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Topic: Circuit Protection with Zener 5,6V (Read 493 times) previous topic - next topic

lalbuque2000

Marcio,


You need to connect the ATMEGA to the connection between the zener diode and the resistor. Your drawing is not connected right.
Make sure you check the ATMEGA pinout, in your drawing you are connecting to pin 8, but the +5Vcc should be pin 7 if you are using the ATMEGA328P 28 pins PDIP package.

I would recommend the use of a LM7805 instead of a zener for several reasons:
1) The voltage drop on the resistor will vary with the current requested by the ATMEGA.
2) The current requested by the ATMEGA will depend on the number of outputs you are using and how much current each output will use.
3) The maximum current supported by the zener will depend on the zener maximum power and the resistor value you are using.

If you really need to use a 5.6V zener, and I am assuming you are using a 1Watt zener (or 1000mW), you need to know the maximum current you need on your ATMEGA to define the resistor value. For instance, if you need 20mA and the input voltage is at most 8V you need a resistor of 100 ohms.

The main problem with the zener circuit is when your input voltage drops to 5V and the ATMEGA still needs 20mA, the voltage drop on the resistor will be ~2V and the ATMEGA will receive only 3V and it will not work.

Using the LM7805 you only need to make sure the input voltage is at least 6V depending on the LM7805 model you are using, consult the datasheet.

Why does your voltage change from 5V to 8V?

Luciano

aweatherguy

Marcio, I think there might be some confusion on what you are trying to protect. The schematic you posted shows pin numbers but does not specify which package. I assumed you have the 28-pin DIP package and are therefore trying to protect the MCU's power supply. If you're using one of the SMT packages then perhaps you're protecting a digital input pin?

If you want a low-dropout regulator you could try the REG-03 from Texas Instruments. It offers 500mA output with only 0.115V dropout -- you would get 5V out with only 5.2V applied to the input. It doesn't get much better than that. It is a little pricey at $6.27 US for onesies. It's also a surface mount (aka SMT) package and I don't know if you have the ability to work with SMT.

The zener idea is what's called a "shunt regulator". Your schematic is not correct -- the output is taken between the resistor and diode. This might be feasible, but only under a limited set of conditions. It is difficult discuss the feasibility of this approach without knowing more about your scenario:

  • What are the maximum and minimum input voltages that will connected?
  • If there are temporary excursions of the input voltage, how large, and how long will they last?
  • What is the acceptable range of voltages on the regulator output?
  • How much current do you need to deliver to the load (MCU and peripherals) at 5V?

Marcio_Marques

Hi Guys,

thank you for your replies! i already decided, i will put one LDO, which aweatherguy adviced, (REG-03) of Texas Instruments. it has a very good Dropout.

i also thought put a MC33269D-5.0 of Arduino board, but it has a 1.0V Dropout, and i needed put 6,0V in Input to get 5,0V in Output.

Thank you for your attention, Any doubt, I'm here.


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