Need help with Non-Inverting amplifier using single supply

Yuvraj Abhimanyu

Joined Jun 10, 2015
15
Hey Guys,

I'm building an Non-invering amplifier using a single 9v supply, and the opamp i'm using in NE5532 from TI.
Now, My question is, in the circuit diagram below, i'm unable to get the output!
I dont understand why!

Plz Help,
Thank you.

P.S- The power supply is 9v, sorry for the wrong value in the circuit diagram.

ericgibbs

Joined Jan 29, 2010
14,171
hi,
With +4.5V on in the inverting input, what would you expect the OPA's Vout is going to be.?

E

BTW: The NE5532 is specified as a dual power supply OPA

Yuvraj Abhimanyu

Joined Jun 10, 2015
15
Last edited:

ericgibbs

Joined Jan 29, 2010
14,171
As the OPA is set for an Inverting gain of 1, the Vout of the OPA would try to go to -4.5V, but as its powered by a single +9V, Vout will tend towards zero.!

Post your asc file and I will post a possible fix.

Yuvraj Abhimanyu

Joined Jun 10, 2015
15
As the OPA is set for an Inverting gain of 1, the Vout of the OPA would try to go to -4.5V, but as its powered by a single +9V, Vout will tend towards zero.!

Post your asc file and I will post a possible fix.
How fail to see how will this be an inverting amplifier?! As far as i no its an non-inverting amplifier because the signal is given to the non inverting side of OPA.

The circuit i've constructed is from the Page No.5, Section 2.1 (plz refer to the link below)
https://courses.cit.cornell.edu/bionb440/datasheets/SingleSupply.pdf

ericgibbs

Joined Jan 29, 2010
14,171
The page #5 circuit is not the same as the circuit you have posted.??

EDIT:
you appear to have the + and - inputs to the OPA crossed over, the +4.5V, should be on the NON inverting and the AC signal should be Capacitor coupled to the Inverting input.???
Also the Gain of the NON inverting input is *2, so the 4.5V should be +2.25V

Attachments

• 25.8 KB Views: 4

Yuvraj Abhimanyu

Joined Jun 10, 2015
15
The page #5 circuit is not the same as the circuit you have posted.??

Please search for Page 5, figure 3 in section 2.1

ericgibbs

Joined Jan 29, 2010
14,171
hi,
I did post that image in post #6. edit.

Look at the attached circuit, is this what you are trying to do.?

Attachments

• 36.5 KB Views: 12

Bordodynov

Joined May 20, 2015
2,938
In this particular case, everything can be made easier.

AnalogKid

Joined Aug 1, 2013
9,489
The circuit i've constructed is from the Page No.5, Section 2.1 (plz refer to the link below)
https://courses.cit.cornell.edu/bionb440/datasheets/SingleSupply.pdf
No, it isn't. Your circuit does not have an input coupling capacitor as shown in the TI document. Your input sinewave is centered about ground. This means that the opamp has 4.5VDC between its two inputs. What you want is 4.5 VDC offset on one input, and zero VDC difference between the two inputs. The coupling cap achieves this.

But your circuit will work only at unity gain. For a more generic AC amplifier that will work at other gains, see:

The last schematic before the comments is this: Non-inverting amplifier using a single voltage rail

C2 ensures that the voltage gain at DC is 1 no matter what the AC gains is, and that there is 0VDC between the inputs. It forms a high-pass filter with R1, so you have to pay attention to that. R3 and R4 set the DC bias at the output, but also load the input, so you have to pay attention to that.

ak

Yuvraj Abhimanyu

Joined Jun 10, 2015
15
hi,
I did post that image in post #6. edit.

Look at the attached circuit, is this what you are trying to do.?
Hi Eric,
I'm trying to build an 'Non-Inverting' Amplifier with variable gain

Yuvraj Abhimanyu

Joined Jun 10, 2015
15
No, it isn't. Your circuit does not have an input coupling capacitor as shown in the TI document. Your input sinewave is centered about ground. This means that the opamp has 4.5VDC between its two inputs. What you want is 4.5 VDC offset on one input, and zero VDC difference between the two inputs. The coupling cap achieves this.
Hi AnalogKid,
The way i presumed it would work is that the 4.5v DC at the inverting terminal will be the new virtual ground. Is this assumption wrong?

And can u explain me the "This means that the opamp has 4.5VDC between its two inputs" part plz?

Yuvraj Abhimanyu

Joined Jun 10, 2015
15
In this particular case, everything can be made easier.

View attachment 87822
I've two questions regarding this circuit diagram.
1. what are the purpose of R4 and R5?
2. without the decoupling capacitors, wont the ac signal go to the DC source?

AnalogKid

Joined Aug 1, 2013
9,489
Hi AnalogKid,
The way i presumed it would work is that the 4.5v DC at the inverting terminal will be the new virtual ground. Is this assumption wrong?
And can u explain me the "This means that the opamp has 4.5VDC between its two inputs" part plz?
Your assumption is not wrong, but it is incomplete. Only one of your two inputs is referenced to your virtual ground. The other is referenced to your real ground.
Your sine source has a DC value of 0V, and R1 is connected to 4.5V. 4.5V - 0V = 4.5V DC between the two inputs.

ak

Yuvraj Abhimanyu

Joined Jun 10, 2015
15
Your assumption is not wrong, but it is incomplete. Only one of your two inputs is referenced to your virtual ground. The other is referenced to your real ground.
Your sine source has a DC value of 0V, and R1 is connected to 4.5V. 4.5V - 0V = 4.5V DC between the two inputs.

ak
Oh i see... Intresting,
Now how is this solved in the circuit tat u've provided ( the circuit u were referring to in the link tat u provided).
now wont the inverting terminal has DC value of 0 and non inverting terminal has DC value of 4.5?

AnalogKid

Joined Aug 1, 2013
9,489
I've two questions regarding this circuit diagram.
1. what are the purpose of R4 and R5?
2. without the decoupling capacitors, wont the ac signal go to the DC source?
Both the 9V DC source and the sinewave source have zero ohm output impedances. So the 9V source sees 200K to "ground" and the center point is 4.5 V. The sinewave source sees 200K to "ground", and the center point is 0.5Vrms. So the opamp + input sees 0.5Vrns riding on 4.5Vdc. The - input sees 4.5Vdc at an impedance of 100K. The DC difference between the two inputs is 0Vdc, and the non-inverting gain is 2 which makes up for the 2:1 attenuator at the + input.

ak

AnalogKid

Joined Aug 1, 2013
9,489
Oh i see... Intresting,
Now how is this solved in the circuit tat u've provided ( the circuit u were referring to in the link tat u provided).
now wont the inverting terminal has DC value of 0 and non inverting terminal has DC value of 4.5?
Nope. Because of the capacitor in the shunt leg, the inverting terminal will float up to whatever voltage it takes to reduce the differential input voltage to zero. That always is the function of negative feedback. So the cap will charge up until the DC voltages at both inputs are the same.

ak

Bordodynov

Joined May 20, 2015
2,938
inverting amplifier:

Yuvraj Abhimanyu

Joined Jun 10, 2015
15
Nope. Because of the capacitor in the shunt leg, the inverting terminal will float up to whatever voltage it takes to reduce the differential input voltage to zero. That always is the function of negative feedback. So the cap will charge up until the DC voltages at both inputs are the same.

ak
hey AK,
I simulated the circuit u were talking about, this is what i got!