LDO replacement for 7805?

Status
Not open for further replies.
Use a voltage regulator with the current rating you require or boost it.

Have you actually tried parralleling them?

I agree that it isn't a good idea in general but I wouldn't totally rule it out. It might not be as Nigel suggests, as one will start to shut down and reduce its voltage slightly, the other might start to take the strain more. I suggest you try it, use a suitable heat sink for the power dissipation and try measuring the input current to each regulator. I might experiment with this myself as I'm interested to see exactly what will happen.
 
I'll Add a heatsink to the D2 and put a fan on it to help throw away the heat. I will try paralleling them and stuff. My Binary clock is almost done, I just gotta hook up the displays.
 
I have recently come across a SAMSUNG component KA78R05 it is a low drop device and it has a 4th pin to shut down in case of overcurrent. the Datasheet is attached.
 
Last edited:
Some regulator datasheets show a schematic where they use a transistor in conjunction with the regulator to provide more current then the regulator itself can supply. Could this be used with a LDO regulator?
 

Attachments

  • Image4.jpg
    220.9 KB · Views: 2,049
  • Image5.jpg
    23.3 KB · Views: 1,709
3v0 said:
Some regulator datasheets show a schematic where they use a transistor in conjunction with the regulator to provide more current then the regulator itself can supply. Could this be used with a LDO regulator?
Yes it can but it needs to have an additional 2nd transistor to provide current limiting.
 
Hero999 said:
Also not that you have to add Vb + Vce at the desired current to the dropout voltage.
I would add only |Vbe| to the drop-out voltage of the regulator, ΔVreg.
And |Vce| = |Vbe| +
ΔVreg when the transistor is on.


EDIT: as far as I know, there are LDO regulators that can source 750 mA or more, as the OP requires.
 
Last edited:
eng1 said:
I would add only |Vbe| to the drop-out voltage of the regulator, ΔVreg.
And |Vce| = |Vbe| +
ΔVreg when the transistor is on.


EDIT: as far as I know, there are LDO regulators that can source 750 mA or more, as the OP requires.
Perhaps there are low current devices as well while the one KA78R05 for e.g., can give 1amp load.
 
mvs sarma said:
Perhaps there are low current devices as well while the one KA78R05 for e.g., can give 1amp load.
Yes, I've read the datasheet you posted.

For my requiremets I usually use the LM2937 (500 mA max). For more current, I have few LM2940's (Io = 1 A max, the drop-out voltage is 0.5 V typ. and 1 V max).
 
eng1 said:
Yes, I've read the datasheet you posted.

For my requiremets I usually use the LM2937 (500 mA max). For more current, I have few LM2940's (Io = 1 A max, the drop-out voltage is 0.5 V typ. and 1 V max).
Well - you are right we don't get so free a supply of these LDOs in small Qty / retail. Incidentally one auctioner provided us with this chip(78R05) and finding comfortable.
when a component is not on hand in 3D, only a 2D document (pdf) is to be enjoyed. Perhaps Digikey will supply and i shall be spending few$+$50 for the international post iteself.
For eample, here i needed CY27EE16ZE to complete an Elektor (5/07) Project (Software Defined Radio) . Hope to get it some day!!
 
If it turns out that you need two regulators to handle the total load, you don't need to hook their outputs in parallel, just use one to supply the LEDs and the other to supply the TTL.
 
Status
Not open for further replies.
Cookies are required to use this site. You must accept them to continue using the site. Learn more…