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Asmyldof
Asmyldof
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Unfortunately there's no time for me to work through the solutions to the issues, so I'm just going to list the issues and leave fixing them to someone else, or your own googling and I'll focus just on the capacitor.

  • You can't use NPN's to switch a positive rail like you do on the left, this will not work properly.
  • You cannot use 390x types to work with half amperes or more.
  • You cannot just put an LED (D8) in series with a heavy load, that will kill the LED or quench the load.
  • A linear voltage regulator, especially an old-fashioned one like the LM317 or LM78xx, needs not just "more than 5V (output voltage)", it needs 2 to 3 volts more than the ouput voltage. So you need more than 7V, possibly for full stability more than 8V.
  • Whether or not you want a resistor depends on the power you will let yourself have as inrush, but more than a few Ohms hardly ever happens in low voltage applications.
  • There's nothing bypassing your pre-charge resistor (which is R13 it seems, not R12), so it's actually just an in-line one.
  • Quite frankly, your schematic is very messy, which makes it hard for anyone to read, and that much less likely for them to try in their free time. Half the time I had for this was spent trying to make sense of it.

Lastly, and separately:
Assuming a perfectly constant current drain (dangerous assumption, but good enough for ball-parking), capacitance required is given by the linear derivation of the time-integrand, this derivation is: dV = (i * dt) / C (anyone is allowed to mathjax that for me for edit-points).

This means that at 0.5 A your capacitor, rating 5*10^-4, will create a voltage drop of:

dV = (0.5 A * 1 s) / 5*10^-4 F = 1000V

For each second.

Or put differently, for 3 seconds, with a drop of no more than 13V - 8V = 5V:

5V = (0.5A * 3s) / C ==> C = (0.5A * 3s) / 5V = 0.3F

And that needs to be 300mF that is also capable of supplying the high currents, not some run-of-the-mill super-cap made to support a 1mA load, with an internal resistance of tens or hundreds of ohms at a specified voltage of 16V.

Unfortunately there's no time for me to work through the solutions to the issues, so I'm just going to list the issues and leave fixing them to someone else, or your own googling and I'll focus just on the capacitor.

  • You can't use NPN's to switch a positive rail like you do on the left, this will not work properly.
  • You cannot use 390x types to work with half amperes or more.
  • You cannot just put an LED (D8) in series with a heavy load, that will kill the LED or quench the load.
  • A linear voltage regulator, especially an old-fashioned one like the LM317 or LM78xx, needs not just "more than 5V (output voltage)", it needs 2 to 3 volts more than the ouput voltage. So you need more than 7V, possibly for full stability more than 8V.
  • Quite frankly, your schematic is very messy, which makes it hard for anyone to read, and that much less likely for them to try in their free time. Half the time I had for this was spent trying to make sense of it.

Lastly, and separately:
Assuming a perfectly constant current drain (dangerous assumption, but good enough for ball-parking), capacitance required is given by the linear derivation of the time-integrand, this derivation is: dV = (i * dt) / C (anyone is allowed to mathjax that for me for edit-points).

This means that at 0.5 A your capacitor, rating 5*10^-4, will create a voltage drop of:

dV = (0.5 A * 1 s) / 5*10^-4 F = 1000V

For each second.

Or put differently, for 3 seconds, with a drop of no more than 13V - 8V = 5V:

5V = (0.5A * 3s) / C ==> C = (0.5A * 3s) / 5V = 0.3F

And that needs to be 300mF that is also capable of supplying the high currents, not some run-of-the-mill super-cap made to support a 1mA load, with an internal resistance of tens or hundreds of ohms at a specified voltage of 16V.

Unfortunately there's no time for me to work through the solutions to the issues, so I'm just going to list the issues and leave fixing them to someone else, or your own googling and I'll focus just on the capacitor.

  • You can't use NPN's to switch a positive rail like you do on the left, this will not work properly.
  • You cannot use 390x types to work with half amperes or more.
  • You cannot just put an LED (D8) in series with a heavy load, that will kill the LED or quench the load.
  • A linear voltage regulator, especially an old-fashioned one like the LM317 or LM78xx, needs not just "more than 5V (output voltage)", it needs 2 to 3 volts more than the ouput voltage. So you need more than 7V, possibly for full stability more than 8V.
  • Whether or not you want a resistor depends on the power you will let yourself have as inrush, but more than a few Ohms hardly ever happens in low voltage applications.
  • There's nothing bypassing your pre-charge resistor (which is R13 it seems, not R12), so it's actually just an in-line one.
  • Quite frankly, your schematic is very messy, which makes it hard for anyone to read, and that much less likely for them to try in their free time. Half the time I had for this was spent trying to make sense of it.

Lastly, and separately:
Assuming a perfectly constant current drain (dangerous assumption, but good enough for ball-parking), capacitance required is given by the linear derivation of the time-integrand, this derivation is: dV = (i * dt) / C (anyone is allowed to mathjax that for me for edit-points).

This means that at 0.5 A your capacitor, rating 5*10^-4, will create a voltage drop of:

dV = (0.5 A * 1 s) / 5*10^-4 F = 1000V

For each second.

Or put differently, for 3 seconds, with a drop of no more than 13V - 8V = 5V:

5V = (0.5A * 3s) / C ==> C = (0.5A * 3s) / 5V = 0.3F

And that needs to be 300mF that is also capable of supplying the high currents, not some run-of-the-mill super-cap made to support a 1mA load, with an internal resistance of tens or hundreds of ohms at a specified voltage of 16V.

Source Link
Asmyldof
Asmyldof
  • 18.5k
  • 2
  • 35
  • 53

Unfortunately there's no time for me to work through the solutions to the issues, so I'm just going to list the issues and leave fixing them to someone else, or your own googling and I'll focus just on the capacitor.

  • You can't use NPN's to switch a positive rail like you do on the left, this will not work properly.
  • You cannot use 390x types to work with half amperes or more.
  • You cannot just put an LED (D8) in series with a heavy load, that will kill the LED or quench the load.
  • A linear voltage regulator, especially an old-fashioned one like the LM317 or LM78xx, needs not just "more than 5V (output voltage)", it needs 2 to 3 volts more than the ouput voltage. So you need more than 7V, possibly for full stability more than 8V.
  • Quite frankly, your schematic is very messy, which makes it hard for anyone to read, and that much less likely for them to try in their free time. Half the time I had for this was spent trying to make sense of it.

Lastly, and separately:
Assuming a perfectly constant current drain (dangerous assumption, but good enough for ball-parking), capacitance required is given by the linear derivation of the time-integrand, this derivation is: dV = (i * dt) / C (anyone is allowed to mathjax that for me for edit-points).

This means that at 0.5 A your capacitor, rating 5*10^-4, will create a voltage drop of:

dV = (0.5 A * 1 s) / 5*10^-4 F = 1000V

For each second.

Or put differently, for 3 seconds, with a drop of no more than 13V - 8V = 5V:

5V = (0.5A * 3s) / C ==> C = (0.5A * 3s) / 5V = 0.3F

And that needs to be 300mF that is also capable of supplying the high currents, not some run-of-the-mill super-cap made to support a 1mA load, with an internal resistance of tens or hundreds of ohms at a specified voltage of 16V.