Channel strip balanced output stage - oscillation pbm

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saint gillis said:
ricardo said:
Have nice 100u electrolytics decoupling the rails near both IC1 & the 1646 but NOT to your CLEAN GND.  Have a separate DIRTY ground for sewage.
Ok! interesting idea!
see http://www.groupdiy.com/index.php?topic=37307.80 from about #41.

ricardo said:
C3 should be reversed.
Why?
Work out the voltages at the +ve & -ve inputs of the 5534 using its input current.

ricardo said:
1/2 a 5532 will do.
Why not a 5534?
That will do too.
 
saint gillis said:
ricardo said:
C3 should be reversed.
Why?
Work out the voltages at the +ve & -ve inputs of the 5534 using its input current.
You mean that maybe it should be reversed, maybe not, right?
You should work this out for ALL your designs .. but for this one C3 should be reversed.
  • The input bias through R1 puts the +ve input of the 5534 at a -ve voltage.  Assuming no DC on the pot, C1 is the right way round.
  • This -ve voltage is reflected on the -ve input on the 5534.  So C3 is the wrong way round
  • The input bias for the -ve input is via R2.  Because it is smaller than R1, there is less voltage across it so the output will be slightly -ve.
  • For small output offset, you need R1 & R2 the same.
 
saint gillis said:
ricardo said:
For small output offset, you need R1 & R2 the same.
Ok so if I lower R1 to 20k, does C3 direction still matters?
Yes.  C3 direction ALWAYS matters.

But making R1 & 2 equal deals with OUTPUT offset.

The +ve & -ve input pins of the 5534 will still be at a -ve DC voltage.

If you won't do the maths for input bias current, why don't you check out any similar 5532/4 circuit and measure the voltages?

If the OPA has PNP inputs like 4580 instead of NPN like 5532/4, you'll find the situation is reversed.  Then C1 will be the wrong way round.
 
saint gillis said:
ricardo said:
  • The input bias through R1 puts the +ve input of the 5534 at a -ve voltage


  • I don't really understand ... excuse me, what is "+ve" , does it refer to a positive voltage or does it mean "non-inverting input" ?

  • I'm not sure what you are asking, but I think I understand Ricardo.

    Opamps that use bipolar transistor input devices will have a base current flowing into or out of both inputs. The device data sheets will list this as input bias current. This current times the DC resistance to ground connected to those inputs will define an input voltage term that adds to other input errors.

    JR
 

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