# input and output resistance of 741 in an inverting amplifier...?!

Discussion in 'Homework Help' started by as21462, Jan 1, 2007.

1. ### as21462 Thread Starter Member

Dec 29, 2006
22
0
Hi guys,...

I really love here.....
I think Im doing better here than talking to my classmates....

next question is how to get the input and output resistance of a 741opamp. the hint to this exercise is to use a load resistance in our suitable simulation circuit.

we are using this opamp in an inverting amplifier...

cheers,
as21462

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The manufacturer's datasheet is the primary source of the information you seek.

It is always helpful to our members if you can post a schematic diagram of the circuit as a supplement to your question.

hgmjr

3. ### as21462 Thread Starter Member

Dec 29, 2006
22
0
Aha,...the thing is that we have to get this resistanc by ourself and compare it with the datasheet.....the schematic of the design would not help here,..because now we have to use another schematic to get the characteristics of this opamp, like offset voltage,openloop gain and input and output resistance..... if still you think the schematic of the design would help I will send it,...

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I see. That is an interesting homework assignment.

Perhaps you can take each of the parameters you are being asked to measure and provide the members with a schematic of the op-amp circuit you plan to use to obtain each of the parameters.

hgmjr

5. ### Distort10n Active Member

Dec 25, 2006
429
1
I have attached a circuit showing how to calculate the output resistance of an op-amp using a resistive load. This is one way and it is rather simple. It saves you from going through the math of calculating Ro. Use this as an example to follow for your 741 assignment.

I will leave it to you to understand the following:

1) Why a large gain and high frequency were used.
2) Why this technique is not advised with CMOS output stages.

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6. ### as21462 Thread Starter Member

Dec 29, 2006
22
0
Wow!...
knight...it is itself another coursework ....but thanks,...

ok,...I have attached the schematic for open loop dc gain....741dcopenloop.doc
and also the schematic for offsetvoltage which im not sure its correct or not......offset.doc
but for the input resistance, no idea yet!!!!

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7. ### Distort10n Active Member

Dec 25, 2006
429
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Both circuits are incorrect.

For the open loop gain, there are several circuits you could use. You could also calculate the open loop gain from the non-ideal gain equation of an inverting amplifier (it is not -Rf/Rg).

The offset voltage circuit is incorrect because think about what will happen on the ouput. You have the non-inverting and inverting pins directly shorted to ground. What do you expect on the output in this configuration and why?

8. ### as21462 Thread Starter Member

Dec 29, 2006
22
0
WOW.... I was happy that I have almost finished!!!!!!!!!!!!!!!!!

open loop gain,....why? it is open loop,and also dc...what is the problem?...should I use any resistors in the input u mean?

offset voltage..............because it has been said when we connect both of inputs to each other, the offset voltage will reveal itslf, I decided to connect them together,...but when I connected them togethe, forexample by connecting them into a voltage source which was 1micro volt the output was different with the time I connect them to a Vsource of 1volt. so I decided to connect them to ground......

9. ### as21462 Thread Starter Member

Dec 29, 2006
22
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look,...I remove the load resistor,..and this is the design I use for openloopdc gain,...and thats the output result...I dont know why that step happens,..it was not happening on the computer I was working on yesterday!!!!!!!!!!!
but anyway,...was the problem just the load resistance?...if yes,...I put it there because it made that unusual step smaller....

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10. ### Distort10n Active Member

Dec 25, 2006
429
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The step goes from 4.03484 and increases by 1µV. I would not worry about it.
You need to tell me what your thoughts are when using your circuit configurations. What parameters are you trying to measure to derive the open loop and offset?

I have attached sample circuits for the open-loop and offset circuits to show why they are incorrect. I have also attached a scope shot of the output from an OPA227 with both inputs directly shorted to ground.

So:

1) What are you using parameter wise to derive the open loop and offset of your op-amp.
2) Do you know why the offset circuit is incorrect?

It is self-evident that we are no longer dealing with ideal amplifiers since we are measuring the finite open-loop gain and offset.

edit..

Here is the link to the Offset scope shot:

http://www.geocities.com/knightofsolamnus/Offset.bmp

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11. ### as21462 Thread Starter Member

Dec 29, 2006
22
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ok,..for the open loop : Vout/(Vpositive-Vnegative) is my open loop gain,...I decided to use the easiest numbers, I mean the difference of one for positive and negative pins of opamp,...I started with Vpositive=2volt and Vnegative=1volt the opamp goes to saturation so I decrease this value until I reached micro and it worked properly so by Vpositive=2microvolt and Vnegative=1microvolt the output is 4.034 volt....and Av=a.034*10^6

for the offset,...I know it mabe goes to saturation,but mycircuit didnt go....and it gave me the answer of 3.83 ... when we connect two input pins of an opamp to ground( actually no difference oltage between them) the output should be zero,...but in real opamp its not zero,...so the output of this circuit is the output offset voltage....I know that n data sheets there is always input offset voltage, so I was thinking by divideing my output to my openloop gain I can get the answer.................

Oh no,...I havent got enough time,..
there are two more exercises!!!!!!!!!!!

12. ### Distort10n Active Member

Dec 25, 2006
429
1
You're not including the affects of offset. Large voltages differentials will not yield good results because they will be multiplied by the open-loop gain. It could be 1V or 2V differential and either result will yield a different value of open-loop gain. Did you happen to check the datasheet to see if your calculations even come close to typical? From what you stated above they do not so that indicates that something is wrong.

Depending on the initial offset (ignoring offset contributed by CMRR, PSRR, input bias and offset currents) it will be multiplied by the open-loop gain when both inputs are simply shorted to ground. It could easily saturate so you can see why this is not a reliable way to test for offset. You can test this by changing supply values.
A +/-5V supply will more than likely saturate to 4.5V depending on what the datasheet says for output swing (headroom). Divide that by your open loop gain. Change to a +/-15V supply rail and the same amplifier will rail to 14.5V and divide that by your open-loop gain. Two different answers.
You are thinking correctly about grounding and dividing the output by a gain. You cannot do this in open-loop, so what's left?

Apr 26, 2005
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14. ### as21462 Thread Starter Member

Dec 29, 2006
22
0
I tried the offset test circuit that u sent to me Joe,.. Vout=19.17mV if I divide it by 1000 (Vios=Vout/1000) it would give me very small input offset voltage= 19.17micro volt...isnt it a bit bizzare!!!

15. ### as21462 Thread Starter Member

Dec 29, 2006
22
0
Wow,...I think I should change the openloopgain seettings to be suited to my opamp,....with the Vss,Vdd 15v....
honestly,I just wrote it to come to upper level in answers,....