# DC offset for LM386why is it there, and how can it be removed?

Discussion in 'General Electronics Chat' started by Rogare, Mar 27, 2012.

1. ### Rogare Thread Starter Member

Mar 9, 2012
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0
Hello! I'm amplifying an audio signal using the LM386 in its "Minimum Parts" configuration, as given on its datasheet, but I'm getting a DC offset of about 1V. The problems with this are that a) it appears to limit my output gain, and b) I believe it will lead to a hum in the output signal.

I'm wondering why this is happening, and what I can do to fix it? (On the datasheet, I read about an output bias of half the supply voltage, but I'm using a supply voltage of about 8 V.)

My specific application is to amplify a short rectangular wave from 00.3 V to 0X V, where X can be set up to about 4 V (half of the supply).

2. ### praondevou AAC Fanatic!

Jul 9, 2011
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Since the input is a square wave from 0 to X it contains DC, right? This DC will appear at the output of the chip (amplified).

Is this a fixed frequency?

3. ### Rogare Thread Starter Member

Mar 9, 2012
78
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Thanks for the reply, praondevou. Sorry, I wasn't clear. The input is a unipolar rectangular wave that oscillates between 0 V and, say, 0.3 V.

0.3 V ------------------------------------
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0 --V ------------------------------------

I'd like to amplify it so that the output oscillates between 0 and, say, 4 V.

4 --V ------------------------------------
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0 --V ------------------------------------

But my response looks like the signal below. Because the max output voltage of the LM386 is half of the supply (8V), I'm stuck at a maximum wave amplitude of 3V, not the 4V I'd wanted.

4 --V ------------------------------------
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1 --V ------------------------------------

4. ### Audioguru Expert

Dec 20, 2007
9,877
1,043
You forgot to say what resistance is the load.
The output cannot go to 0V and it cannot go to the supply voltage.
There is a graph in the datasheet that shows a maximum output of 5.5V peak-to-peak when the supply is 8V and the load is 8 ohms. That is only 2.75V peak.
The graph shows a max output of 7V peak-to-peak when the supply is 8V and the load is infinity. That is only 3.5V peak.

Of course the output has a voltage loss. The schematic of the LM386 shows emitter-follower output transistors like nearly all audio amplifiers.

EDIT:
I forgot to say that with an 8V supply and a 0V input, the output is +4V, not 0V. But its output cannot go down to 0V anyway, even when its input voltage is negative.

Last edited: Mar 27, 2012
5. ### Rogare Thread Starter Member

Mar 9, 2012
78
0
OK, I think I follow. So, what's happening in my bottom graph (in the post above) is that the output wave is centering itself between 0 and the supply voltage. ($V_{out}=V_{in}*gain+\frac{1}{2}*V_{supply}$.)

I'm looking to get the output above 3.5V (up to 4 or higher) and have no DC offset: just 0 up to ~4 and then back down again ($V_{out}=V_{in}*gain$). Am I better using a different chip? Any recommendations would be terrific.

6. ### Ron H AAC Fanatic!

Apr 14, 2005
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What are you driving that requires direct coupling?

7. ### Rogare Thread Starter Member

Mar 9, 2012
78
0
Hi Ron, the load is an earphone.

8. ### Ron H AAC Fanatic!

Apr 14, 2005
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Why don't you use a DC blocking cap, as recommended in the datasheet?

9. ### Audioguru Expert

Dec 20, 2007
9,877
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You are feeding DC into the earphone which offsets its diaphragm which will probably cause distortion. The DC might overheat the earphone or burn it out.

Sound transducers are fed AC, not modulated DC like you are planning. the LM386 and every other modern amplifier that uses a single supply voltage have an output coupling capacitor that blocks DC but passes the AC.

With an 8V supply and a 50 ohm load, the maximum output from an LM386 is about +7.2V and the minimum is about 0.8V. If you couple to the earphone with a series capacitor then the earphone gets a maximum of plus 3.2V and minus 3.2V which is symmetrical and is 6.4V peak-to-peak. The maximum power in the earphone will be 102mW into 50 ohms.

10. ### Rogare Thread Starter Member

Mar 9, 2012
78
0
I'm looking for a chip (or configuration of this chip) that could allow a peak output voltage up to the supply voltage.

From what I understand, even with a capacitor at the output, I'd still then be limited by an amplitude of $\frac{V_{supply}}{2}$. Is that right?

11. ### Ron H AAC Fanatic!

Apr 14, 2005
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Are you putting audio through this amplifier/headphones? Even with an ideal single rail-to-rail amplifier (LM386 isn't), you will never get peak-to-peak output of more than the supply voltage.

12. ### Rogare Thread Starter Member

Mar 9, 2012
78
0
Thanks, Audioguru. A short, positive rectangular burst will give an audio click, which is what this circuit will be amplifying. I see what you mean about adding in a capacitor to get rid of the DC offset, but because my signal is positive only, I end up losing out on 3.2 V of potential gain. Anyways, thanks for all the helpI think for now I may just try and double the supply voltage!

13. ### #12 Expert

Nov 30, 2010
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You are wrong about losing voltage range when adding a capacitor. A coupling capacitor blocks DC but it doesn't block sudden changes in the DC.

14. ### Ron H AAC Fanatic!

Apr 14, 2005
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If you are wanting to only send an occasional positive pulse to the headphones, then an amplifier whose output is biased at or near ground will give you more range, as you surmise.

15. ### Audioguru Expert

Dec 20, 2007
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1,043
No.
That is DC. You do not want DC, you want AC through a coupling capacitor.
You want an output peak-to-peak AC voltage swing that is the entire supply voltage. No amplifier is perfect like that.

Not if you increase the supply voltage so the output voltage swing is 8V peak-to-peak which is what you want.

You want 8V peak-to-peak into 50 ohms. The graph in the datasheet of the LM386 shows that the supply voltage must be 9.5V.

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16. ### Audioguru Expert

Dec 20, 2007
9,877
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Why are you making a "CLICK" (or thud) in an earphone?
Wouldn't a blast from a rock group or from a symphony orchestra sound more interesting?
Maybe use the sound from a car horn. A truck air-horn? A train whistle? A ship's fog horn?