Chip with simple program for Toy

As a hobbyist, you should build your first power supply!  At 1.5A,
you can easily get this with a either a linear regulator or you can opt
for a switching regulator.  In either case, the parts count is low.  A
quick google search should get you some part numbers from which
to start.

Sounds great, there are just a few problems. I'm 17 and my parents are
not hobbyist. Anything that plugs into the wall terrifies them. I'm
already going to have to plug this "proffessionaly" made power supply
into a surge protector so I have a remote switch of sorts. The second
I opened the case to get at the PCB my parents swore they'd never let
me plug it in. It's funny really. I'd feel so much safer plugging in
my own creation than some mass produced peice of junk from China,
which they do on a daily basis.

If you still wish to use the computer power supply, load it with an
incandescent bulb.  A 6 volt bulb on the 5 volt output is not a problem.

sounds good. I'll pick one up from radio shack or something. Thanks
for the help!

 

[Jim Thompson]

On Sun, 8 Jun 2008 18:21:45 +0000 (UTC), sert <jerry@hotmail.com>
wrote:

Let's assume a nonlinear capacitor with c(v)=dq/dv. The q-v
characteristic of the capacitor is given by the formula

q=3*(v^3)

Let's also assume that it's charged to 1V and then connected to
a 0.2 Ohm linear ohmic resistance to discharge.

I tried to write the equations for this simple circuit and then
simulate the behaviour by approximating through the forward
Euler method.

My results are horrendous. First let's see the equations I used:

Using v as the state variable:

dv/dt = -0.556/v

Using q as the state variable:

dq/dt = -3.467*(q^.333), if q>0
dq/dt = 3.467*[(-q)^.333], if q<0

After simulating the circuit I get an oscillating behaviour with
the capacitor never discharging. What did I do wrong? Can
someone make a better analysis of the circuit in question?

Too many twists and turns.

Fundamentals...

i = d/dt(c•v) =>

i = c•dv/dt + v•dc/dt

Try manipulating from there, adding in c as a function of v. I
suspect you inserted too many artificial restraints.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

America: Land of the Free, Because of the Brave

 

[Helmut Sennewald]

"sert" <jerry@hotmail.com> schrieb im Newsbeitrag
news:Xns9AB7D951B938Djtjdfjdfjnbj@147.102.222.230...

Let's assume a nonlinear capacitor with c(v)=dq/dv. The q-v
characteristic of the capacitor is given by the formula

q=3*(v^3)

Let's also assume that it's charged to 1V and then connected to
a 0.2 Ohm linear ohmic resistance to discharge.

I tried to write the equations for this simple circuit and then
simulate the behaviour by approximating through the forward
Euler method.

My results are horrendous. First let's see the equations I used:

Using v as the state variable:

dv/dt = -0.556/v

Using q as the state variable:

dq/dt = -3.467*(q^.333), if q>0
dq/dt = 3.467*[(-q)^.333], if q<0

After simulating the circuit I get an oscillating behaviour with
the capacitor never discharging. What did I do wrong? Can
someone make a better analysis of the circuit in question?

Hello,

You can have drawn the circuit and the result plotted
in 5 minutes with LTspice.
Is it required that you write your own program?

Best regards,
Helmut

 

[John Larkin]

On Sun, 8 Jun 2008 20:56:16 +0200, "Helmut Sennewald"
<helmutsennewald@t-online.de> wrote:

"sert" <jerry@hotmail.com> schrieb im Newsbeitrag
news:Xns9AB7D951B938Djtjdfjdfjnbj@147.102.222.230...
Let's assume a nonlinear capacitor with c(v)=dq/dv. The q-v
characteristic of the capacitor is given by the formula

q=3*(v^3)

Let's also assume that it's charged to 1V and then connected to
a 0.2 Ohm linear ohmic resistance to discharge.

I tried to write the equations for this simple circuit and then
simulate the behaviour by approximating through the forward
Euler method.

