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diagram

This is essentially my circuit that SHOULD get 9 v from the barrel jack connector and if that cuts off, draw the current from the battery. Basically the beginning is a simple battery backup system I found, but I find it to be rather silly since it draws current also from the battery draining it and whilst it can be used (with a resistor) to charge a battery, it doesn't have overcharge protection, so before redesign that bit, the beginning of the circuit is essentially a placeholder.

Essentially the next bit uses an opamp as a buffer to create a virtual ground to be used for the main circuit and then two 500 mA linear voltage regulators to regulate the post-diode ± 4.2 V currents to ± 3.3 V currents.

The datasheets for the regulators are:

https://www.analog.com/media/en/technical-documentation/data-sheets/ADP1715_1716.pdf (for the positive rail)

https://www.analog.com/media/en/technical-documentation/data-sheets/ADP7185.pdf (for the negative rail)

What this circuit will eventually power is a microcontroller (connected to the +3.3v rail and virtual ground) and several analog sensors (for which I need the negative rail) and feed the result after rectification to the microcontroller.

Is there some huge fails in there I have missed?

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Update:

From the great comments I modified the circuit further, dropping the battery backup since it's not really needed at this stage. I went along with the BJT transistor suggestion, since I thought it would be the more cost-effective option than a specific opamp (the versions of the jellybean transistors I chose should be able to carry 600 mA according to their datasheets).

Circuit #2

I mean... how hard can it be to create a regulated ± 3.3 V from a single 9 V source?

I also considered using REC5-1205DRW/H4/A/SMD (https://recom-power.com/pdf/Econoline/REC5-RW.pdf) as a costly but straightforward alternative given that it's also been rated for medical devices, but it would probably be a "tad" overblown and it too would require also a negative source voltage and to acquire that I would need to drive the bastard with 18-36 volts in total, which I think would be really overdoing it.

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    \$\begingroup\$ I’m a little concerned the TL072 is only spec’ed to supply about 10mA and you’re asking it to sink/source the net of all your loads, which may very well exceed that. May have to add boost transistors to shoulder the current load. Also, if the incoming 9V happens to be a little less than the battery voltage (a new battery may be 9.5V) then the battery will source some current even when power running from jack. May want a 2nd diode on the diode path or a Schottky on the incoming power to make that less likely. And the 1uF from the battery is probably not needed. \$\endgroup\$
    – td127
    Feb 21, 2022 at 0:37
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    \$\begingroup\$ First off, your C? doesn't appear to be connected correctly. Take a close look at it. Next, I see a \$9\:\text{V}\$ battery in the circuit. This suggests to me an alkaline, which is the most common version and found everywhere around the world. The Energizer Max at about US$2.50 each has a datasheet that specifies two different end-point voltages: \$4.8\:\text{V}\$ and \$6.0\:\text{V}\$. \$\endgroup\$
    – jonk
    Feb 21, 2022 at 6:02
  • \$\begingroup\$ It would be wiser to plan for these in your design, or else provide us with a specification of a different end-point voltage for your design. Also take note that while the datasheet provides some high-current bar chart values, the actual curves provided don't anticipate currents anywhere near those you suggest in your question. (Don't forget to note that the curves do not specify 100% operation. Instead they specify something closer to about 5% usage with a long relaxation time between uses. Continuous duty isn't even shown there!) You really need to tell us what you expect from the battery! \$\endgroup\$
    – jonk
    Feb 21, 2022 at 6:03
  • \$\begingroup\$ The biggest drawers of current would be a) Raspberry Pi Pico up to 100 mA (most likely a lot less) and b) 2.4" TFT screen which would be also up to 100 mA. The sensors are biosignal sensors (pulse, EMG and galvanic skin response - ie a biofeedback device) and their current consumption would be probably 50 mA max in total. The battery part isn't actually essential, but possibly I'll implement a rechargeable cell system of somesort to provide the possibility of use without an electrical socket. Of course I could also skip the battery bit and leave it to a later iteration, which would be wiser \$\endgroup\$ Feb 21, 2022 at 14:23

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There appear to be a major problem with your circuit. You mention that the two regulators are capable to deliver as much as 500mA but you do not mention how much current you will be drawing from the regulators. Let's hypothesize that the Negative regulator deliver no current at all and that the positive regulator must supply 500mA. Then the problem is within the poor TL072 that must suffer to sync as much as 500mA + the current necessary to maintain the ADP1716ARMZ.

The TL072 is not capable of such high current. At best it could supply much less, in the order of 10 Volts onto 2k resistor, 5mA.

What you need is another type of OpAmp or you need to increase the output capacity of the TL072 with some BJT transistors. Something like this could do the job. Just make sure the transistors are of sufficient power to handle 600mA at 5 volts.

enter image description here

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  • \$\begingroup\$ Good point. The TL072 was essentially a placeholder opamp for me to simulate the sensor part of the circuit and hadn't really thought about what opamps to use in the final design, but your point is certainly valid and hadn't thought about the current requirements for the power supply level opamp. \$\endgroup\$ Feb 21, 2022 at 14:15

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