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I'm planning to use a TL084 JFET quad op-amp with each op-amp configured as a voltage follower (datasheet). Some questions:

  1. Does Vcc+ have to equal -Vcc-? i.e. can I set them to Vcc+=15V, Vcc-=-8V?
  2. I'm planning to run the op-amp in a non-inverting follower configuration with the input V+ between -4V and 12V. The input common-mode voltage range says -12,15 V. Does this mean I can output any voltage between -4V and 12V in this configuration?
  3. The output short-circuit current is listed at 60 mA. Can I safely assume that my voltage follower can source ~60 mA of current?
  4. Should I be adding any passive components to this circuit to make sure it is stable?
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    \$\begingroup\$ Look at the diagram: ti.com/general/docs/…. It looks like you can source but cannot sink 60 ma. \$\endgroup\$ – ilkhd Aug 10 '15 at 18:09
  • \$\begingroup\$ @ilkhd When Vout (and therefore V+) is positive, I would be sourcing current, right? I'm not worried about sourcing too much current when I'm outputting a negative voltage. \$\endgroup\$ – nick_name Aug 10 '15 at 18:14
  • \$\begingroup\$ Still, 60ma is at the very end of the current range. Safer assumption is 50ma. \$\endgroup\$ – ilkhd Aug 10 '15 at 18:18
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    \$\begingroup\$ Judging by equivalent schematics - no, V- can by any. \$\endgroup\$ – ilkhd Aug 10 '15 at 18:23
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    \$\begingroup\$ Btw TL084 has offset, not convenient for making a follower. \$\endgroup\$ – ilkhd Aug 10 '15 at 18:34
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  1. No, no reason for them to be equal, so long as you observe the maximum (and minimum recommended) total voltage and keep inputs within the common-mode range.

  2. The minimum common mode range is +/-11V with a +/-15V supply. With supplies +15 and -8 you will have a common mode range of +11/-4. The input will work in this range, in the follower configuration.

  3. Certainly not. The current is guaranteed to be less than 60mA. It's a minimum of only 10mA and that's with a 15V supply. Use the output swing numbers. With a +/-15V supply it can supply +/-10V to a 2K load (5mA). It might be a bit less with a 22V supply compared to 30V, but probably in that range.

  4. Voltage followers are prone to oscillation if the load is capacitive, otherwise this amplifier is compensated to be unity-gain (the way you are using it) stable.

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  • \$\begingroup\$ So, looking at Figure 6 in the datasheet above--how can I interpret this? Let's say my load is 500Ω and I'm using a voltage follower configuration--what would the max output voltage be? \$\endgroup\$ – nick_name Aug 10 '15 at 19:35
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    \$\begingroup\$ That looks like typical behavior so I would not use it for guarantees, but it looks like maybe (typically) +6/-4V into 500R (or a bit more) with +12/-8 supplies. That's +12mA/-8mA. \$\endgroup\$ – Spehro Pefhany Aug 10 '15 at 20:52
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First, there is no need for symmetric power supplies. However, if you do go this route, your output voltage level limits will also be asymmetric. If you look at the data sheet V(opp), you'll see that the voltage swings are only guaranteed to come within about 5 volts of the power supply, so for a +15, -8 volt setup, your output swing (with about a 10k load) is only guaranteed to be +10 to -3. You may well get better than this, but if you don't you have no complaints.

Your 60 mA is the short-circuit current. That means that if the output is grounded the chip will only draw a maximum of 60 mA. It does not mean that you can provide 60 mA into a load. If you look at Figure 6 you'll get an idea of how much load you can provide and still have the IC working correctly.

Your input common mode voltage is in much the same boat as your output voltage. You need to pay attention to the guaranteed limit, not the typical. In this case, Vcm is about 4 volts less than supply voltages. So for 15/8 you should not count on better than 11 volts to -4. If you do take the typical numbers as your guide, 15 to -5 would be a better choice. But again, if this doesn't work you have no one to complain to.

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