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If we have an AC coupled amplifier like the one below, i've read and been told it needs a DC path to ground to allow input bias currents to flow, though I just can't get my head around this, I bias is drawn from ground? physically how does this work?

AC coupled amplifier

With reference to the circuit below the bias current seems to be drawn up from ground, fi no input is provided where is the driving source? is this related to internal biasing and therefore the power rail is the source?

analysis circuit for calculating effects of Ibias and Voffset

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Indeed Ib+ in the 2nd drawing is indeed flowing out of the ground.

There is nothing strange about that but you need to realize that in order to draw a current "out of the ground" and through a resistor (or an opamp input), you would need a negative voltage.

So the opamps used in the circuits you show need a negative supply rail assuming that the opamps have a biasing current which flows into the opamp.

There are also opamps which have a biasing current flowing out of the inputs (yep, sounds weird!) for example the LM324. This opamp is designed so that it can work without the negative supply voltage but still work with its input voltages close to the ground voltage.

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  1. \$I_{BIAS}\$ is either passed through R1 to ground or, supplied from ground via R1. Either way depends on the type of op-amp and the power supply arrangement (dual or single).
  2. The driving source is the output pin of the op-amp. It will adopt a voltage that satisfies the requirement to drive the bias current through resistor R2 whilst maintaining Vin- and Vin+ at precisely the same level.
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  • \$\begingroup\$ damn that makes a lot of sense, can anyone else confirm this to be the case so I can accept my dude's answer and move on with life? \$\endgroup\$ – Andrew Davis Feb 7 at 14:02
  • \$\begingroup\$ You can wait a day for other answer of course. It's no big deal (providing you do come back!) \$\endgroup\$ – Andy aka Feb 7 at 14:03
  • \$\begingroup\$ @AndrewDavis Yup, I can confirm that to be the case. \$\endgroup\$ – Neil_UK Feb 7 at 14:26

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