Mathmatical analysis of power transfer on resistive power line

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.

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

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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

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

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.

--
Tony Williams.

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.

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.

-- 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

Posting from Google?  See http://cfaj.freeshell.org/google/

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

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:

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.

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.

Ok. Kevin G Rhoads in another post, (Load-line --), has 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

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.