My results are horrendous. First let's see the equations I used:

Using v as the state variable:

dv/dt = -0.556/v

Using q as the state variable:

dq/dt = -3.467*(q^.333), if q>0
dq/dt = 3.467*[(-q)^.333], if q<0

After simulating the circuit I get an oscillating behaviour with
the capacitor never discharging. What did I do wrong? Can
someone make a better analysis of the circuit in question?


Hello,

You can have drawn the circuit and the result plotted
in 5 minutes with LTspice.
Is it required that you write your own program?

Best regards,
Helmut

How do you set up the nonlinear capacitor in LT Spice?

John

 

[Eeyore]

sert wrote:

Let's assume a nonlinear capacitor

Why ? Are you trying to prove EEStor is a fraud ?

I doubt you'd need too much evidence to prove that. The behaviour of
hi-K ceramic dielectrics is very well understood.

Graham

 

[Jim Thompson]

On Sun, 8 Jun 2008 19:40:46 +0000 (UTC), sert <jerry@hotmail.com>
wrote:

Jim Thompson <To-Email-Use-The-Envelope-Icon@My-Web-Site.com
wrote in news:85ao441vohrgf00boagq6lclq2ih99degm@4ax.com:

Too many twists and turns.

Fundamentals...

i = d/dt(c*v) =

i = c*dv/dt + v*dc/dt

Try manipulating from there, adding in c as a function of
v. I suspect you inserted too many artificial restraints.


(Your post came up a bit garbled so I have it fixed above.)

I believe what you wrote is wrong, it is simply:

i = c*dv/dt

That equation is in the book, too.

Nope! The general equation is... i = d/dt(C*v)

Likewise for inductors it's... v = d/dt(L*i)

Don't dispute the MIT grad who found this stuff fascinating ;-)

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

America: Land of the Free, Because of the Brave

 

[Jim Thompson]

On Sun, 8 Jun 2008 19:19:54 +0000 (UTC), sert <jerry@hotmail.com>
wrote:

"Helmut Sennewald" <helmutsennewald@t-online.de> wrote in
news:g2ha0d$kg0$03$1@news.t-online.com:

You can have drawn the circuit and the result plotted
in 5 minutes with LTspice.
Is it required that you write your own program?


Actually it was an exercise from a book. After getting the
suspicious results I tried to do what you suggest but I couldn't
find a model for a nonlinear capacitor of the parameters I
specified (or any nonlinear capacitor.)

If you can help me model the circuit in LTSpice I'd be grateful.

Look up device "C"

In virtually _any_ flavor of Spice, terms can be added to the model
card to make it have a first and second order temperature sensitivity,
and a first and second order voltage sensitivity.

If your version of Spice supports behavioral modeling you can have
almost any kind of non-linear behavior.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

America: Land of the Free, Because of the Brave

 

[John Popelish]

John Larkin wrote:

How do you set up the nonlinear capacitor in LT Spice?

Copied from LTspice help file for component capacitor:

(begin excerpt)

There is also a general nonlinear capacitor available.
Instead of specifying the capacitance, one writes an
expression for the charge.

LTspice will compile this expression and symbolically
differentiate it with respect to all the variables, finding
the partial derivative's that correspond to capacitances.

Syntax: Cnnn n1 n2 Q=<expression> [ic=<value>] [m=<value>]

There is a special variable, x, that means the voltage
across the device. Therefore, a 100pF constant capacitance
can be written as

Cnnn n1 n2 Q=100p*x

A capacitance with an abrupt change from 100p to 300p at
zero volts can be written as

Cnnn n1 n2 Q=x*if(x<0,100p,300p)

This device is useful for rapidly evaluating the behavior of
a new a hypothetical charge model for, e.g., a transistor.

(end excerpt)


--
Regards,

John Popelish

 

[Phil Hobbs]

sert wrote:

Let's assume a nonlinear capacitor with c(v)=dq/dv. The q-v
characteristic of the capacitor is given by the formula

q=3*(v^3)

Let's also assume that it's charged to 1V and then connected to
a 0.2 Ohm linear ohmic resistance to discharge.

I tried to write the equations for this simple circuit and then
simulate the behaviour by approximating through the forward
Euler method.

