Compensation capacitors

[throbscottle]

I'm struggling somewhat to get my head around frequency compensation caps.

Scenario 1. If I have a resistive potential divider of several resistors which can be connected at different nodes to the load (an op-amp + input) using reed switches.
Scenario 2. If I have a resistive potential divider where there are only 2 resistors, but one of them can be switched for different values by reed switches. The common point of the switches is connected to the load (again an op-amp + input)

The top resistor is 9M in scenario 1 or 10M in scenario 2

I know each resistor has some inductance and some capacitance, and that the switches have some inductance and capacitance, and the load circuit has some capacitance (and resistance high enough to ignore). I also know that the ratio of reactance should be the same as the ratio of resistance.

Now, given that I'm somewhat optimistically hoping for a -3db point around 1MHz, how in 7 hells do I work out what caps I should be putting in? Is it as simple (eg if the divider is 10:1) as putting in capacitance of say (50 - parasitic)pF on the lower part of the divider and a 5pF trimmer on the top part? (which is already giving horrendous drop in top resistor's impedance at 10KHz) Or do I have to do complicated maths?

edit: maybe I should just be putting a trimmer in the bottom half

 

[alec_t]

Is it as simple (eg if the divider is 10:1) as putting in capacitance of say (50 - parasitic)pF on the lower part of the divider and a 5pF trimmer on the top part?

That's the way I'd do it. My 'scope has something along those lines. I think the principle is to swamp the unknown parasitic capacitance on the upper part with a known capacitance (as per your ~5pF) to make the division ratio more predictable. Inevitably the impedance drops.

 

[throbscottle]

Thanks for the quick response! I think I'm going to play with just putting trimmers on the lower part first, see what I get. At least I have a 'scope now so I can see where my guesswork is going! (For "trimmer" here it may just be 2 bits of bent wire. This is going to be interesting...

 

[alec_t]

Good luck with that. For a 10:1 divider, stray capacitance will obviously have 10 x the effect on the higher resistance arm than on the lower arm. If you are tweaking the lower arm, can you be sure that the upper arm stray is constant?

 

[crutschow]

Normally in a resistive divider attenuator (such as a scope probe) you are compensating for the stray load capacitance (such as the scope input which is typically 1M ohm and about 50pF) so there is a single compensation capacitor across the high value resistor (10M ohm for a 10:1 scope probe). This capacitor is adjustable (trimmer type) to optimize the square-wave response and its value is about 1/10 of the scope input capacitance.

 

[throbscottle]

Ok did a little experiment with 2 resistors on a bit of perfboard. Only signal source I have at the moment is my 'scopes calibrator 1KHz square wave - not high enough for stray capacitance to have a noticeable effect, so I added in a 22pF cap. Top resistor 10M, bottom resistor I tried 1M, 100K and 10K. Trimmer goes down to a couple of pF. Set trimmer for the 1M, then tried other resistors, observed increasing peaks on rising edges as the flat parts of the waveform got lower. Got distracted with something else then. So it proves I need separate capacitors for each alternative lower resistor, and these are going to need to be trimmable.

Alec, that's a good point. It didn't occur to me it might vary - I'm relying on mechanical stability as it is. The idea was that the existing stray capacitance across that resistor would be so miniscule that a low value trimmer on the lower resistor would be large by comparison. Too cold, late and cluttered to do the messing around with bits of wire.

 

[unclejed613]

http://cds.linear.com/docs/en/application-note/an47fa.pdf

there's a lot of info in this pdf about compensation, including a reprint of a Tektronix app note about how probe compensation works. there's also a lot of info about amplifier stability, and a chater at the end about Murphy's Law.....

 

[throbscottle]

Thank you unclejed, what an incredibly informative document!

 

[MikeMl]

For compensation of a voltage divider R1 over R2, make XC2 =XC1*R2/R1

Look at the sim. The 10:1 probe is compensated when the trimmer is 20pF for a total shunt capacitance at the bottom of the divider of 20+50+20= 90pF.

 

[throbscottle]

Thank you - I have looked at a few descriptions of these things now and they just make my head go round and round (and sometimes they talk about numbers with J in them) - that is a nice and simple formula I can use. Now, I don't have a probe lead to worry about and my amp's input capacitance is only 5pF, so I should be looking at a requirement of 0.5pF across the 10M resistor for a 10:1 divider - ok that's just silly. So if I put a 2pF capacitor there needs extra 15pF on bottom, ignoring the capacitance of the resistors for now. For 100:1 divider needs extra 195pF and 1000:1 needs whopping extra 1.995nF. Still gives me 79K at 1MHz. Pretty sure this will go horribly pear shaped when I try it out. Better dig out the reed switches and build the prototype.

