,  then all current goes in that cap through the ESR of the cap. If you take two caps, you still have 1A total current. But now divided over 2 caps. So less current per cap and less current through each ESR

The problem is you have DC behaviour and AC behaviour.
For DC the inrush current is dictated by the capacitance and limitted by the ESR ( the RC time) once the inrush current is passed and the voltage over the cap is max only the ESR dictates the ripple current dissipation. So both caps will supply current to the circuit ( buffer), only limmited by there stored energy and ESR.
A small capacitance with low ESR will be empty during a surge while the big one still is delivering current, but if the small one is empty the big one has to supply the extra. Total current draw stays the same, dissipation can alter for good or worse. If the big one has lower ESR the dissipation will become lower if the smal one is empty. Because all current now goes through a lower ESR.

But at AC ( what you measure) the current through a cap is dictated by the reactance and ESR. If the high ESR cap has a low reactance at that frequency, most current will flow through that cap because of the much lower reactance and a small part through the second cap with low esr and high reactance.

So the dissipation has to be calculated per frequency and per cap. And in this case will be higher if only one cap was used.( all current then goes through the high ESR)

If the low reactance cap has a low esr and the other a high esr and high reactance the dissipation will be lower as in the first example but total dissipation wil be less because only a small part of the current goes through the high esr cap.

If frequency goes up, reactance of both becomes so small, the ESR is going to dominate and in that case they behave as two parallel resistors.

In your plot you see this happen. The ESR dives down at the end.

Today I took 4 MKT caps parallel. The ESR of one single cap was much higher as the 4 parallel.

1 philips 10 % 250V MKT 2.2 uF, measures 0.371 Ohm at 1KHz and 0.078 Ohm at 100 KHz.

The 4 caps parallel including extra copper to connect them, so extra ESR :
At 1 KHz  i measure 0.097 Ohm ( if you correct for the extra copper 1/4th of the ESR)
At 100 KHz i measure 0.020 Ohm, ( again about 1/4th of the ESR of one cap)

Edit: hope it now makes more sense.

[The Electrician]

I sort of see what you mean, but I'm still confused. Using 2 equal caps gives me half the ESR, but the frequency formula turns negative so I'm not sure what to do in that case. (Both caps 1uF, ESR 0.01 gives ESRparr 0,005 but a maximum frequency of Sqrt[-2.5E14])
I probably have to draw the complex vectors to see what happens here...

The behavior of the ESR of a pair of paralleled caps is somewhat counterintuitive, to say the least!  The result depends on just how different the ESR of one cap is from the other, and the difference in the capacitances too.

Further investigation shows that the frequency formula you're referring to only gives a useful result if the ESR of C1 is less than (C2*R2)/(2*C1+C2).  To understand this see the attached images.  In these images and in the formulas, R1 is the ESR of C1 and R2 is the ESR of C2.

In the first image, the red horizontal line is the value of the ESR of C1.  The ESR of C1 is less than (C2*R2)/(2*C1+C2) in this case.  The blue curve is the ESR of the parallel combination of the two caps.  Just above the graphs is listed the values of the capacitances and the ESRs.  Note that at low frequencies, the ESR of the combination is GREATER than the smallest ESR, but there is a crossover frequency, above which the combination ESR is LOWER than the smallest ESR.  The frequency formula is giving that crossover frequency.

In the second image the ESR of C1 is greater than (C2*R2)/(2*C1+C2).  In this case, there is no crossover of the parallel ESR with the ESR of C1; the parallel ESR is less that the ESR of C1 for all frequencies and the frequency formula fails.

The lesson here is that paralleling a good cap with a crappy cap doesn't always crappify the good cap, even at low frequencies.  You just have to be sure that the crappy cap is enough crappier than the other one, and the crossover frequency is high enough for your purpose.  I did not realize this myself; this shows the value of a proper analysis.

[The Electrician]

I'm curious Leco, in that first image you say
let z1=r1 + Xc1. Is it because you are using the complex reactance that we dont need to do the sum of squares... ie 1/j = -j, which takes the 90deg phase shift and direction into account?

The answer is yes.  i find the arithmetic much easier if the computations are done with complex numbers.  I also find that doing things this way avoids mistakes.

For this particular problem, we specifically wanted the real part of the impedance (that's what ESR is), and doing all the computations with complex numbers made it easy to extract the real part.

