You should also notice that these resistors are connected directly across the base-emitter junctions of their respective transistor, Q1 or Q2, and this means that the voltage across R3 or R5 is capped at around 0.7V. Referring now to your schematic, the two resistors R3 and R5 are sat squarely in the path of any current (over and above its normal operating current) that the opamp is sourcing or sinking via its output. Again, the key thing to notice is that this "sunk" current now emerges from the opamp's negative power terminal: In that scenario, current now enters the output, and the opamp is said to be sinking current.
If, however, the load is attempting to "pull up" the output potential, the opamp counters by switching on Q2 instead. The key point to realise is that all this extra current comes in via the opamp's positive supply terminal: The opamp output is said to be sourcing current. When the load connected to an opamp's output is trying to "pull down" that output to some lower potential, then the opamp's upper output transistor (Q1 in my diagram below) will switch on in an attempt to counter this. To understand the roles of R3 and R5, start by understanding the output stage of a typical opamp, in particular what happens when an opamp is sourcing current to a load, and how that changes when it is sinking current from the load. Thus I am unable to say anything about what R4 actually does, except that Andy aka's idea of protecting the opamp input is more likely than mine.
I agree with Andy aka, but I'm not yet sure how R4 could protect the opamp's input from voltages at the output that exceed its acceptable input range.
R4 is to balance the opamp's input bias currents, as explained in this Andy aka has pointed out that the input bias current is so small that my inital thought about R4 is incorrect.
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