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how do I draw a load line on a log plot?

Thanks everybody for your great responses to my question about diodes and a simple resistor/diode series circuit. I'm glad to discover that calculating diode current based on an arbitrary resistor is NOT a simple task and involves calculus and approximation equations, so my ego is intact. Anyways...

I like the 'load line' method of determining the diode current by drawing a the resistors load line on the same plot as the diodes V/I curve and where they intersect you have the diode current. BUT, my diode plot is LOG and I cant draw a load line on it with a ruler. Any suggestions?

Thanks!

Asa

Reply to
acannell
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Calculate and plot a bunch of points for the load line, and connect them with a curved line.

Reply to
redbelly

Is it a log log plot, i.e. both axis are log...in which case the plot for the resistor is still a straight line...

if not, then I think your best bet is to convert the plot to linear, so that the resistor load line is straight and the diode line is exponential.

Mark

Reply to
Mark

Bend your ruler.

-- Paul Hovnanian mailto: snipped-for-privacy@Hovnanian.com

------------------------------------------------------------------ Madness takes its toll. Please have exact change.

Reply to
Paul Hovnanian P.E.

Hum, actually if you just enter all the constants into the Eber-Molls (or is that just concerning transistors?) equation, and equate that with the circuit conditions, i.e., Vin = Vr + Vd and currents are equal, it's just algebra to solve.

What they said, plug and chug is probably the easiest way... not the happiest, but heck, you'll have to plug and chug anyway to graph either, so you might as well!

Tim

-- Deep Fryer: a very philosophical monk. Website:

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Reply to
Tim Williams

Just enter the equations in a spreadsheet. Make a graph and zoom in as needed on the intersection point.

--Mac

Reply to
Mac

You'll have to read off some values and create a new graph.

Frankly I don't understand why you're bothered about the exact If.

A first aproximation to diode current using a sensible value for Vf ( 0.7 V for small signal - 1V for medium power etc ) invariably suffices.

As noted by others, if you're that bothered about precise diode current you likely have a marginal design.

Graham

Reply to
Pooh Bear

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