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ygg-it 29th April 2008 03:21 PM

Cancelling DC Offset in a bipolar Opamp
I have two non inverting bipolar opamps connecting in series.
Total gain = 8 (4*2)

Audio input signal go into the (+) input as for a classical non inverting configuration using an RC network

C= 4u7F connected between the the source and the (+) input

R= 22K connected between the (+) input and GND

I have feed the plus (+) input of the first opamp with a DC signal connecting a 1,6 M resistor to a wiper of a 500K trimmer connected between the regulated -15V and +15V power supply.

I have regulated the trimmer to get null offset at the end of the last amplifier.

It works very well, even if I have to wait till 5 minutes since the offset adjustment is very sensible to temperature. (starting from 50mV went to aroun 1-2 mV after 5 minutes)

I would like to know from you if this solution is good or there are any hidden disadvantages...... I'm doing something wrong?
Is the same to feed the (+) input or better to feed the (-) input?
Are the components value ok?

The purpose is HIFI headphone amplifier

AndrewT 30th April 2008 10:14 AM

a variation of 50mV to 2mV of output offset sounds too much from cold to warm.
I think there must be a better way.

ygg-it 30th April 2008 02:05 PM

My chip is AD826

My feedback resistor is 3,3k||10k
My input gnd resistor is ten times (22k)

From the datasheet the input bias current vs temperature go from 2.5uA to 3.3uA in the range 20C - 60 (max temperature in the case)

Just in the first stage this generate a Voffset from 200mV to 258mV that would be multiplied by two in the second stage (400mV to 516mV)
Of course my servo circuit is stable with temperature so you get this big variance at the output during powering on....

jcx 30th April 2008 03:07 PM

the plus and minus input resistance to DC potential should be equalized for minimum Vos drift with input bias current changes

ygg-it 30th April 2008 04:15 PM

Yes I know but I want to stay with the 22k / 4,7uF roll off

I'm just asking if my servo DC is correct, i'm not looking into other values solution since this is the one that works better.

AndrewT 30th April 2008 04:32 PM

where is the servo?

ygg-it 30th April 2008 11:08 PM

2 Attachment(s)
For servo I intend feed the (+) input with a 1,6Mohm resistor connected to the wiper of a 500k potentiometer

AndrewT 1st May 2008 07:30 AM

that's an input bias setting/trimmer not a servo.

Your DC resistances on the inverting and non-inverting inputs of both opamps are not matched.
22k//1M6 should be equal to 3k3//10k
100r should be equal to 1k1//1k1.
Change your feedback resistances and your input resistances to get these to match.

I don't like your input attenuator. Better to have a switchable gain.

ygg-it 2nd May 2008 07:04 AM

>that's an input bias setting/trimmer not a servo

yes, you are right, sorry for the miswording

>Change your feedback resistances and your input resistances to get these to matchC

already done, offset went to spare millivolt, but the RC with lower C and higher R sounds much better. So I would like to stay with the bias setting.

Again: I'm just asking is for you is ok the solution to feed the not inverting input (so have no impact on the feedback circuit)

>I don't like your input attenuator. Better to have a switchable gain

The switchable gain would change part of the feedback network, so the offset voltage would change too as a result. In addition you would overload such a vulnerable circuit area with extra capacitances and wiring

cliffforrest 2nd May 2008 10:14 AM

There is no need to use a 500K pot.

10K to 50K is fine, but 10K will bleed 3mA from the supplies.

In general keep all impences low (other things being equal) to minimise noise pickup, layout problems.

EDIT: And I should have added that you are injecting any PSU noise into the op-amp input, so a 20K pot with a cap to ground on the wiper to filter this out may be wise. Not too big or it will make adustments difficult.


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