# Mathmatical analysis of power transfer on resistive power line

Discussion in 'Electronic Design' started by Mook Johnson, Apr 24, 2007.

1. ### Mook JohnsonGuest

I am trying to do an analysis but have hit a wall.

Here is the circuit

VSupply ---------/\/\/\/------------ Vout @ X watts load
R

The Vsupply is a power supply whose output voltage follows this equation.
Vsupply = 500 + Iout * R

Iout is the supply output current and R is the resistance between the source
R is between 200 and 500 ohms depending on the installation.

The load is a switching power supply (costant power) that consumes 50 - 400W
and wants a nominal input voltage of 500V.

The intention is that as the power consumption goes up the input current
goes up and the supply voltage goes up at the same time to maintian the
supply input voltage at ~500V. These loads move slowly so there is plenty of
time for the source to respond to load changes.

I'd like to setup some equations to see what happens of the R value used in
the surface equation is different from the actual resistance between source
and load. How far off can they before the compensation causes an error of +
or - 100V at the supply input.

Seems like a straight forward problem but its got me twisted up somewhat.

Can any of your guys with sharp math skills help me out.

thanks

2. ### John PopelishGuest

Rearrange this to solve for Vout.
According to the above equation, it doesn't matter what the
resistance or R is.
And according to the equation, gets it.
100V=R*Iout is the limit on R if the maximum voltage
allowance for R's drop is 100 volts.
To me, also.

3. ### Tim WescottGuest

Think of it this way: Your Vsupply is just a Thevinin (sp?) voltage
source, with Rs nominally equal to -R. So your voltage at the load is
just Vl = Vs - Iout * (Rs + R). When Rs = -R, Vl = Vs - Iout * (R - R)
= Vs.

At this point it should be easy to vary Rs and R and try things out at
maximum Iout.

--

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

4. ### Mook JohnsonGuest

100V=R*Iout is the limit on R if the maximum voltage allowance for R's

You have to consider that the surface voltage is constantly adjusting as the
load current goes up so the voltage across R can be greater than 100V.
These are the numbers I get from Excel by sucessivie guestimation.

I used a power level of 600 watts instead of 500 and checked R at sever
spots between 100 and 500 ohms
Calr R is the number used for calculating the source voltage (Vin) = 500 + I
* Rcalc. Vo = Vin - actualR * I. I was set to 600W/500V for the nominal
case, 600W/400V for the low case and 600W/600V for the high case at each
resistance analyzed. The calc R was manually adjusted to give the Vo for
that case.

This tells me what I need to know but I'd like to understand the equations
behind these guys.

Vin current calc R actual R Vo Watts
620 1.2 100 100 500 600
549.875 1.5 33.25 100 399.875 599.8125
700 1 200 100 600 600

Vin current calc R actual R Vo Watts
740 1.2 200 200 500 600
699.875 1.5 133.25 200 399.875 599.8125
800 1 300 200 600 600

Vin current calc R actual R Vo Watts
860 1.2 300 300 500 600
850.025 1.5 233.35 300 400.025 600.0375
900 1 400 300 600 600

Vin current calc R actual R Vo Watts
980 1.2 400 400 500 600
999.875 1.5 333.25 400 399.875 599.8125
1000 1 500 400 600 600

Vin current calc R actual R Vo Watts
1100 1.2 500 500 500 600
1149.875 1.5 433.25 500 399.875 599.8125
1100 1 600 500 600 600

5. ### Tony WilliamsGuest

I can develop a Vout sum, but have difficulty in
interpreting it. It looks like there are always
two stable states? It goes like this........

Vsupply = 500 + Iout.R1 where R1 is the theoretical
value of the line resistance. So call R2 the actual
value of the line resistance.

Vout = 500 + Iout.R1 - Iout.R2.

Also P = Vout.Iout, or Iout = P/Vout.

Vout = 500 + R1.P/Vout - R2.P/Vout.

Vout^2 = 500.Vout + P.(R1 - R2).

Vout^2 - 500.Vout - P.(R1 - R2) = 0

That's a quadratic equation, (a.x^2 + b.x + c) = 0,
where in this case a= 1, b= -500 and c= -P.(R1 - R2).

-b (+/-) sqrt(b^2 - 4ac)
The roots of x = --------------------------
2a

500 (+/-) sqrt(500^2 + 4P(R1 - R2))
Similarly, Vout = ----------------------------------
2

In theory, the max and min values should be plugged
into the expresion to get a range for 4P(R1 - R2).

But here's where I run into confusion.... it's that
damned '(+/-)', which says that there are two possible
answers for Vout for every value of P(R1 - R2).

Are there two stable states for Vout? I don't know.

That's a quadratic equation, (a.x^2 + b.x + c) = 0,

Since b < 0, -b >0 so avoid the +/- by

-b + sqrt(b^2 - 4ac)
First root, x1 = --------------------------
2a

2c
Second root, x2 = -------------------------
-b + sqrt(b^2 - 4ac)

Since a > 0 and c < 0, x1 is positive, x2 is negative.
Now ask yourself, "Which one(s) make sense?" ....