My results are horrendous. First let's see the equations I used:

Using v as the state variable:

dv/dt = -0.556/v

Using q as the state variable:

dq/dt = -3.467*(q^.333), if q>0
dq/dt = 3.467*[(-q)^.333], if q<0

After simulating the circuit I get an oscillating behaviour with
the capacitor never discharging. What did I do wrong? Can
someone make a better analysis of the circuit in question?

The Euler method and its relatives (such as the Milne
predictor-corrector) have real trouble with decaying exponential-ish
things like this. The problem is that the difference scheme has an
extraneous solution (the oscillating one) that dominates the decaying
exponential.

You need a better solver--a simple one that works OK for this is the
classical fourth-order Runge-Kutta scheme, or with a bit more work you
can code up the Adams-Bashforth-Moulton predictor-corrector (my personal
fave). You can look them up on the web, or find them in most elementary
numerical analysis books.

Cheers,

Phil Hobbs

 

[Tim Wescott]

sert wrote:

Let's assume a nonlinear capacitor with c(v)=dq/dv. The q-v
characteristic of the capacitor is given by the formula

q=3*(v^3)

Let's also assume that it's charged to 1V and then connected to
a 0.2 Ohm linear ohmic resistance to discharge.

I tried to write the equations for this simple circuit and then
simulate the behaviour by approximating through the forward
Euler method.

My results are horrendous. First let's see the equations I used:

Using v as the state variable:

dv/dt = -0.556/v

Using q as the state variable:

dq/dt = -3.467*(q^.333), if q>0
dq/dt = 3.467*[(-q)^.333], if q<0

After simulating the circuit I get an oscillating behaviour with
the capacitor never discharging. What did I do wrong? Can
someone make a better analysis of the circuit in question?

After all that extra crap you have to simulate this?

First, you don't have to go past q = 3 * v^3, and dq/dt = -v/0.2 ohms.
This gives you your dv/dt = -(5/9)(1/v).

Then, you have an ordinary separable differential equation, and you
don't have to simulate it. A few minutes of quality time with a
2nd-year differential equations book (you did save yours, didn't you?)
gets you

v^2 = -(10/9 V^2/sec)t + C, with C = 1V^2.

So v = sqrt(1V^2 - (10/9 V^2/sec)t)

Couldn't be easier.

You can do a similar exercise using q as the state variable. I didn't,
because solving one first-order nonlinear problem keeps the really nerdy
part of me happy for about a year, so unless I'm getting paid for it I
don't do more than that.

I suspect that you are running into trouble with your simulation because
with v being a perfect square root of time, your effective gain is going
to infinity as the charge goes to zero. Modifying your model to a more
easily calculated

q = (some constant)*v + (some other constant)*v^3

would alleviate your simulation problems, and would probably be more
realistic, too, in that you'd model the inevitable capacitance formed by
the end conductors and whatever space is between them.

(note that you didn't have to solve the whole differential equation to
see the infinite gain; it's buried in your dq/dt = -(something)*q^(1/3),
which goes to infinity as q goes to zero).

--

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

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Helmut Sennewald]

"sert" <jerry@hotmail.com> schrieb im Newsbeitrag
news:Xns9AB7E32E8D917jtjdfjdfjnbj@147.102.222.230...

"Helmut Sennewald" <helmutsennewald@t-online.de> wrote in
news:g2ha0d$kg0$03$1@news.t-online.com:

You can have drawn the circuit and the result plotted
in 5 minutes with LTspice.
Is it required that you write your own program?


Actually it was an exercise from a book. After getting the
suspicious results I tried to do what you suggest but I couldn't
find a model for a nonlinear capacitor of the parameters I
specified (or any nonlinear capacitor.)

If you can help me model the circuit in LTSpice I'd be grateful.

Hello,

Save the text below into a file named test.asc .
Open it with LTspice and press RUN.

