Need some help with 2 problems

[Wonker]

1. What is the total resistance of this network? This is a Parallel circuit right?
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2.How do I figure out the gain of this Jung Multi Loop Preamp? I have no clue what this is?
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I don't need the answers...just a really good walk through on how to get them.

Thanks

[cbarrow7625]

In #1, is the 10 ohm resistor connected to ground or is the top signal in & the bottom (through the 10 ohm resistor) signal out? I assume it is a single ended signal and not balanced.

If the 10 Ohm resistor is the output, then the total series resitance in the circuit is just under 168 Ohms. In this scenario it is basically a series circuit with some elements in parallel with each other. the first 50 ohm is in parallel with the 1.5k reisitor. That = approx 48 Ohms. The way it is drawn, the next 50ohm is shunted out of the circuit altogether, ignore it. 100 Ohms in series = 148 Ohms total so far. 10 ohm + 10 Ohm (=20 Ohms) in parallel with a 20 ohm resistor. 20 // 20 = 10 ohms. Up to 158 Ohms. Another 10 ohms in series brings it up to 168 ohms.

Where is that from? Looks really odd. A larer clip of the schematic surrounding it would give better perspective on what was trying to be accomplished.

I could take a guess on #2 but I'd probably be wrong. What is the 2nd op amp? Another standard non-inverting stage or what? It only shows a single input. Looks more like a digital inverter.

[Wonker]

Originally Posted by cbarrow7625

Where is that from? Looks really odd. A larer clip of the schematic surrounding it would give better perspective on what was trying to be accomplished.

This is all they give you. I'm taking an open book pre test...and having a very hard time with it.

Thanks

[cbarrow7625]

Good for you, using all of your resources. I thought that the diagrams looked susiciously like textbook stuff. In that case, my assesssment of #1 is almost certainly correct & it is 168.387 Ohms to be exact.

Good luck with the other one.

[cbarrow7625]

I don't know why, but I just looked up the answer for #2 (Google is an awsome thing). In a Jung multi-loop configuration, the inner loop is set to a very high gain to increase linearity & reduce distortion. The overall gain for the stage is set by the outer loop. In this case, the outer loop gain is figured just like any other non-inverting stage. It has an overall gain of "11".

I hope that is correct (insert disclaimer here).

[Wonker]

Thanks for the help.

I have to do a test like this in the morning as part 2 of a job interview. This is what they gave me to prepare.

I'm freaking out

[cbarrow7625]

Good Luck. There are tons calculators & written calculations on the web for all this stuff that can be really useful to brush up on your skills.

Just search for "op amp calculator" or "resistor network calculator" or "Thevenin / Norton Equivalent circuit calculator", etc. and you'll probably find more help than you know what to do with.

[Enzo]

I'd agree at 168.4 ohms. Nitpicking here: you can;t extend the calculation to six digits, none of your terms have that many digits.

But the problem is basically simple, and they are finding out if you know how to calculate series and parallel resistances. Do you know how?

Look at the problem, the 50 ohm resistor is shorted across, so ignore it. The path through the network is through a 50 ohm and 1500 ohm in parallel (48.39 ohms), then through 100 ohms. Next through a paralleled 20 ohms and two 10 ohms in series. The two tens make 20 ohms, and that in parallel with the other 20 ohms makes 10 ohms. And finally another 10 ohms in series.

SO you have 48.39 , 100, 10, and 10 ohms in series. Adds up to 168.39, which we have to trim to four significant digits - 168.4

[Chuck H]

Hmmm...

I get 121.5

[Chuck H]

50 parallel with 1.5 is 1.456, omit the second 50 because of the jumper. Then the 100. Then 10 + 10 = 20 parallel with another 20 = 10. Then another 10.

1.456 + 100 + 10 + 10 = 121.456

Chuck

[Robert M. Martinelli]

Chuck,
no disrespect intended, but you probably missed that the 1.5 resistor has a K at its right, so paralleling it with the 50 Ohms resistor you get 48.38 Ohms.

The entire network's equivalent resistance is 168.38.... Ohms IMHO.

Cheers

Bob

[Chuck H]

OK, good. I'd much rather be wrong than living in bizarro world. I thought everyone had gone funny. Glad it was just me.

Chuck

[Enzo]

Right, 1.5k - 1500 ohms.

[Don Moose]

This is why I agitate for the 1k5 style notation. Harder to lose the decimal's position.

[Chuck H]

I'm with Don. Though I never even considered it before

Actually, the way the 1.5k is marked is a bit goofy. Probably intentional since although there's nothing else it could mean it would be easy to misread on a glance. Which I did. So I guess I failed the test so to speak.

Of course it should have been indicated like this:

[SyncTronX]

Originally Posted by cbarrow7625

(snippage) #2 In a Jung multi-loop configuration, the inner loop is set to a very high gain to increase linearity & reduce distortion.

The overall gain for the stage is set by the outer loop. In this case, the outer loop gain is figured just like any other non-inverting stage. It has an overall gain of "11".

Using this calculator:

Non-Inverting Op-Amp Resistor Calculator

I entered the following information.
R1 - 4 K ohms 1k + 3K
R2 = 1013 K ohms, 1M, +3k + 10K
V2 = 0
V1 = 1

Result:
Gain = 0
R2 = Nan
R4 = Nan