1
\$\begingroup\$

Why does the output of the opamp not rail to VDD or VSS when the positive feedback factor (2/3) is larger than the negative feedback factor (1/2)?

enter image description here

Version 4
SHEET 1 880 680
WIRE 0 32 -224 32
WIRE 320 32 80 32
WIRE 176 144 176 128
WIRE -352 160 -448 160
WIRE -224 160 -224 32
WIRE -224 160 -272 160
WIRE -176 160 -224 160
WIRE 0 160 -96 160
WIRE 144 160 0 160
WIRE 320 176 320 32
WIRE 320 176 208 176
WIRE 0 192 0 160
WIRE 144 192 80 192
WIRE -448 208 -448 160
WIRE 176 224 176 208
WIRE 0 304 0 272
WIRE -448 336 -448 288
WIRE 80 384 80 192
WIRE 144 384 80 384
WIRE 320 384 320 176
WIRE 320 384 224 384
WIRE 80 416 80 384
WIRE 80 528 80 496
FLAG 0 304 0
FLAG 80 528 0
FLAG -448 336 0
FLAG 320 176 out
FLAG 176 128 0
FLAG 176 304 0
DATAFLAG 48 160 ""
DATAFLAG 80 272 ""
DATAFLAG -144 32 ""
DATAFLAG 256 32 ""
SYMBOL res 64 400 R0
SYMATTR InstName R1
SYMATTR Value 3k
SYMBOL res 240 368 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 3k
SYMBOL res 16 288 R180
WINDOW 0 36 76 Left 2
WINDOW 3 36 40 Left 2
SYMATTR InstName R3
SYMATTR Value 2k
SYMBOL res -192 176 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 0 56 VBottom 2
SYMATTR InstName R4
SYMATTR Value 2k
SYMBOL res -16 48 R270
WINDOW 0 32 56 VTop 2
WINDOW 3 0 56 VBottom 2
SYMATTR InstName R5
SYMATTR Value 1k
SYMBOL res -256 144 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R6
SYMATTR Value 4k
SYMBOL voltage -448 192 R0
SYMATTR InstName V1
SYMATTR Value 8
SYMBOL OpAmps\\OP77 176 240 M180
WINDOW 0 -36 131 Left 2
SYMATTR InstName U1
SYMBOL voltage 176 208 R0
SYMATTR InstName V2
SYMATTR Value 30
TEXT -456 384 Left 2 !.op
\$\endgroup\$
5
  • \$\begingroup\$ Looks like the gain of your system is \$A_v=\frac12\$ to me. And in this particular case (odd coincidence) \$R_5\$ is carrying no current at all. Why don't you write with a little more detail about what you are seeing? \$\endgroup\$
    – periblepsis
    Commented Jun 15 at 6:05
  • \$\begingroup\$ @periblepsis yes you're right. So does that mean the positive feedback can be ignored here? I don't know what else could I write. \$\endgroup\$
    – hana
    Commented Jun 15 at 6:05
  • \$\begingroup\$ Yes, because \$R_1\,R_4-R_2\,R_3=0\$. If that were not true, then you'd be in a different situation. But as it is, \$R_5\$ will never have any current. May as well remove it. \$\endgroup\$
    – periblepsis
    Commented Jun 15 at 6:10
  • \$\begingroup\$ @periblepsis Thanks, I just changed it and it seems like you're right. \$\endgroup\$
    – hana
    Commented Jun 15 at 6:26
  • 2
    \$\begingroup\$ You know that \$V_{-}=V_{+}\$ (ideal opamp). You know that \$V_-=\frac12 V_{\text{out}}\$. And you know that the shared node of \$R_4\$, \$R_5\$, and \$R_6\$ is \$V_x=2 V_+=2 V_-=2\left(\frac12 V{_\text{out}}\right)=V{_\text{out}}\$. So both sides of \$R_5\$ are always at the same potential. So no current in \$R_5\$. \$R_6\$ can be ideally any non-zero value. Doesn't matter. And \$R_5\$ can be removed. \$\endgroup\$
    – periblepsis
    Commented Jun 15 at 6:27

1 Answer 1

Reset to default
2
\$\begingroup\$

The negative feedback (output to inverting input) is 0.5 and the positive feedback (output to non-inverting input) is actually 1/3, and 1/3 < 0.5.

To see this either reduce the input voltage to zero or analyze a the change in the voltage at the non-inverting input for a change at the op-amp output.

The overall gain from V1 is 0.5.

The OP77 is not a rail-to-rail output anyway, so you’ll not get the full swing out (see the datasheet for typical swing numbers) but it will work as an amplifier rather than a comparator.

\$\endgroup\$
3
  • \$\begingroup\$ You're right about OP77 not a rail-to-rail but the positive feedback is R6||(R3+R4)/[R6||(R3+R4)+R5] = 2/3. \$\endgroup\$
    – hana
    Commented Jun 15 at 6:25
  • \$\begingroup\$ It is (R6||(R4+R3)/(R6||(R4+R3) + R5))*R3/(R3+R4) = 1/3 . Your calculation correctly gives the voltage at the R6-R4 junction, which is further divided by two before it gets to the non-inverting input. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jun 15 at 6:31
  • 1
    \$\begingroup\$ I see, you're right then. I forgot the voltage divider by R3 and R4. \$\endgroup\$
    – hana
    Commented Jun 15 at 6:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.