[The Electrician]

But at AC ( what you measure) the current through a cap is dictated by the reactance.

It isn't dictated by the reactance alone; what about the ESR?  I know you are aware that the ESR is involved, but in this sentence you forgot to include it.  This statement makes it sound like the reactance alone dictates the current.

If the high ESR cap has a low reactance at that frequency most current will flow through that one and a small part through the cap with high esr and high reactance. So the dissipation has to be calculated per frequency and per cap. And in this case wil be lower if only one cap was used.

In the first part of the first sentence you refer to "the high ESR cap" as though the other cap will NOT be a "high ESR cap", but then later in the sentence you are apparently referring to the second cap, and you call it "the cap with high esr".  I find this confusing.

I am aware that English is not your native language.  With all due respect, and with no intention to be insulting, I must say that sometimes your sentence structure is not clear, and you seem to be saying (or not saying) things that you don't intend to say, or you fail to say what you really intend to say.

I have visited your web site, and I'm impressed by (and envious of) your collection of instruments.  You have obviously spent a lot of time experimenting with measurements of capacitors and inductors, and I enjoy reading your results.  I hope you post more, but sometimes I have difficulty understanding what you're saying.   

Sometimes I disagree with what you say, but sometimes when I appear to disagree, the reason is that I'm trying to clarify something you have said that wan't clear.

These results are an example of what I find unclear.  I think they would have been better presented in the form of a chart.

Today I took 4 MKT caps parallel. The ESR of one was much higher as the 4.
1 MKT 2.2 uF At 1KHz 0.371 Ohm, at 100 KHz 0.078 Ohm
The 4 the same parallel including extra copper to connect them, so extra ESR :
1 KHz  0.097 Ohm ( if you correct for the extra copper 1/4th of the ESR)
At 100 KHz 0.020 Ohm, again about 1/4th of the ESR of one cap.

At any rate, in my post above, I extend my analysis and show fully how the ESR of the parallel combination depends on frequency and on the relative values of the ESR and capacitance of the two capacitors.

[The Electrician]

I was looking at the expression for the parallel ESR of two caps, and decided to see what it becomes at very low frequencies.

The attachment show limit formulas for the value of the parallel ESR as the frequency goes lower and lower.

The first expression is for different cap and ESR values.  The second expression is for identical cap values, but possibly different ESRs.

[The Electrician]

It's always good to have experimental verification of a mathematical result that is counterintuitive.

I picked out a couple of mylar capacitors with ESR versus frequency curves as nearly identical as possible.  The first image shows three sweeps of the two .47 uF caps.  The top two curves are the ESR of the two caps individually.  The bottom curve is the ESR of the two caps in parallel.

The second image shows the same three sweeps as the first image, plus a sweep of one of the caps with an additional 4 ohm resistor in series with it.  The top curve is of the cap with 4 ohms in series.

The condition for the existence of a parallel ESR that is greater than the ESR of the better cap is that R1 is less than (C2*R2)/(2*C1+C2).  If we set C2=C1, this expression becomes R1<R2/3.  So if we have caps with the same capacitance, with ESRs of R1 and R2, R2 must be greater than 3*R1 for a parallel ESR greater than the ESR of the better cap to exist.

The experimental results are not quite straightforward because the ESR of these capacitors varies with frequency.  But, if we can find a frequency where R2 is greater than 3*R1, we should see the ESR of the parallel combination being greater than the ESR of one cap alone.

The third image shows the previous curves plus one new one.  The new curve is the ESR of a parallel combination of one of the caps with a 4 ohm resistor in series and the other cap without any additional resistance in series.  This is to show the effect of a crappy cap (the one with an additional 4 ohms in series) in parallel with a good (sort of) cap.

The ESR of the combination starts out less than the ESR of a single cap, but at about 1.13 kHz, the ESR of the combination becomes greater than the ESR of a single cap, until at about 500 kHz it is again less than the ESR of a single cap.

Note that at the frequency of marker A, the ESR of the cap with the 4 ohm resistor in series is about 5.8 ohms.  At that same frequency, the ESR of a single cap is about 1.78 ohms.  That is about the 3 to 1 ratio the mathematics predicts should be where the crappy cap can cause the parallel ESR to be greater than the ESR of the "good" cap, and indeed that is where the ESR of the combination becomes greater than the ESR of the "better" cap.