Finding the Capacitance from Lissajour plot info?

[Wayne]

If I have the Phase then say 40deg and the magnitude say 10 then how can I
convert that into capasitance.
I have the following circuit:

___SIGGEN_____
Ś Ś
Ś--/\/\/\/\-----Ś Ś---Ś
R_shunt Cx
Ś Ś
----V?----

I am measuring the voltage drop accross R_shunt and comparing that with the
siggen so I have V_R_shunt - SIGGEN = Z and on a scope I can measure
the phase. In other words I have the equiverlant info as you would find
when creating a Lissajour plot.
How can I find the capacitance of Cx with the info I have here?

Cheers

Wayne

 

[Tim Wescott]

Wayne wrote:

If I have the Phase then say 40deg and the magnitude say 10 then how can I
convert that into capasitance.
I have the following circuit:

___SIGGEN_____
Ś Ś
Ś--/\/\/\/\-----Ś Ś---Ś
R_shunt Cx
Ś Ś
----V?----

I am measuring the voltage drop accross R_shunt and comparing that with the
siggen so I have V_R_shunt - SIGGEN = Z and on a scope I can measure
the phase. In other words I have the equiverlant info as you would find
when creating a Lissajour plot.
How can I find the capacitance of Cx with the info I have here?

Cheers

Wayne



It's not clear from your schematic or your text -- you have a signal

generator that is connected to a resistor and capacitor in series, and
the resistor is grounded? Or is the capacitor grounded?

The following answers assume the resistor is grounded.

Method 1:

Get the RMS voltages of both the signal and the resistor voltage.
Ignore phase, which means that you assume the capacitor is purely
reactive. Then

V_c^2 + V_r^2 = V_s^2,

where V_c is the capacitor voltage, V_r is the resistor voltage and V_s
is the signal generator voltage. This implies that V_c = sqrt(V_s^2 -
V_r^2).

Now, the current is I = V_r/R, and the capacitive reactance is X_c =
V_c/I, so you get X_c = R*sqrt(V_s^2 - V_r^2)/V_r. The capacitance can
be found from the capacitive reactance and frequency, C = 1/(2*pi*X_c).

Method 2:

Assume that the capacitor isn't purely reactive. Measure the amplitude
and phase of V_r respective to V_s, and solve for the capacitive
impedance Z_c (note that you have to use complex arithmetic):

V_r
Z_c = -----------.
V_s - V_r

Z_c will, in general, be complex, so it will be Z_c = R_c - jX_c, where
R_c is the capacitor's equivalent series resistance and X_c is the
capacitive reactance. If Z_c is purely imaginary all is well and good.
If Z_c has a significant resistive component then it is up to you to
decide if this is measurement error or the fault of the capacitor, and
whether it is best modeled as a series resistance, a parallel
resistance, or something more complicated.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[Wayne]

Thanks Tim.

The resistor is in series with the capacitor and the signal gen is across
one lead of the capacitor and one lead of the resistor. In other word, if
they were both resistors it would be a voltage divider. The resistor or
the capacitor can be ground.
Forgive me for asking a dumb question but why are V_c^2 + V_r^2 = V_s^2 to
the power of 2?
If this was a voltage divider then you would have V?=Vs(R/R+Rc)?

In method 2 could you give me an example with the following figures?

Vs=10v p-p
Vr=4v p-p
Phase = 45deg (with respect Vs)

Cheers

Wayne


Coul
"Tim Wescott" <tim@wescottnospamdesign.com> wrote in message
news:10ldvo2errd1829@corp.supernews.com...

Wayne wrote:

If I have the Phase then say 40deg and the magnitude say 10 then how can
I
convert that into capasitance.
I have the following circuit:

___SIGGEN_____
Ś Ś
Ś--/\/\/\/\-----Ś Ś---Ś
R_shunt Cx
Ś Ś
----V?----

I am measuring the voltage drop accross R_shunt and comparing that with
the
siggen so I have V_R_shunt - SIGGEN = Z and on a scope I can measure
the phase. In other words I have the equiverlant info as you would find
when creating a Lissajour plot.
How can I find the capacitance of Cx with the info I have here?

Cheers

Wayne



It's not clear from your schematic or your text -- you have a signal
generator that is connected to a resistor and capacitor in series, and the
resistor is grounded? Or is the capacitor grounded?

The following answers assume the resistor is grounded.

