LM7805 5V regulator Minimum input voltage and diode protection

[superfly501]

I'm looking at 5V LM7805 regulators data sheets to use with my MCU and most versions of the LM7805 seem to have a minimum input voltage from 7-8V, but looking at the $50 robot here, it is using 6.0V Nimh batteries with the LM7805. Why can he use 6.0V when the minimum for LM7805 is labelled at 7-8V?

I've never used diodes in the past with voltage regulators, but is it a good idea to put them in for protection? I have some 1N5818 30V/1A Schottky diodes, would I shove one in series with the input voltage and another one with the output voltage?

[aaa3a]

u can use diode to cut about 0.7 from your 6 volt battery which gives 5.3 volt and i think your $50 MCU can stand with that this example can help you

Creating a +5V power supply from a 6V battery

 

The digital chips that you will be using with your lab kit require a +5V power supply. However, +5V batteries are not available. In this part you will convert a +6V battery into a +5V power supply by using a diode.

 

A diode is an electronic device that allows current to flow in only one direction. As it turns out, common diodes produce a voltage drop of around 1V (0.7V, actually) when current flows through them. We will use the voltage drop across a  diode in order to change a 6V supply into a 5V supply. You will build the circuit shown below using a 6V battery, your breadboard, and one diode.
Diodes are polarity-sensitive - the two leads are not interchangeable.  The symbol for a diode is shown below, along with a picture of the physical device.  Note that the cathode portion of the diode has a dark band on it.  In the symbol, the cathode is indicated by the point of the triangle. 
               

Build your circuit as follows (see diagram below):
1. Place your diode between hole 2g and the leftmost hole of the row labeled '+'. Make sure that the black band is on the side nearest the '+'.

2. Connect a wire from the bottom, right hole labeled '+' to the top, right hole labeled '+'.

3. Connect a wire from the bottom, right hole labeled '-' to the top, right hole labeled '-'.

4. Connect a wire from the bottom, left hole labeled '-' to the GND terminal of your 6V battery.

5. Connect a wire from hole 2f to the +6V terminal of your battery.

 

When you have hooked up the circuit,  you’ve created a power supply capable of supplying the standard digital levels of 5 Volts and 0 Volts, known as Vdd  (or Vcc) and GND. All of the holes along the rows labeled '+' are at 5V and all of the holes along the rows labeled '-' are at GND.


i hope that solved your problem if that's true kindly click helpful for me

regards

[superfly501]

Hey aaa3a,

Thanks for the response. I appreciate the effort to make the diagram for me, but that's not exactly what I'm looking for. I apologize if I was unclear in my original post, which I have edited now. I meant to say why the $50 robot uses 6.0V for the LM7805 voltage regulator, when the minimum input voltage stated by data sheets for most variations of the LM7805 is 7-8V. Also I was inquiring on using diodes with the LM7805, not on using diodes as a method of voltage regulation. I been wanting to to thank people, but I have no idea where that button to raise people's helpful count is.

[aaa3a]

the electronic parts has some tolerance and there are  some regulators can accept 6volts as input test that on a breadboard first and measure it by voltmeter and u will see it works about using diode as protection i see it's good idea but take care it will cut vdd=0.7 volt from your output volt and i think it's exaggerated to use diode for protection because the voltage regulator will not harm your MCU if it is fry but use it if you want or you may use small resistor as fuse

to make me helpful click under my name on the left helpful and that's not a must but if my reply really help u you can do that


thanx 

[Soeren]

Hi,

I'm looking at 5V LM7805 regulators data sheets to use with my MCU and most versions of the LM7805 seem to have a minimum input voltage from 7-8V, but looking at the $50 robot here, it is using 6.0V Nimh batteries with the LM7805. Why can he use 6.0V when the minimum for LM7805 is labelled at 7-8V?

Oh, it gets worse...
a 6V NiMH is 5 cells of (nominal) 1.2V/cell.
A NiMH cell charges to 1.40V..1.45V and is considered flat when discharged to a voltage of (depending on discharge current) 0.9V to 1.1V (larger discharge currents means a lower End-Of-Discharge [EOD] voltage).
This means that a 5 cell battery is around 7.0V..7.25V right out of the charger, but down to 4.5V..5.5V at EOD (when flat).

LDO (Low Drop Out) regulators can be found with drop out voltages as low as a few hundred mV, while 0.5V..1.0V is probably more common, so the right type can maintain a 5V output down to around 5.3V input and this will probably let you use around 80..90% of the capacity (depending very much on discharge current).

A shunt regulator (eg. a zener regulator, whether boosted or not, or an integrated shunt regulator) will be able to maintain 5V out right down to virtually 5V in (it works by shunting the over-voltage).

Switching regulators can maintain 5V out, with an input varying more than a 5 cell battery and they are much more efficient than a linear regulator (which have a loss of: voltage drop multiplied by output current).


Some controllers will run happily from eg. 2V to 5V and can thus be used with a 3-cell NiMH with no regulation at all.

I've never used diodes in the past with voltage regulators, but is it a good idea to put them in for protection? I have some 1N5818 30V/1A Schottky diodes, would I shove one in series with the input voltage and another one with the output voltage?

The best (as in minimal loss) way to apply a diode for reverse voltage protection is to use a fuse inline with the positive line and then the diode over the supply with the anode to ground and cathode to the positive line ("reverse biased"). with correct polarity nothing happens, but if you reverse the polarity, the diode conducts, shorting the supply and blowing the fuse (the fuse keep the reverse voltage from reaching dangerous levels (for the downstream circuitry) and the fuse protects the diode from shorting out permanently.

It's a very good solution for careful people (people who never reversed polarity on something or other, are people with a very limited build history, if any at all, so better safe than sorry), but if you're sloppy about polarity, it can get expensive in replacement fuses

If you use polarized connectors (making reverse polarity impossible), you have no use for that diode.


A different use of diodes with integrated voltage regulator circuits is for protecting the regulator against the backwards discharge of eg. large filter capacitors on the output side, while the input has gone to 0V, but that's not your everyday circuit.

[superfly501]

Hey Soeren, thanks very helpful as usual.

Is there a reason why LM7805 is more popular than using diodes as voltage regulator for MCUs, well according to my experiences? Everybody online and at my engineering labs seem to use LM7805 regulators. The downsides of diodes I would guess are the output voltage changes as the batteries discharge and diodes seem to have a lower current rating.

The best (as in minimal loss) way to apply a diode for reverse voltage protection is to use a fuse inline with the positive line and then the diode over the supply with the anode to ground and cathode to the positive line ("reverse biased"). with correct polarity nothing happens, but if you reverse the polarity, the diode conducts, shorting the supply and blowing the fuse (the fuse keep the reverse voltage from reaching dangerous levels (for the downstream circuitry) and the fuse protects the diode from shorting out permanently.

Would you ever suggest using low resistance resistors, say 1/4 W and 10ohm, as fuses? Let's say a circuit that could draw up 3-5 A? With such a small voltage drop across it, I don't see it passing the power rating until there is a short. Also why can't we simply use a diode inline with the power source instead of making the said fuse circuit? Wouldn't something like a schottky have a very high breakdown voltage, preventing a reverse current from going through anyway?

A different use of diodes with integrated voltage regulator circuits is for protecting the regulator against the backwards discharge of eg. large filter capacitors on the output side, while the input has gone to 0V, but that's not your everyday circuit.

I've heard that high load changes can cause current to reverse, which is why diodes can be helpful, is that what you are referring to?