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wdcw 12th April 2011 03:31 AM

Input Resistance
 
Hi everyone,

I'm having a bit of trouble trying to calculate the input resistance of a circuit.

The question is:

When a 100Mv signal with a 10k internal resistance is connected to the input of the amplifier the input voltage Vi is 67mV.
The output voltage Vo with no load connected is 1V.
The output voltage with a 1k load is 625mV.

Find the input & output resistance of the amplifier?

Could someone help with this?

Cheers

Conrad Hoffman 12th April 2011 04:32 AM

For the output, draw a voltage divider with two resistors, bottom one with a value of 1k grounded. Pretend the upper one is inside the amp. The upper one still has 1V applied. The midpoint is .625V. Using ohms law, figure out the current through the lower resistor- you know the value and the voltage across it. That same current flows through the upper resistor. Since you know the voltage at both ends (1 and .625) you know the voltage across it. Ohms law again gives you the value, which is the output resistance of the amp. You do the input in basically the same way. For inputs it's actually easier to use a pot and adjust it for 1/2 value on the output compared to the pot shorted or not present. At 1/2 value the input resistance is equal to the pot value.

wdcw 12th April 2011 04:40 AM

Thanks Conrad Hoffman,

I'll give it a go.

Thanks again

Cheers

Elvee 12th April 2011 09:18 AM

You and Conrad implicitly assume a purely resistive input impedance, which is not necessarily the case: often, there also is a significant capacitive component.

In this case, the resistance obtained with this method will yield incorrect results, with theoretically up to 40% error.

To check that the impedance is purely resistive, you can make the test at two frequencies, 100Hz and 10KHz f.e.
If you find identical values, then it's fine.

DF96 12th April 2011 11:14 AM

This looks like a student homework assignment, not a real world problem.

j beede 13th April 2011 06:25 AM

Quote:

Originally Posted by DF96 (Post 2535690)
This looks like a student homework assignment, not a real world problem.

...your point being?

godfrey 13th April 2011 09:39 AM

Quote:

Originally Posted by j beede (Post 2536749)
...your point being?

It's safe to assume simple resistive impedances.

DF96 13th April 2011 10:45 AM

Yes, homework problems don't have stray capacitance (not until next year, anyway!) and it is difficult to attach a meter to a textbook to measure it at different frequencies.

SY 13th April 2011 10:49 AM

Quote:

Originally Posted by j beede (Post 2536749)
...your point being?

That putting up one's homework problems on an internet forum, then passing off the answers as one's own is cheating. Cheating the instructors, cheating the school, cheating the other students, and most importantly, cheating oneself.

dchisholm 13th April 2011 07:48 PM

Quote:

Originally Posted by SY (Post 2536936)
That putting up one's homework problems on an internet forum, then passing off the answers as one's own is cheating. Cheating the instructors, cheating the school, cheating the other students, and most importantly, cheating oneself.

As somebody who has spent significant time on each side of a classroom podium . . . I don't entirely agree with you. (In fact, this afternoon or tomorrow I will be grading a lab assignment that includes estimating the output resistance of a linear power supply regulator using output voltage measurements under various loads.)

In practice, knowing "where" to find a solution (or, a method for a solution) is nearly as commendable as knowing "how" to find a solution. Although Tom Lehrer poked fun at the process in "Lobachevsky" (see Lobachevsky Lyrics, and also Lobachevsky video), in fact this same process is at the heart of the "apprentice" system of mastering a trade: the aspiring craftsman starts by performing insignificant portions of a task, but progressively takes on larger portions as the master craftsman guides, instructs, and demonstrates. In that context, nobody accuses the master craftsman of "doing the apprentice's work for him".

Conrad Hoffman's
reply to the original post isn't bad. He outlined a method, rather than simply stating a formula without explanation. He got bogged down in a verbal explanation when a diagram or two would have been much more effective. ("A picture is worth 1x10^3 words." - but NOTHING beats HARDWARE!)

I would have asked the original poster to show his work as far as he could go with the problem - perhaps by posting a schematic of the problem as he understands it. Hopefully, that wouldn't lead all the way back to "What is a Thevenin equivalent source?", but if it does . . . then that's where we start. (It's never too late to learn.)

In my experience, the response to that approach is either no response; or "OK - here's what I have, and this is where I don't see the next step"; or, "Just gimme the answer. This is for my Senior Project, and I ain't got time to go through all that fundamentals crap.". From that point, Forum members are pretty good at formulating appropriate responses.

"WDCW" (the original poster), would you share a little about yourself and your course of study?

Dale

p.s. - since the original post showed an Australian flag icon, it was rather obvious that this is a homework question. If it had been from certain other parts of the world, the question may have been from an engineer whose company had just been selected as an out-source for a U.S. or European company that had just awarded bonuses to the managers who fired its engineering staff.


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