Method 1:

Get the RMS voltages of both the signal and the resistor voltage. Ignore
phase, which means that you assume the capacitor is purely reactive. Then

V_c^2 + V_r^2 = V_s^2,

where V_c is the capacitor voltage, V_r is the resistor voltage and V_s is
the signal generator voltage. This implies that V_c = sqrt(V_s^2 -
V_r^2).

Now, the current is I = V_r/R, and the capacitive reactance is X_c =
V_c/I, so you get X_c = R*sqrt(V_s^2 - V_r^2)/V_r. The capacitance can be
found from the capacitive reactance and frequency, C = 1/(2*pi*X_c).

Method 2:

Assume that the capacitor isn't purely reactive. Measure the amplitude
and phase of V_r respective to V_s, and solve for the capacitive impedance
Z_c (note that you have to use complex arithmetic):

V_r
Z_c = -----------.
V_s - V_r

Z_c will, in general, be complex, so it will be Z_c = R_c - jX_c, where
R_c is the capacitor's equivalent series resistance and X_c is the
capacitive reactance. If Z_c is purely imaginary all is well and good. If
Z_c has a significant resistive component then it is up to you to decide
if this is measurement error or the fault of the capacitor, and whether it
is best modeled as a series resistance, a parallel resistance, or
something more complicated.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[Fred Bloggs]

Wayne wrote:

Thanks Tim.

The resistor is in series with the capacitor and the signal gen is across
one lead of the capacitor and one lead of the resistor. In other word, if
they were both resistors it would be a voltage divider. The resistor or
the capacitor can be ground.
Forgive me for asking a dumb question but why are V_c^2 + V_r^2 = V_s^2 to
the power of 2?
If this was a voltage divider then you would have V?=Vs(R/R+Rc)?

In method 2 could you give me an example with the following figures?

Vs=10v p-p
Vr=4v p-p
Phase = 45deg (with respect Vs)

Cheers

Wayne


Coul
"Tim Wescott" <tim@wescottnospamdesign.com> wrote in message
news:10ldvo2errd1829@corp.supernews.com...

Wayne wrote:


If I have the Phase then say 40deg and the magnitude say 10 then how can
I
convert that into capasitance.
I have the following circuit:

___SIGGEN_____
Ś Ś
Ś--/\/\/\/\-----Ś Ś---Ś
R_shunt Cx
Ś Ś
----V?----

I am measuring the voltage drop accross R_shunt and comparing that with
the
siggen so I have V_R_shunt - SIGGEN = Z and on a scope I can measure
the phase. In other words I have the equiverlant info as you would find
when creating a Lissajour plot.
How can I find the capacitance of Cx with the info I have here?

Cheers

Wayne




It's not clear from your schematic or your text -- you have a signal
generator that is connected to a resistor and capacitor in series, and the
resistor is grounded? Or is the capacitor grounded?

The following answers assume the resistor is grounded.

Method 1:

Get the RMS voltages of both the signal and the resistor voltage. Ignore
phase, which means that you assume the capacitor is purely reactive. Then

V_c^2 + V_r^2 = V_s^2,

where V_c is the capacitor voltage, V_r is the resistor voltage and V_s is
the signal generator voltage. This implies that V_c = sqrt(V_s^2 -
V_r^2).

Now, the current is I = V_r/R, and the capacitive reactance is X_c =
V_c/I, so you get X_c = R*sqrt(V_s^2 - V_r^2)/V_r. The capacitance can be
found from the capacitive reactance and frequency, C = 1/(2*pi*X_c).

Method 2:

Assume that the capacitor isn't purely reactive. Measure the amplitude
and phase of V_r respective to V_s, and solve for the capacitive impedance
Z_c (note that you have to use complex arithmetic):

V_r
Z_c = -----------.
V_s - V_r

Z_c will, in general, be complex, so it will be Z_c = R_c - jX_c, where
R_c is the capacitor's equivalent series resistance and X_c is the
capacitive reactance. If Z_c is purely imaginary all is well and good. If
Z_c has a significant resistive component then it is up to you to decide
if this is measurement error or the fault of the capacitor, and whether it
is best modeled as a series resistance, a parallel resistance, or
something more complicated.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Adjust the signal generator frequency (lower) until Vr=0.707*Vs= which
makes Vc=Vr and therefore Cx= 1/(2*pi*Freq*Rshunt).

