Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Will this regulator work for my project?

Status
Not open for further replies.

alanbacklund87

New Member
I've got a project I'm working on where my PIC10f200 Microcontroller MUST be controlled by power supply that gives off around 12 to 13v. I've successfully programmed my PIC10F200 using the 5v power that is supplied by my PicKit2 programmer but now it must work in my application.

My question is, will the78L05 voltage regulator successfully drop the 13v supply to a safe 5v for the VDD pin? If this will not work, what does everyone use?
 
Do you know how much current your PIC10F200 uses in you application? High current LEDs? etc
The 78L05 is low power. The data sheet will give a current rating.
Watts=Volts X Current
Volts = 13-5=8 For the LM78L05.
Data sheet will give a temperature rise/watt for that package. I like to keep my parts to under 25C temp rise form heating.
I did not give you an answer but a direction.
 
My PIC10F200 is using right at about 45ma and the 2 LED's that I am currently powering are drawing about 95ma a piece

Total Current draw on the circuit is 235ma
 
Last edited:
can't i just use the 78L05 to power my PIC10F200 and then my transistors and led's can just be powered directly off of the 13v source using 10k ohm resistors on the transistors and the appropriate value for the leds?
 
8 volts across the LM78L04.
45mA to power the PIC, (LED current from 13 volts)
Watts = 360mW
200C/watt (Assuming TO-92)
80C temp rise
Room temp=25C
Die temperature = 100C Max temp is 150C.
I think that is too hot.
 
Am i able to limit the current of the 13v on the input pin of the LM78L04 with a resistor to reduce the heat? would that work?
 
Last edited:
1) you could use a resistor.
2) you could use a 4.7 volt zener.
3) LM7805
I like 3 best and 1 least.
 
I'm currently working with a LM7805C but just don't like the size. I ordered a few of the 78L05 just to see if they will work for my application.

The LM7805c works great and does not produce any heat with a 470ohm resistor. When I remove the resistor there is warmth that i can BARELY feel to the touch. It appears that the resistor is doing the trick and does not affect my PIC10F200.

When I up the resistor to a 1k, the sequence of events that my PIC10F200 is supposed to perform fails slightly.

I'm hoping that when I get the 78L05 in the mail and test it, the PIC10F200 will perform the way it should even with a resistor. I guess I'll just wait and see. Considering the current draw is just 45ma, the 100ma rating for the 78L05 should work just fine.
 
Last edited:
Most PIC have a way to slow down the OSC. 1/4, 1/16, etc. I think the 10F200 only runs 4mhz.
45ma seem high.
We are using 10Fxxx and we put the part in to sleep mode much of the time. Check for event, sleep, check for event, sleep.
 
i thought it seemed high as well considering the tech sheet says that it should draw 20ma. My meter is only a $30 craftsman so really it could just be an unreliable reading.

I'd rather it read high than low

I'm new to all of this so I really don't know what your talking about when you say "check for event, sleep, etc..."
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top