non inverting amplifier

[dcac]

Hi I have just started studying about amplifiers (op amps). It has been two days I cannot figure out how can you get an amplification if you have the same size inputs.

Let say I have V1 input of 5v and open loop amplification is very high, so that the gain of all circuit is determined by G=1/B passive attenuator (negative feedback). So if I want to have G=100 I would need two resistors with ratio 1:99 or lets pick 1k and 99k ohms.

So V2 is determined by negative feedback. V2= Vout x 1k/(1k+99k)=[Vin x G] x 1k/(1k+99k)=

=5 x 100 x 0.01=5V.
So the V2=V1 when the negative feedback is used. I do not understand how can op amp can amplify a signal when two inputs are the same:
V1=V2 then Vin is V1-V2=0 so 0 x G equals 0 v output.

Thank you for help

 

[Harald Kapp]

Your design has several flaws:

1) There is no feedback. the negative input of the opamp is tied to GND. It should be tied to the connection 99R<>1R

2) V- is not an input. The input voltage is applied between V+ and GND

To understand the behaviour of the circuit, assume (V+-V-)=0V (input voltage differential is negligible). What output volatge do you need to make the voltage at the output of the voltage divider (1R, 99R) the same as the input voltage? Voilá!

 

[dcac]

Harald Kapp, thanks for spotting the mistake in diagram. If you have an input the same as feedback so how can you get any output at all?

 

[Harald Kapp]

Because the feedback is 1/100 of the output? What is the output so the feedback is the same value as the input?

 

[dcac]

Hi am trying to theory into practice. I am getting a problem to get right amplification.
I have 274mV as V+ and 1/4 negative feedback so gain is 4. When I measure with multimeter V+ =274mV, but output is 6.92V and V-=1.71V. Where the mistake could be?
Supply voltage is 8.4V.

 

[dcac]

Harald Kapp, the output would be 500v?

 

[Harald Kapp]

Harald Kapp, the output would be 500v?

For an input voltaeg of 5V:
In theory: yes
In practice: no

The output is limited by the supply voltage of the amplifier and the internal voltage drops. It can never exceeed the supply voltage.

I have 274mV as V+ and 1/4 negative feedback so gain is 4. When I measure with multimeter V+ =274mV, but output is 6.92V and V-=1.71V. Where the mistake could be?
Supply voltage is 8.4V.

sounds like your circuit still is not correct. Show me the actual circuit including the connections of the input voltage and the measurement points used. 6.92/1.71=4, so your gain is set right. How did you connect Vin?

 

[Harald Kapp]

Btw: you should not edit your old posts and change the content. The new image in your first post shows the corrected circuit according to my first answer. It will misslead others who will not be able to reproduce why I answered that way in the first place. Better post corrections in a new post, so everybody will be able to follow the development of the thread.

 

[dcac]

this is the circuit I made all the readings measured by voltmeter are marked with red. I do not understand why my negative feedback is really negative while it should be equal to the v+.
Please ask any question you need. Thank you for your help

 

[(*steve*)]

dcac -- you need to show ALL the connectins, not just some of them.

I presume there is some connection between the -ve of the right hand side 9V battery and the -ve of the LHS battery?

If there is not, you are effectively leaving the non-inverting pin open since voltages are measured between 2 points and only one of them has any connection to your circuit.

 

[dcac]

Steve, thank you for spotting the mistake and you was right I left the ve- pin disconnected.
Here is what I get when measure the p.d.:
v+=283 mV
Vout=7.83V
Vfeeback=1.95V

The Vout is wrong, However, the issue goes away when I use only one battery.
Here is the circuit using two batteries and photo pictures of the real circuit:

The circuit with only one battery:

 

[dcac]

sorry for doubling the message. I don't know how to delete the message.

 

[Harald Kapp]

In your top circuit you still need to connect the "-" poles of the two batteries. Without this connection, there is no reference for the amplifier with respect to the input signal.
That's why your bottom circuit is fine.

P.S. I deleted the double post.

 

[(*steve*)]

Have we asked for the part number of the op-amp and discovered if it is specified to have its inputs so close to the negative rail?

 

[dcac]

The part number is named L272M Dual power operational amplifier
here are pin outs:

 

[Harald Kapp]

The application notes in the datasheet operate the amplifier with inputs near ground. Since the internal circuit diagram shows a pnp input stage, this is plausible. I couldn't find an explicit statement with regard to min. input voltage in the datasheet.

 

[dcac]

thank you for answering so quickly my questions. Is there is away of marking the thread solved?