It's using the charge equation (x=voltage)

q=3*x*x*x

This is a capacitor with the voltage dependent
capcitance of

C = dq/dv = 9*x*x

Best regards,
Helmut


Version 4
SHEET 1 880 680
WIRE 96 48 16 48
WIRE 192 48 96 48
WIRE 192 80 192 48
WIRE 16 96 16 48
WIRE 16 192 16 160
WIRE 192 192 192 160
FLAG 16 192 0
FLAG 192 192 0
FLAG 96 48 vc
SYMBOL cap 0 96 R0
SYMATTR InstName C1
SYMATTR Value q=3*x*x*x
SYMBOL res 176 64 R0
SYMATTR InstName R1
SYMATTR Value 0.2
TEXT 16 -8 Left 0 !..ic V(vc)=1
TEXT 16 -40 Left 0 !.tran 0 1 0 1m

 

[Jim Thompson]

On Sun, 8 Jun 2008 21:07:33 +0000 (UTC), Tim Woodall
<devnull@woodall.me.uk> wrote:

God said, "div D = rho, div B = 0, curl E = - @B/@t, curl H = J + @D/@t,"
and there was light.

http://tjw.hn.org/ http://www.locofungus.btinternet.co.uk/

My sweat shirt says, "And God said, 'Let there be light'..."

Then the div/curl expressions.

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

America: Land of the Free, Because of the Brave

 

[Fred Bloggs]

Let's assume a nonlinear capacitor with c(v)=dq/dv. The q-v
characteristic of the capacitor is given by the formula

q=3*(v^3)

Let's also assume that it's charged to 1V and then connected to
a 0.2 Ohm linear ohmic resistance to discharge.

I tried to write the equations for this simple circuit and then
simulate the behaviour by approximating through the forward
Euler method.

My results are horrendous. First let's see the equations I used:

Using v as the state variable:

dv/dt = -0.556/v

Using q as the state variable:

dq/dt = -3.467*(q^.333), if q>0
dq/dt = 3.467*[(-q)^.333], if q<0

After simulating the circuit I get an oscillating behaviour with
the capacitor never discharging. What did I do wrong? Can
someone make a better analysis of the circuit in question?

View in a fixed-width font such as Courier.


.
.
.
.
. i
. ->
. .----------. v= i x R
. | ^ |
. | | | d 3
. | | | and i= - -- (3 x v )
. --- / dt
. C --- v \ R
. | / 2 dv
. | | \ so v = -9 x R x v x --
. | | | dt
. | | |
. '----------.
. or dt= -9 x R x v x dv
.
.
.
. t v(t)
. / /
. from which | dt = | -9 x R x v x dv
. / /
. 0 v(0)
.
.
. --------------
. 9 2 | 2
. or t = - - x R x (v - 1) from which v = | 1 - ----- x t
. 2 \| 9 x R
.
.

 

[Tim Wescott]

Tim Woodall wrote:

On Sun, 08 Jun 2008 13:27:35 -0700,
Tim Wescott <tim@seemywebsite.com> wrote:
sert wrote:
Let's assume a nonlinear capacitor with c(v)=dq/dv. The q-v
characteristic of the capacitor is given by the formula

q=3*(v^3)

Let's also assume that it's charged to 1V and then connected to
a 0.2 Ohm linear ohmic resistance to discharge.

I tried to write the equations for this simple circuit and then
simulate the behaviour by approximating through the forward
Euler method.

My results are horrendous. First let's see the equations I used:

Using v as the state variable:

dv/dt = -0.556/v

Using q as the state variable:

dq/dt = -3.467*(q^.333), if q>0
dq/dt = 3.467*[(-q)^.333], if q<0

After simulating the circuit I get an oscillating behaviour with
the capacitor never discharging. What did I do wrong? Can
someone make a better analysis of the circuit in question?
After all that extra crap you have to simulate this?

First, you don't have to go past q = 3 * v^3, and dq/dt = -v/0.2 ohms.
This gives you your dv/dt = -(5/9)(1/v).

Then, you have an ordinary separable differential equation, and you
don't have to simulate it. A few minutes of quality time with a
2nd-year differential equations book (you did save yours, didn't you?)
gets you

v^2 = -(10/9 V^2/sec)t + C, with C = 1V^2.

