Hi,
I have a circuit with an Atmega328p microcontroller, the input voltage is 5Volts, but I want, when the input voltage goes up, eg: 6, 7 or 8Volts, the microcontroller and the remaining circuit just stick with the 5.6Volts.
I have a zener of 5.6Volts, connected to the 5Volts of the input, and a resistance of 1K to the mass.
The circuit is attached
the problem is that it is not working, does anyone know what the problem is, or how can I solve it?
thank you.
The Zener should be between 5V and GND with no resistor between. It is not super good protection - but I must admit smell of burning electronics warned me something is wrong before any damage was caused (except my burned fingers, the Zener was HOT, I guess it was close to dying).
hmm all right, but if I don't put the resistor, the zener will explode as soon as I raise the input voltage to 7 or 8 volts.
Yes. It will protect only from short transients or if the power source is weak. Or it may blow fuse (activate polyfuse).
so, what´s the best choice to protect against overvoltages and overload?
I am no pro, I have only read part of a book about this topic! But it all depends on source of the overvoltage. Is it only brief power source overshoot (on power up?) or direct lightining strike? Can you afford to blow fuse protecting your electronics (use fuse and "crowbar circuit") or does the electronics need to continue to work uninterrupted?
Marcio_Marques:
Hi,I have a circuit with an Atmega328p microcontroller, the input voltage is 5Volts, but I want, when the input voltage goes up, eg: 6, 7 or 8Volts, the microcontroller and the remaining circuit just stick with the 5.6Volts.
I have a zener of 5.6Volts, connected to the 5Volts of the input, and a resistance of 1K to the mass.
The circuit is attached
the problem is that it is not working, does anyone know what the problem is, or how can I solve it?
thank you.
If the input voltage is that variable you should use a regulator to give 5V. Probably a low-dropout
regulator.
What is the 5V source that sometimes 8V?
You might consider using a voltage regulator instead. Check the schematic for any Arduino board to see an example. Attached is a crop from the Uno schematic showing the voltage regulator. Note that IC1 and IC2 are overlapping footprints and only one of them can be installed on any one board. The UNO boards I have been populated with the IC2 option.
This will "protect" the MCU from inputs as high as 20V -- depending on how much current you draw in the overall system. The highest sustained input voltage that is safe will depend on how much power is dissipated in the voltage regulator and how you've mounted it for heat dissipation. Any more on that topic is probably beyond the scope of this post.
I'm not seeing you attached picture...
Are you talking about the power supply or signals (inputs)?
For signals, the ATmega chip has protection diodes (I think they are mainly intended to protect against static discharge). I'm not sure what the current rating is... I'm pretty sure you'd be safe with a 10K resistor but I'm not sure about a 1K... Probably 1K is enough...
Have a look at the “Ruggeduino”. - their web site shows the protection circuits they use on inputs/outputs etc . The idea of using a voltage regulator on an input is a good one and I’ve seen it used before.
Zener diode protection is basically useless.
Voltage limit for an Arduino pin is not 5volt.
It's VCC + 0.5volt (and GND - 0.5volt).
If the Arduino is off, then pin limit is 0.5volt, making a 5.6volt zener useless.
Arduino pins have internal clamp diodes. They might be ok for occasional protection if current is limited to 1mA.
External schottky clamp diodes (with current limiting resistors) could be another option.
Tell us what you have connected to the Arduino pins.
And why you connect this to three pins.
You might not need all of this.
Leo..
hammy:
Have a look at the “Ruggeduino”. - their web site shows the protection circuits they use on inputs/outputs etc . The idea of using a voltage regulator on an input is a good one and I’ve seen it used before.
The Ruggeduino pin protection makes the mistake of assuming that the Arduino is always on.
Schottky diodes between pin and VCC and pin and ground, and a big 5volt TVS diode across VCC would have been better.
That said, I have seen people phantom-powering an Uno (~50mA) throught the internal pin protection diodes, and not kill it.
Leo..
Thanks for your replies, I already put the zener as it is in the attachment, and I already added a voltage regulator LM7805, but when I apply 5v in the LM7805 input, it only gives me output 2v, only when I put 6V in the input, it puts 5v in the output.
Its a power input, not a signal input or output.
I think the best way is the Arduino uno schematic, with voltage regulator. But I'm working with DIP package
Sorry, my attachment not work. But is simple, is just one zener with 1k resistor connected to a 5V power line.
Different regulators have a different dropout voltage. You need to base the regulator on how much current you need, the size package, and heatsinking. Some LDO regulators require very little overhead voltage, which is more efficient since that voltage is turned to heat. A quick look on digikey shows a 2V dropout for the 7805 at 1A. I saw a TO-92 5VLDO regulator with a .6V dropout, but it was also limited to 100mA.
What exactly is the power source? Min/max voltages?
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:
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
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:
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.
