OP AMP power supply

  • Thread starter neg_ion13
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  • #1
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Main Question or Discussion Point

Hello,

I am trying to design a foundation for an audio preamp that is powered by a single 9V battery. I am successful so far except the single power supply part. I am using the voltage divider scheme to get a +4.5V and a -4.5V reference voltages. With no load the supply voltages are close to even but when I add a load of say 1K the voltage shifts at the supply pins. The plus increases and the neg sides decreases. Do this mean the positive side is pulling more current and therefore dropping more voltage? The set up is 1mV AC input to non-inverting side with negative feedback for A = 100.
 

Answers and Replies

  • #2
4,662
5
How much current do you need at the midpoint between 0 and 9 volts? You could use two equal series resistors (e.g., 10k ohm) and a LM324 (or LM358) as a voltage follower to get a stable voltage. It can source/sink up to 8 milliamps.

Bob S
 
  • #3
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Thanks for the suggestion. I would to keep the IC count low as I am running on a 9V battery. I'm not sure what current I need yet as I have not considered the next stage of the amplifier. The main problem I am having is; due to this voltage offset at the supply pins my negative half of the signal is clipping to early as I adjust the gain. What would cause the supply levels to shift like that? I am using 2 10k resistor for the voltage divider with ground referenced in the middle of the V divider and the voltage + and - at either end of the divider.
 
  • #4
berkeman
Mentor
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Hello,

I am trying to design a foundation for an audio preamp that is powered by a single 9V battery. I am successful so far except the single power supply part. I am using the voltage divider scheme to get a +4.5V and a -4.5V reference voltages. With no load the supply voltages are close to even but when I add a load of say 1K the voltage shifts at the supply pins. The plus increases and the neg sides decreases. Do this mean the positive side is pulling more current and therefore dropping more voltage? The set up is 1mV AC input to non-inverting side with negative feedback for A = 100.
Using 2 9V batteries would probably be cleaner. Or invert the 9V with a DC-DC to give you the negative rail.
 
  • #5
1,762
59
Hello,

I am trying to design a foundation for an audio preamp that is powered by a single 9V battery. I am successful so far except the single power supply part. I am using the voltage divider scheme to get a +4.5V and a -4.5V reference voltages. With no load the supply voltages are close to even but when I add a load of say 1K the voltage shifts at the supply pins. The plus increases and the neg sides decreases. Do this mean the positive side is pulling more current and therefore dropping more voltage? The set up is 1mV AC input to non-inverting side with negative feedback for A = 100.
What value are the resistors in your voltage divider?

If one side of your load is connected to the output of the opamp, where is the other side of your load connected?
 
  • #6
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As I said in my first reply 2 10ks. The other side of the load is connected to ground.
 
  • #7
1,762
59
Sorry, I missed your reference to the 2 10Ks.

Where is ground, at the junction of your voltage divider or -9V?
 
  • #8
30
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At the junction of my voltage divider. So I have +4.5 at the top, ground ref in middle and -4.5 at bottom. Also, I made sure my op amp output offset was zero volts with the non-inverting input connected to ground.
 

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