So v = sqrt(1V^2 - (10/9 V^2/sec)t)

Couldn't be easier.

tim@feynman:~$ maxima

Maxima 5.10.0 http://maxima.sourceforge.net
Using Lisp GNU Common Lisp (GCL) GCL 2.6.7 (aka GCL)
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
This is a development version of Maxima. The function bug_report()
provides bug reporting information.
(%i1) 'diff(v,t) = -5/(9*v);
dv 5
(%o1) -- = - ---
dt 9 v
(%i2) ode2(%,v,t);
2
9 v
(%o2) - ---- = t + %c
10
(%i3)


That's easier ;-)


Tim.


Only if you already have Maxima (or some commercial variant) loaded.

Besides, every once in a while you need to do it by hand, just to remind
your self that it _can_ be done.

--

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

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[C. Nick Kruzer]

Hi ngbud,

I have a computer power supply maximum 185 Watts load. It is a switching
power supply which needs to have a load to stabilize before start up. I
have read that the load should be roughly a minimum 5% of total
capacity. [In your case... 90W×5% = 4.5W]

If that is applied to the 12v output it would be 0.375 Amp and for 5v
output it would be 0.9 Amp. I don't know if that is correct, you should
check my figures, I have ADHD and am not very intelligent.

The load resistance I use on my 185W power supply is two 6v incandescent
light bulbs in parallel connected to the 5v output. I arrived at that
amount by watching the PS fan after adding the various loads, as
explained in the instructions/guide below:

This guide, which I found online years ago, is for my particular power
supply, but it can be used to get an idea of how to hook up your PS
resistors..

1) Make sure the on/off switch is off.

2) If there is a switch for selecting either U.S. Standard 115 volts AC
or European 230 volts AC make sure it is set accordingly.

3) Connect a load resistor from +5V to ground or from +12V to ground.

4) While looking at the fan turn on the power supply. If everything is
hooked up correctly the fan should spin and run at a constant speed. If
the fan doesn't turn on or it turns on briefly and then shuts off there
is something wrong, check to see if the load resistor is properly
connected. If the fan pulses continuously, turning on, off, on, off then
it is almost working, but the load resistor needs to be smaller.

I found that using incandescent bulbs as the resistors worked well. I
didn't have to calculate resistance or worry about power-heat-
dissipation values. The bulbs I used were like the 12v courtesy lamps
found in automobiles, except they were 6v. Circuit designs found in old
electronics~electricity experimenter books frequently use incandescent
bulbs as resistors and indicators.

I'm sorry I don't have a reference URL for the instructions above. I
only have author and date: Max Davis 10/00.

My Power Supply: Compaq # 172417-002 [172432-001] Purchased from
http://www.allelectronics.com/ Catalog # PS-185

I am not an expert in electronics. Please check my information and
calculations before applying them to your situation.

insula

 

[google@woodall.me.uk]

On Jun 9, 12:50 am, Tim Wescott <t...@seemywebsite.com> wrote:

Only if you already have Maxima (or some commercial variant) loaded.

Besides, every once in a while you need to do it by hand, just to remind
your self that it _can_ be done.

True, but you have to also use maxima occasionally or you forget how

to drive it. I started with:

(%i1) 'diff(v(t),t) = -1/(9*R*v(t));
d 1
(%o1) -- (v(t)) = - ------
dt 9 v(t) R
(%i2) desolve(%,v(t));
1
9 v(0) R - laplace(----, t, lvar)
v(t)
(%o2) v(t) = ilt(---------------------------------, lvar, t)
9 lvar R


"That CAN'T be right, I think there should be a square root in there
somewhere ... well it can be but I don't think it's what the OP was
expecting."

Goes and looks in manual. "Oh, of course, it's ode2 for ordinary
differential equations"

Tim.

 

[Helmut Sennewald]

"Fred Bloggs" <nospam@nospam.com> schrieb im Newsbeitrag
news:484C6008.5070809@nospam.com...

