Debating on using LM2937 or 7805

[MrDEB]

I think I would be safe reguardless which one but would it really make any difference. Just curious
Disreguard the win tag as it is comming off (9v input to pic = smoke)
going to insert 2n2222 to enable the phototransistor section (when dark = windows come on.
the PIC is using PWM

 

[Hero999]

If you're going to use a wall wart then the LM7805 is fine, if you're using a battery, go with the LM2937.

 

[MrDEB]

using a wall wart
thats what I kinda figured, the 7805

 

[MrDEB]

according to this thread I should use the LDO regulator?

No current limiting required; the circuit will draw only the current that it needs.

You will be wasting a lot of power in the 7805. Say your PIC board draws 100mA which means it requires 5*0.1=0.5W. The voltage drop across the 7805 will be 12-5 = 7V. The same 100mA flows through it, so it will be dissipating 7*0.1 = 0.7W, which means it requires heatsinking, making the overall efficiency 0.5/(0.5+0.7), or 41%.

It would be better to start with a 6V battery, and use a low-dropout regulator (not a 7805). Another way to up the efficiency is to use a switching regulator like a 34063, which would up the overall efficiency to ~90%, meaning that you would only be drawing ~46mA from the battery (instead of 100mA).

 

[MrDEB]

link to source
https://www.electro-tech-online.com/threads/pic-power-system.105144/

 

[Electronworks]

If you are using a wall cube input, you only need to worry about heat. Efficiency is not an issue. Work out (or measure) how much current your circuit takes and calculate the power dissipation from:

P = (Vin - Vout) x current

If this results in P being greater than about 0.5W, your linear regulator will need a heat sink.

Either way, an LDO + heatsink will be a much simpler design than an 34063 switched mode design (as outline in a previous post).

In terms of performance, there will be no difference between a 2937 and a 7805, although the 7805 will be cheaper

 

[Hero999]

What's the supply voltage?

The LDO doesn't make any difference to the power dissipation for a given input-output voltage at a certain current, it just reduces the minimum operating voltage.

If you're using a 9V battery, you want it to work until the battery voltage drops to 6V so a LDO is required.

If you're powering this off a 12V automotive system when go for the LM7805 which will only work down to 7V.

 

[MrDEB]

Using a wall wart 9v dc output.
Have several to choose from with different current capabilities.
Need to caculate total current but it can't be much (a PIC with 8 LEDs running under PWM, a phototransistor. Thats about it (see schematic)
Not real happy with the phototransistor section. Feel there has to be a better methoid to perform same function
DOOR OPEN WINDOW LEDs DISABLED
DOOR CLOSED WINDOW LEDs ENABLED but only in darkness.
pretty simple

 

[Hero999]

Then go for the LM7805.

The filter capacitor on a typical wall wart is often undersized, so add a 1000µF to 2200µF capacitor to the input of the LM7805.

Why not just use a regulated 5V wall wart so you don't have to worry about adding a regulator?

 

[MrDEB]

I have several 9v dc wal warts and being that I am powering a PIC I want to be guaranteed no morte than 5 volts.
Besides this application is going to a friends house that is "OFF GRID"

 

[Electronworks]

an unregulated wall cube + 5V regulator will be cheaper than a regulated wall cube