Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

LM324 supply voltage

Status
Not open for further replies.

jimkess

Member level 4
Joined
Dec 22, 2016
Messages
76
Helped
0
Reputation
0
Reaction score
0
Trophy points
6
Activity points
650
How to select voltage supply value for LM324 op-amp? what is the relation between output voltage of op-amp and the supply voltage? I am doing a schematic that has supplied 12V to the op-amp, but i want to use 5V, what will be the difference?

Also can i connect the ground of the 5V chip IC and ground of 12V supply to the same ground? making sure here

thanks
 

How to select voltage supply value for LM324 op-amp? what is the relation between output voltage of op-amp and the supply voltage? I am doing a schematic that has supplied 12V to the op-amp, but i want to use 5V, what will be the difference?

Also can i connect the ground of the 5V chip IC and ground of 12V supply to the same ground? making sure here

thanks
Let's (gasp) let the spec sheet tell us:

Capture.PNG

Yes, you can (and should) connect both grounds together.
 

just to make sure, does LM324 also amplify ac signal? the question arose since LM324 does not require -ve voltage supply rail, so will the output be rectified or will be output just be ac signal?

i tested the circuit in proteus and there was no clippering of the the negative side but in real i am getting clipped signal(rectified signal) with only positive side.
 

Hi,

For sure LM324 can amplify AC signal.
But output swing (and input volateg range) is limited by the supply voltage rails (minus dropout voltages).

Show your proteus circuit with waveforms.

Klaus
 

The usual solution for a single supply OP amplifying AC is to couple it through a capacitor and bias the amplifier output to Vcc/2, or in case of LM324 a bit lower to center of the useable output swing.

Needless to say that you are allowed to operate LM324 with dual supply if it serves your circuit purpose.
 
the ckt/waveform is attached.

the question remains then, after 10-bit ADC, the output are digital values from 0 to 1024. when these values are plotted the signal is showing clipped ac signal(i guess). should it be like this? how do one plot ac signal from the ADC values showing both negative and positive swing?

1.gif2.gif
 

Hi,

You circuit shows an integrater OPAMP circuit, where the DC offset is not defined. It may be anywhere.

On a real circuit you may find that the output either drifts to (and saturates at) the positive rail or the negative rail.

You need to stabilize your DC operating point.

Klaus
 

Hi,

You circuit shows an integrater OPAMP circuit, where the DC offset is not defined. It may be anywhere.

On a real circuit you may find that the output either drifts to (and saturates at) the positive rail or the negative rail.

You need to stabilize your DC operating point.

Klaus
i am actually not amplifying the input but integrating. for this circuit how do i create the DC offset? or DC bias it is called i have come to know.

edit:
is the example bias circuit good for this integrator?
3.gif

i am also not sure whether i need ac coupled or dc coupled circuit, how do i decide that? i need to know amplitude of the input signal at the output...i think i need dc coupled circuit
 
Last edited:

Hi

how do i create the DC offset?

At first you need to specify where you want your output DC point.

You have an AC input signal, therefore I assume you want a symmetrical output swing. (This usually calls for a dual power supply)
This means DC bias is at VCC/2.

There are several ways to achieve this.
Instead of connecting +IN to GND ... use a symmetrical voltage divider maybe 10k/10k between VCC and GND. Connect the center of the voltage divider to +IN.
Then you need a DC feedback from output to -IN.
Use a high ohmic resistor 1M or even higher.
This makes the integrator "non ideal". Indeed it creates a low pass filter (related to the AC input) with a low cutoff frequency.
fc = 1 / ( 2 * Pi * R * C)
(the area below fc is for DC stabilizing, the area above fc is the integrating area)

It depends whether your AC source has a low impedance DC path to GND or not.
I assume it has a low impedance DC path to GND.
(Now I can´t read your device values in the picture for the integrating Ci and Ri)
Use th same value as Ri and connect it from -IN to VCC.

That´s it.

Klaus
 

used your setting, but i am getting a straight line dc signal at the output, there is no ac signal at the output or the amplitude at the output is too small.

used 12Vsupply, with the voltage divider i fed center 6V into +IN according to your advice. After that, I connected the 1M(10M also) in the feedback path from out to -IN. Checked various frequencies for fc but there is nothing at the output. Next, removed the resistor in feedback path, there is output with decreased amplitude. But as you say, it sounds like the resistor in feedback is mandatory for correct dc bias??

so overall with the design you said, there is ac signal at the output.
 

Hi,

Show the simulated curcuit.

Klaus
 

here is the circuit and oscilloscope output.

1.gif2.gif

input voltage 50mV and frequency tested at 50Hz and 1kHz, same result(dc output)
 

Hi,

"You need a DC feedback.."
For this you need to connect the resistor in parallel to your feedback capacitor.
(In your case DC is still blocked by the series capacitor)

"Use the same value as Ri and connect it from -IN to VCC..."
This is missing in your circuit.

The quality of the picture is still that bad, one can not read any values.

Klaus
 

is the circuit and waveforms good now?

1.jpg2.jpg
 

Hi,

The circuit is like recommended.
The waveform like expected ... by me.
Hopefully you expected the same.

Klaus
 
the o/p is as expected. but i have yet to understand the the feedback res and the res between -ve in and power supply.

thanks
 

Hi,

feedback.
A capacitor blocks DC. This means there is no DC feedback. = not possible to stabilize the DC operating point.

RES:
First see that both resistors at +IN create half the supply voltage. -IN = VCC/2
You have to do the same for +IN: 10k from signals source (to GND) and 10k to VCC. So if the input signal is 0V, then you have VCC/2 at -IN
Then the Opamp circuit is in stable, regulating state. V_out = VCC/2.

Klaus
 
why was the -IN connected to Vcc with resistor 10kOhm? why was this resistor value selected? I checked other biasing circuit but didn't produce result like yours, any book recommendation?
 

Hi,

from post#9:
(Now I can´t read your device values in the picture for the integrating Ci and Ri)
Use th same value as Ri and connect it from -IN to VCC.

Klaus
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top