Let's assume a nonlinear capacitor with c(v)=dq/dv. The q-v
characteristic of the capacitor is given by the formula q=3*(v^3)

Let's also assume that it's charged to 1V and then connected to a 0.2 Ohm
linear ohmic resistance to discharge.

I tried to write the equations for this simple circuit and then simulate
the behaviour by approximating through the forward Euler method.

My results are horrendous. First let's see the equations I used:

Using v as the state variable:

dv/dt = -0.556/v

Using q as the state variable:

dq/dt = -3.467*(q^.333), if q>0
dq/dt = 3.467*[(-q)^.333], if q<0

After simulating the circuit I get an oscillating behaviour with the
capacitor never discharging. What did I do wrong? Can someone make a
better analysis of the circuit in question?

View in a fixed-width font such as Courier.


.
.
.
.
. i
. -
. .----------. v= i x R
. | ^ |
. | | | d 3
. | | | and i= - -- (3 x v )
. --- / dt
. C --- v \ R
. | / 2 dv
. | | \ so v = -9 x R x v x --
. | | | dt
. | | |
. '----------.
. or dt= -9 x R x v x dv
.
.
.
. t v(t)
. / /
. from which | dt = | -9 x R x v x dv
. / /
. 0 v(0)
.
.
. --------------
. 9 2 | 2
. or t = - - x R x (v - 1) from which v = | 1 - ----- x t
. 2 \| 9 x R
.

Hello Fred,

Thanks for this calculation. It's perfect.
It's exactly the same curve which I see in LTspice.

Best regards,
Helmut

 

[Tim Wescott]

sert wrote:

Fred Bloggs <nospam@nospam.com> wrote in
news:484C6008.5070809@nospam.com:

View in a fixed-width font such as Courier.
.
--------------
. 9 2 | 2
. or t = - - x R x (v - 1) from which v = | 1 -
----- x t . 2
\| 9 x R .
.



Thanks. But there's something unusual going on; the capacitor is
discharging completely. A linear capacitor, as we know, never
discharges completely.

Also, for values of t over a certain threshold the formula is
undefined.

Let's assume that the capacitor will discharge completely as the
formula suggests. Surely the time t will continue (time will not
stop!) but the formula doesn't actually give v=0 for any t after
the discharging, it's simply undefined.

Why? Is there some other non-obvious solution to the DE?

A linear capacitor coupled with a linear resistor never discharges
completely.

The formula itself may go undefined once the time exceeds the discharge
time, but that just tells you that you need to tack another formula on
to it, piecewise, namely v = 0 for all t > (discharge time).

--

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

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

Thanks Nick. BTW, good name ;-). There was a lot of great information
there. I'm going to see if i can find some 6v bulbs this week and try
it out.

 

[Stephen J. Rush]

On Thu, 16 Nov 2006 20:27:05 -0800, The Flavored Coffee Guy wrote:

default wrote:
On 16 Nov 2006 02:45:21 -0800, "The Flavored Coffee Guy"
elgersmad@rock.com> wrote:

But, a
Wimshurst Generator, will freewheel until all of the intertia of the
spin has be degraded by the friction of bearings and brushes have
brought it to a stop.

Aren't you overlooking another source of friction? The plates in a
Wimshurst machine rotate in air. Air adds drag.

Do Wimshurst machines work in vacuum?

No, I didn't overlook the areodynamics of the situation, I am still
looking for equations to determin these and other factors, and I have
the same question you do, Would a Wimshurst Generator Work in a Vacuum?
That is a good question.

This machine uses razor blades instead of brushes, and there is no
mechanical contact. All of the power is taken off of the disk by glow
discharge, or corona discharge.
http://www.coe.ufrj.br/~acmq/karlb.jpg

Therefore, even the thought of friction concerning brushes can be
eliminated in the Voss Machine, if that method of using sharp edges is
put to use to collect or deposit charges, and that reduces friction to
the bearings.

Not just friction. The power delivered by any generator comes from its
prime mover. In Wimshurst and similar electrostatic machines, the output
power is so low that the load it imposes on the prime mover goes
unnoticed, swamped by friction losses.

You can't win.
You can't break even.
You can't even quit the game.