Thread replies: 6
Thread images: 2
Thread images: 2
Anonymous
Operational amplifier, homework 2016-11-26 20:24:31 Post No. 226440
[Report] Image search: [Google]
Operational amplifier, homework 2016-11-26 20:24:31 Post No. 226440
[Report] Image search: [Google]
File: homework.png (13KB, 640x300px) Image search:
[Google]
13KB, 640x300px
I'm humbly requesting some aid in this exercise.
The question is at what value of U1 does the red/yellow LED activate.
I did a simple division of the voltage and got 8.9V as the 'pivot', and over/above that as activating red/yellow. My mate went to a study group and got the answer 4.8V as the 'pivot', which is more likely to be correct. But I really don't get how.
Anyone here that know much about electronics that can lend a hand?
>>
I mean that 4.8V is probably correct since there was a bunch of people at the study group and 4.8V seemed to be the consensus. But how?
>>
File: FmYjcHV.gif (1024KB, 218x228px) Image search:
[Google]
1024KB, 218x228px
Damn, I got it now, the resistances are parallel, not in sequence...
>>
>resistances are parallel
No, its a series voltage divider(10K + 4.7K) that is completely independant. The question is:
>What is the voltage at the + input?
To find the voltage at the +:
>what proportion of the voltage (15V) does the 4.7K resistor drop?
R/Rtotal gives you the proportion, then multiply by the total voltage.
V = (R / Rtotal) * Vapplied
V = 4.7 / 14.7 * 15v
V = 4.796
>>
>>226440
Best resource for how op amps work.
http://chrisgammell.com/how-does-an-op-amp-work-part-1/
>>
>>226440
This really isn't about the OP Amp at all, it's more just seeing if you know what a voltage divider is.
Thread posts: 6
Thread images: 2
Thread images: 2