 

[John Woodgate]

I read in sci.electronics.design that Wayne <nospam-mail@wlawson.com>
wrote (in <swE5d.672$Ue2.441@newsfe6-gui.ntli.net&gt about 'Finding the
Capacitance from Lissajour plot info?', on Sun, 26 Sep 2004:

The resistor is in series with the capacitor and the signal gen is
across one lead of the capacitor and one lead of the resistor. In other
word, if they were both resistors it would be a voltage divider. The
resistor or the capacitor can be ground. Forgive me for asking a dumb
question but why are V_c^2 + V_r^2 = V_s^2 to the power of 2?

Because the two voltages have a phase difference of 90 degrees (or, in
practice, very nearly; the capacitor won't be perfectly loss-free).

If this
was a voltage divider then you would have V?=Vs(R/R+Rc)?

In method 2 could you give me an example with the following figures?

Vs=10v p-p
Vr=4v p-p
Phase = 45deg (with respect Vs)

It's helpful to use a 'phasor diagram' to see what is going on. The
current I is in phase with the voltage across the resistor, and the
voltage across the capacitor is at 90 degrees to the current. So our
diagram is (*use Courier font*):

/|
/ |
/ |
/ |
Vs = 10 / |
/ | Vc =?
/ |
/ |
/ |
/ 45 |
/__________| ----------> Current I
Vr = 4 V

Now, I hope you can see that this is not, in theory, possible. The angle
at the apex of the triangle must be 45 degrees as well and the triangle
must be isosceles. So Vc must be 4 V as well, and then by Pythagoras
(which is the same as the squared values in Tim's equation), Vs =
sqrt(4^2 + 4^2) = sqrt(32) = 5.66 V.

One practical explanation, if your figures are correct, is that the
capacitor is very lossy, thus having a lot of resistance of its own.
This add to the horizontal side of the triangle (without altering the 4
V across the resistor). Another 3.07 V brings the figure right. The Vc
must also be 7.02 V, of course, because of the 45 degrees. We then get
by Pythagoras, Vs = sqrt(7.07^2 + 7.07^2) = sqrt(100) = 10.

It's a very poor capacitor if this is true. There are other
possibilities. What frequency are you working at and what are the
resistor and capacitor values? Have you actually measured 10 V at the
signal generator, or is that just what the output control says? If the
signal generator source impedance isn't very much smaller than the
impedance of your RC circuit, the Vs value will be way off, indeed it
may well be 5.66 V!
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Wayne]

What frequency are you working at and what are the resistor and capacitor
values?
I am working from 1Hz to 1Mz. This is a psudo capacitor, in that I mean

that I am measuring the capacitance of a cell - such as a batery and trying
to model it in the same way as a capacitor. So there will be no XL factors.

Have you actually measured 10 V at the signal generator, or is that just
what the output control says?
The voltage can be anything from 0.01V to 10V. I have measured it with a
scope.

Wayne




If the signal generator source impedance isn't very much smaller than the
impedance of your RC circuit, the Vs value will be way off, indeed it
may well be 5.66 V!
--

"John Woodgate" <jmw@jmwa.demon.contraspam.yuk> wrote in message
news:yDapWhAq8xVBFwAe@jmwa.demon.co.uk...

I read in sci.electronics.design that Wayne <nospam-mail@wlawson.com
wrote (in <swE5d.672$Ue2.441@newsfe6-gui.ntli.net&gt about 'Finding the
Capacitance from Lissajour plot info?', on Sun, 26 Sep 2004:
The resistor is in series with the capacitor and the signal gen is
across one lead of the capacitor and one lead of the resistor. In other
word, if they were both resistors it would be a voltage divider. The
resistor or the capacitor can be ground. Forgive me for asking a dumb
question but why are V_c^2 + V_r^2 = V_s^2 to the power of 2?

Because the two voltages have a phase difference of 90 degrees (or, in
practice, very nearly; the capacitor won't be perfectly loss-free).

If this
was a voltage divider then you would have V?=Vs(R/R+Rc)?

In method 2 could you give me an example with the following figures?

Vs=10v p-p
Vr=4v p-p
Phase = 45deg (with respect Vs)

It's helpful to use a 'phasor diagram' to see what is going on. The
current I is in phase with the voltage across the resistor, and the
voltage across the capacitor is at 90 degrees to the current. So our
diagram is (*use Courier font*):

/|
/ |
/ |
/ |
Vs = 10 / |
/ | Vc =?
/ |
/ |
/ |
/ 45 |
/__________| ----------> Current I
Vr = 4 V

Now, I hope you can see that this is not, in theory, possible. The angle
at the apex of the triangle must be 45 degrees as well and the triangle
must be isosceles. So Vc must be 4 V as well, and then by Pythagoras
(which is the same as the squared values in Tim's equation), Vs =
sqrt(4^2 + 4^2) = sqrt(32) = 5.66 V.