 

[crutschow]

.............................. Now, I don't have a probe lead to worry about and my amp's input capacitance is only 5pF, so I should be looking at a requirement of 0.5pF across the 10M resistor for a 10:1 divider - ok that's just silly. ........................

That's not silly. You can generate such small capacitances with a gimmick capacitor, for example. For 0.5pF you would likely need only a turn of so. You adjust the capacitance by the amount of twist.

 

[throbscottle]

Hmm, well, how about that!

 

[throbscottle]

No need for the gimmick. I built the circuit shown on a piece of perfboard, used a 1-10pF trimmer across the 10M resistor and trimmed it to match the 1M resistor. Used 5% resistors, not the precision ones shown. Load was my 'scopes x10 probe. Added the other caps in based on measured value of the trimmer, didn't need to adjust them. Relay coils will be connected to transistor collectors eventually, using dip-switches for this experiment.

Trouble is my only ceramic trimmer is knackered now and I have to buy one

Pleased with the result, didn't think it would be this simple! Thanks folks!

 

[MrAl]

Hello,

I derived the upper capacitor for the "probe" compensation and basically came up with the same result. When R1 is the upper divider resistor and C1 is in parallel with R1, and R2 is the lower divider resistor and C2 is in parallel with R2, we can find the upper cap C1 with:
C1=(C2*R2)/R1

but another more direct way of looking at it is to look at it as actually two different dividers, one made of resistors (R1 and R2) and one made of capacitors (C1 and C2).

If a voltage were to appear across the divider R1 and R2 with no capacitors, we would want to see a voltage across R2 that is equal to R2/(R1+R2). So if R1 was 9 ohms and R2 was 1 ohm then we'd want to see 1 volt across R2 if there was 10 volts applied.
Well the same has to hold for the capacitors, looking at the applied voltage as if it was an AC voltage or a transient. The voltage divider voltage across C2 we then want to see is:
C1/(C1+C2)

so with C1=1 Farad and C2=9 Farads, we get the same voltage: 1 volt across C2 with 10 volts applied.

Note the slight difference in the two equations though. One has R2 in the numerator and one has C1 in the numerator, although the denominators are still adding together both component values.
Also note that the C1 cap is small compared to C2, whereas R1 is large compared to R2. That's because the impedance of each element is related to the inverse of the capacitance so we see an 'opposite' effect for caps than we do for resistors.

So since the two equations must come out to the same ratio, we can equate them and that allows us to simplify it to:
C2*R2=C1*R1

and this is easily solved for either one of the four values knowing the other three.

The "compensation" capacitor would be C1, the upper cap, which is used to compensate for the fact that C2 is in the circuit naturally as a shunt capacitance because of factors we can not control. The impedance does get a little messed up that's true, but that's life, and it wont be a pure shunt capacitance presented to the load because the higher the frequency (and the faster the rise time) the faster the unfavorable transient current expires, and also the caps are never perfect and always have at least some small series resistance.

 

[throbscottle]

As always a complete and clear explanation from you, MrAl

 

[MrAl]

Well thank you throbscottle, i like to try to add a little depth to the discussions which i think makes the circuits more interesting.

I should have also added that there is also some line inductance in series with R1 so the transient current surge should be a little less anyway, although we always need to know the equivalent total shunt capacitance for a probe as well as the resistance (like "1 Megohm, 10pf" or something like that).

 

[throbscottle]

After this discussion I realised that the trimmer actually needs to be rated at at least 1KV since that is the intended max voltage for the meter, and I don't trust the "gimmick" idea at that voltage. The highest voltage ones (that don't cost a small fortune) I can find are rated 500VDC, so I was wondering, if I make this a series arrangement with a fixed HV capacitor of smaller value (3.3pF for now) than what the trimmer will be set to, this will protect the trimmer I know you are not **supposed** to put capacitors in series to increase the working voltage like this, but it should be ok since the smaller, fixed capacitor will bear the brunt of it.

I also thought I could make a capacitor using a little bit of d/s pcb and adjust the value by trimming some copper off - is this a better solution?

