🚀 go-pugleaf

Re: steuer questions

 [STUFF ABOUT OPAMP GAIN DELETED]

	Ok what you had was a classical negative feedback amplifier
like so. When connected in this manner you can think of the op-amp
						  as a device which
	Vi	      |\			  will adjust its 	
	--------------|+ \  			 to make the inputs
                      |    \            	 the same. That is	
	              |      >----------Vo 	 with an applied 
		      |	   /      |		 Vi of 1V the output
	         -----|-  /       |	 will be adjusted such that 
		|     |/          |	 the resistor division supplies
                |	R2	  |	 one volt to the neg pin.
		|------/\/\/\-----
		|			    Notice that the current 
		< R1			 through R2 virtually all goes
		>			through R1 since the input 
		|		impedence of the amplifier is large.
		|		If you do a bit of simple math here
		GND		You will find the gain = (R2+R1)/R1

				examples
				If R1=R2 the gain = (1+1)/1 = 2
				If R1 = 1K & R2 = 4K
					 The gain = (1+4)/1 = 5
	      
	Notice, when keeping R1 constant, increasing R2 increases the
	gain. Conversely decreasing R2 makes it less to a minimum of
	one when R2 = 0. In both your examples R1 was
	very large indeed being the input impedence of the device itself.
	In the first case R2 was nearly 0 (wire resistance) therefore
	you had (K + 0)/K where K was the imput impedence. ie K/K = 1
	In the second example you made R2 not zero but still very much
	smaller than the input impedence therefore the gain did not
	seem to change appreciably. If you had made R2 large enough you
	would have had an increase in gain as it started to come within
	orders of magnitude of the devices input impedence. By this time
	though the size of the impedences would have made the circuit
	very susseptible to noise.

--
       >|  "Only average people            | Trevor J German BSc    |<
       >|        never make mistakes."     | NCR, Waterloo, Ontario |<
       >| tjgerman@53iss6.Waterloo.NCR.COM | Canada.                |<