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Problem: I designed this circuit for a project, but we have to use a biased power supply of +15v and -15v for the amplifiers.

I'm a bit confused about how this linear equation (V2 = 20(V1) + 5) would work if all the amplifiers used a biased power supply of +15v and -15v.

V1 = -0.25v to 0.25v AC V2 = 0v to 10v DC


Question: will this cut off the 20v output in the inverting Amplifier and change the equation to V2 = 15(V1) + 5.


Picture of the old circuit and the new circuit below enter image description here


An Amplifier biased using +15v and -15v power supply

enter image description here

problem solved enter image description here


Any answer would be greatly appreciated.

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    \$\begingroup\$ The circuit in your upper diagram makes no sense at all - why would you buffer and attenuate a -15V dc level only to then add it to an amplified version of vin - the lower of the two 10k resistors at the summing junction should be 30k then feed -15V straight in. You also don't appear to state what the original supply voltage was. \$\endgroup\$ – Andy aka May 24 '15 at 14:21
  • \$\begingroup\$ well I thought it would help to get rid of a loading effect. My teacher said it looked alright but she could be wrong. So should I get rid of the buffer? \$\endgroup\$ – Omuse May 24 '15 at 14:49
  • \$\begingroup\$ Oh sorry I didn't realize you answered that. \$\endgroup\$ – Omuse May 24 '15 at 14:53
  • \$\begingroup\$ Oh it's -0.25v to 0.25v AC as the input and 0v to 10v as the output. \$\endgroup\$ – Omuse May 24 '15 at 14:59
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Your circuit using 30k to -15 will work just fine, as long as you keep in mind the output limits. Depending on the op amp and the output load, an amp using +/- 15 volts will only be able to drive its output to somewhere in the range of +/- 10 volts to +/- 15 volts. Look up "rail to rail op amp".

So, if you're using an op amp with a +/- 10 volt output range, it will only respond to inputs in the range of -1 to +0.5 volts. For an op amp with +/- 15 volt capability, your input must be limited to the range -1.5 to +1.0 volts.

If you ever get to building circuits for real, this is not a good idea. Power supply levels are not usually held to high precision, and the voltage can vary with the loads being driven. For theoretical exercises like this it's OK, but in the real world can get you into all sorts of trouble.

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  • \$\begingroup\$ Thanks, that's really important info thanks for telling me. \$\endgroup\$ – Omuse May 26 '15 at 7:12
  • \$\begingroup\$ I tried simulating the new circuit in Multisim but I get a straight line for the output when it should be 0v to 10v AC. I posted the picture of the circuit above. \$\endgroup\$ – Omuse May 26 '15 at 10:30
  • \$\begingroup\$ nevermind I fixed it \$\endgroup\$ – Omuse May 26 '15 at 10:43
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Your power supply voltage will not affect the gain of the amplifier, but it will affect the voltage swing for the amplifier output. An opamp cannot output beyond its rails, and many opamps cannot approach the rails - see the opamp's specification for output voltage swing for details.

Since you're deriving the 5v offset from the power supply, you will need to adjust those resistors to give 1/3 of 15V instead of 1/4 of 20V.

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  • \$\begingroup\$ Okay, so that also means I don't need a buffing amplifier? I only added it to get rid of a loading effect from the voltage divider. \$\endgroup\$ – Omuse May 24 '15 at 15:01
  • \$\begingroup\$ But I didn't know a summing amp could also work like that.So I can change the lower resistor to 30k ohms? \$\endgroup\$ – Omuse May 24 '15 at 15:02
  • \$\begingroup\$ I tried simulating the new circuit in Multisim but I get a straight line for the output when it should be 0v to 10v AC. \$\endgroup\$ – Omuse May 26 '15 at 10:28
  • \$\begingroup\$ I posted a pic above of the problem. \$\endgroup\$ – Omuse May 26 '15 at 10:29
  • \$\begingroup\$ nevermind I fixed it. \$\endgroup\$ – Omuse May 26 '15 at 10:43

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