 

[rumpfy]

Yes.
Mr AL has pointed out that the compensation components are not always 'pure'. This problem gets worse when you try to get high attenuation values in a 1 stage attenuator like you are doing. With increasingly high values of series resistor, they are traditionally manufactured as a thin deposit of conductive material and then it is cut into a spiral, which adds a lot of inductance. This kind of attenuator can be made to work well enough up to say 1 mHz and with real care up to say 10 mHz, but over that, you really need to distribute the capacitance along the resistive bits. Tektronix make attenuators up to 10 mHz from discrete parts, but when the bandwidth is say 60 mHz, the attenuators are make as subassemblies.
On one occasion I had to make a 60:1 attenuator to measure the anode voltage on a valve line output stage (6000 Volt with a bandwidth of 500 kHz) and this was not too hard. The series section was divided into 10 sections and this allowed suitable values of capacitance. As the divide ratio increases, the capacitor(s) in the top section become smaller. On another occasion, I needed a attenuator of 1000;1 with a bandwidth of 50 kHz to attenuate a 10 kV pulse. The series resistor I used was a 200 megohm thing that had heaps of inductance. The best bandwidth was a achieved by wrapping a copper foil around the resistor and spaced from it. It was all a bit suck it and see, but your query about "can I use some pcb material" is the kind of flexible thinking you need to have in solving these attenuator problems.
Pulse testing tells you VERY quickly about the bandwidth of your design.
Hope this helps.

 

[MrAl]

After this discussion I realised that the trimmer actually needs to be rated at at least 1KV since that is the intended max voltage for the meter, and I don't trust the "gimmick" idea at that voltage. The highest voltage ones (that don't cost a small fortune) I can find are rated 500VDC, so I was wondering, if I make this a series arrangement with a fixed HV capacitor of smaller value (3.3pF for now) than what the trimmer will be set to, this will protect the trimmer I know you are not **supposed** to put capacitors in series to increase the working voltage like this, but it should be ok since the smaller, fixed capacitor will bear the brunt of it.

I also thought I could make a capacitor using a little bit of d/s pcb and adjust the value by trimming some copper off - is this a better solution?

Hi there,

If i understand you correctly you want to put two caps in series in order to raise the voltage rating of the original cap. Well, this works to some degree but there is a catch. The catch is that in order to double the voltage rating (and that is usually the goal) the two capacitor values have to be the same. So if you have a 1uf 100v cap in series with a 1uf 100v cap you might get by with using them up to 200v in theory, but limiting that to 150v would be a better idea. But for this discussion lets go by pure theory for now for and remember to derate by some amount like 25 percent of the total.

So with 1uf and 1uf we roughly double the voltage rating. But this 'works' to some degree because we had both caps the same value and that means the voltage divides equally between the two caps. But if we had different values then the voltage would not divided equally between the two caps so the rating would not roughly double like it did before. In fact, the new rating would highly depend on the two values as well as their individual voltage ratings.

With 1uf 100v and 2uf 100v for example, two thirds of the voltage appears across the 1uf cap so that means we end up with only one third across the 2uf cap. This means we'd only gain about 33 percent of the original rating, which in this case is only 33 volts. So the rating with the two together would only amount to about 133 volts, and that's before any safety derating. With two 1uf caps or two 2uf caps we doubled the rating which is a 100 percent increase, but with one with a value twice as high we only went up by 33 percent. So you see that it quite different.

For two caps one with a lower value CL and one with a higher value CH the increase factor 'A' is:
A=CL/(CL+CH)

so the total increase is 1+A.

The example was 1uf and 2uf, so there CL=1 and CH=2 (same units) and so A comes out to:
A=1/(1+2)=1/3

so the total increase is:
1+1/3=4/3

and that works out to about 133 percent. Derating that by about 25 percent we get:
133*0.75=99.75 percent, which is still around 100 volts.

But it's always best to get the right rating to begin with. Spend a little more and it should last a long time.

 

[rumpfy]

You are building an attenuator, and for every series resistor there is a corresponding voltage drop. If you have trouble selecting a suitable series resistor, then split it into smaller values and increase the shunt capacitor across each one.
So if you want a attenuator with a 10:1 ratio, feeding into a 1 megohm load in parallel with 50 pF, and you do it with two 4.5 megohm resistors in series, then you get the attenuation is 1/ (4.5 + 4.5 + 1 ) = 1/10. The 50 pF load capacitance gives a time constant of 1 x 50 = 50 micro sec. The time constant of the series bit is 50 microsec, so the shunt capacitance is 5 pF. If you split the series R into two x 4.5 megohm, the shunt capacitance for each resistor is 10 pF. The voltage rating should be 450 volt; which is a lot. So split the series R into 3, with each resistor being 3 megohm and each shunt being 15 pF. The voltage rating then is 300 volt which is more manageable. You can keep going with this approach but you need to consider the shunt capacitance at each step if you want a good bandwidth and you get up to 8 or so series resistors.