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I am looking to make a very accurate steady 5V power supply that can be switched on and off using the digital IO pins on an Arduino. I planned to use a 9V power supply connected to a 2N2222 transistor switched on/off using the Arduino and the emitter pin of the 2N2222 connected to an LM7085 to get a regulated 5V (see image). Simulations say I won't get 5V out but I am not sure why.

Any help would be much appreciated.

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Edited

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R Doyle
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    Welcome! _"very accurate"_ != 7805. How much accuracy do you need in numeric terms? Have you considered a linear regulator with an enable pin to simplify things? – winny Feb 20 '23 at 17:20
  • @winny Probably on the order of 1mV. Is there a better option than an LM7085? No, I hadn't I didn't know they existed. Would have any suggestions? – R Doyle Feb 20 '23 at 17:24
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    1 mV accuracy is more precise than any integrated linear regulator I've ever seen, but who knows. If you go looking for 7805 alternatives anyway, you may as well select one with a shutdown pin. – ocrdu Feb 20 '23 at 17:34
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    Close to 0 % chance on +-1 mV happening. Product recommendations are off-topic here but here are 1751 available 5 V regulators with enable pin to limit your search/point you in the right direction: https://www.digikey.com/en/products/filter/power-management-pmic/voltage-regulators-linear/699?s=N4IgjCBcpg7LBOKoDGUBmBDANgZwKYA0IA9lANogBMALAnBALrEAOALlCAMpsBOAlgDsA5iAC%2BxMAA4pABmQg0kLHiKkKIAMw0aYPSGYh2nHgJHixYoA – winny Feb 20 '23 at 17:41
  • Ok thank you for your help – R Doyle Feb 20 '23 at 17:45
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    This sounds like an XY problem to me. What are you planning on doing with this 5 +/- 0.001 V power supply? How much current does it need to supply? Does it need to be stable over temperature changes? What transient response do you need? – vir Feb 20 '23 at 17:46
  • @vir It is going to supply 4 voltage-dividing circuits with thermistors for temperature measurement. So the more accurate the input voltage is the more accurate my measured voltages will be. Current shouldn't be an issue I don't think. Stable enough to temperature, given the goal. I'd say the response could be long enough I will only be taking measurements every few minutes but the whole point is to turn the thermistor circuit on and off with the Arduino so the thermistors don't self-heat. So maybe then the transient response needs to be short? – R Doyle Feb 21 '23 at 09:54
  • Not a knock on you but this is very much an XY problem. The Arduino has an AREF pin that is specifically designed to deal with this situation. Tie the positive supply voltage of your dividers to AREF and the internal ADC will use that as the "top" of the conversion. Also, it's likely that the internal ADC has 10 bits of resolution which means that each "step" is 4.9 mV with a 5 V full scale. – vir Feb 21 '23 at 18:56

1 Answers1

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Simulations say I won't get 5V out but I am not sure why.

The NPN BJT is acting as an emitter follower and this means that the emitter voltage must be lower than the base voltage. Given that the base voltage is 5 volts, I estimate that the emitter voltage will be about 4.3 volts and no linear regulator in the universe is going to conjure up 5 volts on its output from 4.3 volts on its input.

Any help would be much appreciated.

I recommend you use a PNP BJT to do the switching and use an NPN to control the PNP. Better still, use a p-channel MOSFET for the switching device. Something like this: -

enter image description here

Modified image from here.

Andy aka
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  • Thanks Andy. Would something like the edited circuit above work? Where V1 is the arduino pin and ground. – R Doyle Feb 20 '23 at 17:16
  • An Arduino could connect to the node called on/off but I don't know what you mean by V1. Grounds need to be common of course. @RDoyle – Andy aka Feb 20 '23 at 17:24
  • I attached a modified version of my original circuit to my original post, which now uses a MOSFET. – R Doyle Feb 20 '23 at 17:26
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    Yes, that will work but, the problem you have is in generating the floating 5 volt control signal. I strongly suggest you use the circuit in my answer because it will directly interface to an MCU like an Arduino. – Andy aka Feb 20 '23 at 17:49
  • there's really no need for so low as 330R, another 10K would work there just as well, and save current.. – Jasen Слава Україні Feb 21 '23 at 05:39
  • @Andyaka Thanks. In your circuit what is the "Load"? – R Doyle Feb 21 '23 at 09:44
  • @RDoyle it will be your 5 volt regulator circuit. [What should I do when someone answers my question](https://electronics.stackexchange.com/help/someone-answers). – Andy aka Feb 21 '23 at 10:23