One practical explanation, if your figures are correct, is that the
capacitor is very lossy, thus having a lot of resistance of its own.
This add to the horizontal side of the triangle (without altering the 4
V across the resistor). Another 3.07 V brings the figure right. The Vc
must also be 7.02 V, of course, because of the 45 degrees. We then get
by Pythagoras, Vs = sqrt(7.07^2 + 7.07^2) = sqrt(100) = 10.

It's a very poor capacitor if this is true. There are other
possibilities. What frequency are you working at and what are the
resistor and capacitor values? Have you actually measured 10 V at the
signal generator, or is that just what the output control says? If the
signal generator source impedance isn't very much smaller than the
impedance of your RC circuit, the Vs value will be way off, indeed it
may well be 5.66 V!
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Tim Wescott]

Wayne wrote:

Thanks Tim.

The resistor is in series with the capacitor and the signal gen is across
one lead of the capacitor and one lead of the resistor. In other word, if
they were both resistors it would be a voltage divider. The resistor or
the capacitor can be ground.
Forgive me for asking a dumb question but why are V_c^2 + V_r^2 = V_s^2 to
the power of 2?
If this was a voltage divider then you would have V?=Vs(R/R+Rc)?

In method 2 could you give me an example with the following figures?

Vs=10v p-p
Vr=4v p-p
Phase = 45deg (with respect Vs)

Cheers

Wayne


Coul
"Tim Wescott" <tim@wescottnospamdesign.com> wrote in message
news:10ldvo2errd1829@corp.supernews.com...

- snip -

The voltages are squared because you know they're 90 degrees out of
phase, which means that you need to use the pythagorean theorem to add
them, with the generator voltage being the "hypotenuse". See any good
text on basic electronics that includes "phasors" (the ARRL handbook
used to go into this, I sure hope they still do -- or get a 30 year old
ARRL handbook).

I was solving dividing by the cap voltage instead of multiplying --
here's the solution with the right formula:

Vs = 10V + j0V
Vr = 2.83V + j2.83V (j = sqrt(-1), 2.83 = 4*sin(45 deg)).

Vc 7.17V - j2.83V
Zc = ----*R = ----------------*R = (0.76 - j1.76)*R.
Vr 2.83V + j2.83V

This indicates a capacitor with quite a large resistive component, or a
failure in your phase measurement. If you use method 1 you get a
capacitive reactance of 9.17*R, which would give you a phase shift of
around 66 degrees.

You may want to double check your measurement or calculations,
particularly if you're like me and you occasionally divide when you
should be multiplying .

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[Tim Wescott]

Wayne wrote:

What frequency are you working at and what are the resistor and capacitor
values?

I am working from 1Hz to 1Mz. This is a psudo capacitor, in that I mean
that I am measuring the capacitance of a cell - such as a batery and trying
to model it in the same way as a capacitor. So there will be no XL factors.

Have you actually measured 10 V at the signal generator, or is that just
what the output control says?
The voltage can be anything from 0.01V to 10V. I have measured it with a
scope.

Wayne




If the signal generator source impedance isn't very much smaller than the
impedance of your RC circuit, the Vs value will be way off, indeed it
may well be 5.66 V!

Ah. In that case, the source impedance of the thing could be quite
lossy, and your measurements may not be that far off.

You'll want to characterize it at a number of different frequencies,
though -- batteries don't look like simple RC combinations (more like a
voltage source in series with an RC which is in series or parallel with
another RC, etc.). Add to that that for most batteries the source
voltage changes a bit and the apparent resistances change a _lot_ with
charge state, and you're in for an interesting modeling problem.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

 

[John Woodgate]

I read in sci.electronics.design that Wayne <nospam-mail@wlawson.com>
wrote (in <7qF5d.716$Ue2.527@newsfe6-gui.ntli.net&gt about 'Finding the
Capacitance from Lissajour plot info?', on Sun, 26 Sep 2004:

So there will be no
XL factors.

What do you mean by 'XL factors'? Inductance?

Your cell may indeed have 3.07 ohms of loss resistance.
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk