Is there a way to measure the equivalent capacitance from a node to ground in a circuit?

Thank you.

I tried to test the circuit below by using captab to measure the capacitance from node A to ground.

What I expected is C1 + C2 (this is C1|| C2). However, the result from cabtab is only C2.

The page doesn't allow me to attach an image now. I will post the image link of the picture below.

The circuit is very simple VDD- C1 -C2 - GND. Where A is the point between C1 and C2 and C1 and C2 are in series.

http://postimg.org/image/up31n9uup/

I suspect you must have picked the detail=nodetoground option rather than the default "node" or the other alternative "nodetonode" option. If you pick "nodetoground" it only includes the capacitance to ground, not the effective capacitance to ground. 

With this netlist:

// Example of various captab detail options

VDD (vdd 0) vsource dc=1.8
C1 (a vdd) capacitor c=1p
C2 (a 0) capacitor c=1p

dc dc
captabDefault info what=captab where=screen
captabNodeToNode info what=captab where=screen detail=nodetonode
captabNodeToGnd info what=captab where=screen detail=nodetoground

Then I get this in the output (note I only posted the relevant part):

**********************************
Capacitance Table `captabDefault':
**********************************
Capacitance values computed in DC analysis `dc' at T = 27 C.

a : a fixed=2 p , variable=0 , sum=2 p .

vdd : vdd fixed=1 p , variable=0 , sum=1 p .

*************************************
Capacitance Table `captabNodeToNode':
*************************************
Capacitance values computed in DC analysis `dc' at T = 27 C.

a : a fixed=2 p , variable=0 , sum=2 p .
a : 0 fixed=1 p , variable=0 , sum=1 p .
a : vdd fixed=1 p , variable=0 , sum=1 p .

vdd : vdd fixed=1 p , variable=0 , sum=1 p .
vdd : 0 fixed=0 , variable=0 , sum=0 .
vdd : a fixed=1 p , variable=0 , sum=1 p .

************************************
Capacitance Table `captabNodeToGnd':
************************************
Capacitance values computed in DC analysis `dc' at T = 27 C.

a : 0 fixed=1 p , variable=0 , sum=1 p .

vdd : 0 fixed=0 , variable=0 , sum=0 .

As you can see from the lines I highlighted in red, you are getting with the a:a lines the total capacitance on a node.

BTW, the attach image problem for me seems to be broken in Safari currently but OK in Firefox. I'll speak with our IT folks who manage the forums to see if it can be addressed (it's happened before and normally needs something resetting I think).

Regards,

Andrew.

Thank you very much for the detailed answer.
You are right that I chose detail=nodetoground.
However, I see that the other options (node or nodetonode) also don't return the
effective capacitance to ground as in the example below.
I have not simulated it yet but from how capacitance is calculated in each case I don't think the result is right.
The circuit below includes a bjt transistor and a Miller capacitor between
base and collector.

Here is the link to the circuit image: http://postimg.org/image/xvekod2wh/

(I tried both Chrome and Firefox but both didn't work now).

I would like to measure the effective capacitance from node A to ground.
What I expect is (A+1)*C where A is the gain of the inverting amplifier.
However, the results that simulator returns will be something as below.
(assuming that the transistor is ideal and there are no parasitic capacitors in the transistor)
1. detail = nodetoground
A : 0 - result: 0
2. detail = node
A : A - result: C
3. detail=nodetonode
A : 0 -result: 0
A : collector - result: C

The result is "right" but it's not doing what you're expecting. It would rather have helped if you had explained at the beginning you were aiming to measure the effect of Miller capacitance, which is not what happens with captab. In essence the Miller effect is telling you the equivalent input impedance caused by an impedance between the input and output of a gain stage - by "effective" I assumed you meant the capacitance to all nodes as if it was referred to ground.

Given that the Miller capacitance will be frequency depended (because the gain will be frequency dependent), this isn't something the simple captab analysis would give you, which is based on all the capacitances in the devices rather than any consideration of the gain (it doesn't run at a particular frequency, so it couldn't do that anyway).

I've never thought about how you would simulate to compute this for all nodes - I'm sure you could simulate it for a specific node by injecting a sharp transition and measuring the effective time constant (or something like that?). Most of the time as a designer, I've just been happy to take advantage of it to allow compensation capacitors to be smaller, rather than worrying about simulating what the equivalent input capacitance would be. 

Regards,

Andrew.

Thank you.

Sometimes, I would like to know the input/output capacitance or equivalent capacitance from a node to ground for some purpose.

As you said, it can be done by injecting a sharp transition and measuring the effective time constant. I have just given it a try. However, how would you inject the sharp transient? The equivalent capacitance will depend on the biasing condition so the injected source should not change the biasing condition as well. I am stuck at this.

As I said, I've never really thought about doing this - usually just knowing the actual capacitance has been enough. 

I would expect that you could connect a source (maybe via a resistor) to the node where the voltage source steps between the bias voltage and a small increment above the bias voltage (in order to not cause a large-signal change in the bias condition). I've not tried doing this though... (and won't have time in the coming week to make this kind of experiment).

Regards,

Andrew.