__________________________________________________________________________

Now, having gone through all that ... Why is OP not using
graphical analysis? This is a classic load-line plot.
You can "what if" the load variations by just plotting
a variety of load resistance slopes. Use Excel to model
the source, make a plot, print it out -- then use a
ruler to draw the various load lines in questions on it.

7. ### whit3rdGuest

Yes, that's exactly right. You need the source to have
negative resistance.

Trouble is, all negative resistance schemes are unstable and
prone to oscillation (in my experience). For a practical device,
it's OK to identify the suitable range of input voltages (like, 400
to 550VDC) for the distal device, and make multiple taps
(or setpoints) for the source (like 500V, 600V, 700V), then
do some digital magic to select a setpoint. There can be
hysteresis in such a system, and that stops oscillation.

In analog terms, latch a current-sense-value, compare it to
the optimum for the setpoint that is now selected, and if
it is different from optimum by more than 0.6 setpoint-steps,
increment (or decrement) to another setpoint. After a few
seconds have passed, and things have settled down
repeat the cycle.

This gives a transfer function with stepwise approximation to
the negative resistance, but NO points on the transfer function
curve actually have a local derivative that supports oscillation.

8. ### Tim WescottGuest

-- snip --

AFAIK just about every mid-budget tape player in the world uses a driver
with an effective source resistance equal to the negative of the capstan
motor's drive resistance. Perhaps they're going to brushless motors ala
CD players, but I wouldn't count on it. As a consequence of the
negative drive the motor is very stiff, and they don't have to shell out
for a speed controlled motor.

But you are correct in being jaundiced -- I wouldn't want to try this
trick in anything but a well-controlled environment, and oscillation on
a 500V power line may be just a tad more serious than oscillation on a
motor drive.

--

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

Yes, that's exactly right. You need the source to have
Not necessarily. Depends on the nature and value of the
negative resistance and value of the load resistance.
For example, Herbert J. Reich's "Functional Circuits
and Oscillators" text discusses this in detail. There
is some, but less material, on it in his "Theory and
Application of Electron Tubes" which can be had in PDF
form from Millet's website:
http://www.pmillett.com/technical_books_online.htm

Just about any negative feedback loop involving analog
electronics will appear to have a negative resistance
when viewed from an appropriately chosen pair of
terminals. That's why there is such a literature on
achieving loop stability for negative feedback loops.

Most analyses will treat the issues in terms of feedback,
rather than in terms of negative resistance. But that
is just a matter of convention and what people are used to.

10. ### whit3rdGuest

Yes, that's exactly right. You need the source to have
negative resistance.

Trouble is, all negative resistance schemes are unstable and
prone to oscillation (in my experience). For a practical device,
it's OK to identify the suitable range of input voltages (like, 400
to 550VDC) for the distal device, and make multiple taps
(or setpoints) for the source (like 500V, 600V, 700V), then
do some digital magic to select a setpoint. There can be
hysteresis in such a system, and that stops oscillation.

In analog terms, latch a current-sense-value, compare it to
the optimum for the setpoint that is now selected, and if
it is different from optimum by more than 0.6 setpoint-steps,
increment (or decrement) to another setpoint. After a few
seconds have passed, and things have settled down
repeat the cycle.

This gives a transfer function with stepwise approximation to
the negative resistance, but NO points on the transfer function
curve actually have a local derivative that supports oscillation.

11. ### Tony WilliamsGuest

stated that only the '+ sqrt..etc' root is reasonable.

500 + sqrt(500^2 + 4P(R1 - R2))
So Vout = ---------------------------------
2

Making... (2.Vout - 500)^2 = 500^2 + 4P(R1 - R2).

This expression allows the, (400-600V), allowed Vout range
to be plugged in, to see the allowed range of P(R1 - R2).

Vout P(R1 - R2) P(R1 - R2)/400W
---- ---------- -----------------

600 60000 +150 ohms
500 0 0
400 -40000 -100

It is only the maximum power that produces appreciable
swings in Vout. The OP's original requirement was 400W
and this produces the example permitted R1-R2 ohm-range.

Just as a reminder, R1 is the estimated average line
resistance, used to set up the positive feedback on
the bench, and R2 is the actual line resistance that
is subsequently encountered in the field.

The OP stated a field range of "200 to 500 ohms", so
set R1= 350 ohms on the bench, and now vary R2.

500 + sqrt(500^2 + 4.400(350 - R2))
Vout = --------------------------------------
2

R2 (350 - R2) Vout
-- ---------- ----

500 -150 300 <-- Vout= Out of spec.
450 -100 400 <-- Vout= In spec, as predicted.
400 -50 456 "
300 +50 537 "
200 +150 600 "

So it looks as though a 200-500 ohm line will not quite
meet the 400-500V range required for Vout, at 400W.

12. ### whit3rdGuest

Sweet; I didn't know that. It has also been used for woofers (linear
motors, similar reasons), and I've toyed with it for inductive
pickups (to make a more favorable L/R time constant).

The application here, though, has a variable-load switching power
supply connected through enough wire that one expects hundreds
of ohms. 200 ohms of 20 gage copper is 6km of wire (3km transmission
line). It isn't a simple matter of compensating a pure-resistor load
with some negative resistance.  