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Topic: Questions regarding Voltage regulators (Read 1 time) previous topic - next topic

Cue2

Jan 20, 2013, 04:14 am Last Edit: Jan 20, 2013, 04:16 am by Cue2 Reason: 1
If a positive voltage is applied to the output pin of a LM7805 voltage regulator while the input pin is left floating (with 10uF capacitor to ground) will this damage the regulator?

If I want a 3.3V supply from say a 9V battery is there any disadvantage in using the output of a 5V regulator as the input of a 3.3V regulator rather than wiring the battery straight to the 3.3V ?

LarryD

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If a positive voltage is applied to the output pin of a LM7805 voltage regulator while the input pin is left floating (with 10uF capacitor to ground) will this damage the regulator?

The only way to answer this is to connect it correctly and see if it still works.

Quote
If I want a 3.3V supply from say a 9V battery is there any disadvantage in using the output of a 5V regulator as the input of a 3.3V regulator rather than wiring the battery straight to the 3.3V ?

I believe you need more than a 2 volt difference between i/p and o/p.

Why wouldn't you just connect the Vin to the i/p of both regulators?
The way you have it in your schematic isn't the same as how you have it wired up!

CrossRoads

Damage may occur:

Texas Instruments:
http://www.ti.com/lit/ds/symlink/lm340-n.pdf (covers 78xx series regulators also)

SHORTING THE REGULATOR INPUT
When using large capacitors at the output of these regulators, a protection diode connected input to output
(Figure 9) may be required if the input is shorted to ground. Without the protection diode, an input short will
cause the input to rapidly approach ground potential, while the output remains near the initial VOUTbecause of the
stored charge in the large output capacitor. The capacitor will then discharge through a large internal input to
output diode and parasitic transistors. If the energy released by the capacitor is large enough, this diode, low
current metal and the regulator will be destroyed. The fast diode in Figure 9 will shunt most of the capacitors
discharge current around the regulator. Generally no protection diode is required for values of output capacitance
? 10 ?F.
RAISING THE OUTPUT VOLTAGE ABOVE THE INPUT VOLTAGE
Since the output of the device does not sink current, forcing the output high can cause damage to internal low
current paths in a manner similar to that just described in the "Shorting the Regulator Input" section.

Solution shown is 1N4002 from output (Anode) to input (cathode)

Fairchild
http://www.fairchildsemi.com/ds/LM/LM7805.pdf
Nothing mentioned

Read the datasheet for other sources to be sure.
Diode from out to in will certainly not hurt.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

SirNickity

I wish my datasheet had said something about that.  I don't know if it's not much concern for the parts I use, or if they just neglected to warn about that particular case.  Nothing I've made so far has suffered, but as I very recently became aware of this practice, all my future projects will incorporate it.

Sometimes it seems that "low parts count" is synonymous with "not very robust".

CrossRoads

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"low parts count" is synonymous with "not very robust"


Agreed.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

BillO

If the input to the regulator is left floating and not grounded accidentally, or attached to a lower voltage source, there should be no problem.  But as CrossRoads noted, a diode sure won't hurt.


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"low parts count" is synonymous with "not very robust"


I am not sure I agree with this.
Facts just don't care if you ignore them.

CrossRoads

Take a look at this for a well protected board

http://ruggedcircuits.com/html/ruggeduino.html
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

BillO

Very nice.

It seems that you cannot power it from the USB cable.  I actually like that idea.
Facts just don't care if you ignore them.

CrossRoads

Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

BillO

Okay, I see it now.  Hmm, almost a little disappointed.
Facts just don't care if you ignore them.

oric_dan

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If I want a 3.3V supply from say a 9V battery is there any disadvantage in using the output of a 5V regulator as the input of a 3.3V regulator rather than wiring the battery straight to the 3.3V ?

First off, as always come up around here, the little 9V batteries don't have enough energy
to run an Arduino for very long. For battery power, a lot of people use NiMH AA-cells, or a
Li pack.

Secondly, the downside to connecting the 3.3V v.reg right to the input voltage is it will
run a little hotter than if daisy-chained off the 5V v.reg. The downside to running off
the 5V buss, however, is now the 5V v.reg must supply the current to the 3.3V v.reg,
so you don't have as much overall current capability, and it will heat up a little more.

On my own boards, I always use large 5V 1 Amp v.regs, and daisy chain to an LDO [low
dropout] 3.3V v.reg. The 1 Amp reg will provide the necessary current for most apps,
and is also physically large enough it won't overheat too badly.



dc42


Quote
If I want a 3.3V supply from say a 9V battery is there any disadvantage in using the output of a 5V regulator as the input of a 3.3V regulator rather than wiring the battery straight to the 3.3V ?

First off, as always come up around here, the little 9V batteries don't have enough energy
to run an Arduino for very long.


True. However, the OP is probably talking about a standalone atmega328 or similar arrangement, otherwise he wouldn't need voltage regulators. A 9V battery can power an atmega328p for quite a long time, if you don't have a USB-to-serial converter taking additional current, and you reduce the clock frequency to 8MHz or lower. What you shouldn't do is run motors, solenoids or other high current devices from a 9V battery.
Formal verification of safety-critical software, software development, and electronic design and prototyping. See http://www.eschertech.com. Please do not ask for unpaid help via PM, use the forum.

oric_dan

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the OP is probably talking about a standalone atmega328 or similar arrangement,

Ha, you read a lot more into his post than I did, LOL. He better not be drawing more than
about 25 mA total load, else the 9V battery won't last very long, cf 500 mA-hr of energy,

http://www.jameco.com/webapp/wcs/stores/servlet/ProductDisplay?freeText=198791&langId=-1&storeId=10001&productId=198791&search_type=jamecoall&catalogId=10001&ddkey=http:StoreCatalogDrillDownView


retrolefty


Quote
the OP is probably talking about a standalone atmega328 or similar arrangement,

Ha, you read a lot more into his post than I did, LOL. He better not be drawing more than
about 25 mA total load, else the 9V battery won't last very long, cf 500 mA-hr of energy,

http://www.jameco.com/webapp/wcs/stores/servlet/ProductDisplay?freeText=198791&langId=-1&storeId=10001&productId=198791&search_type=jamecoall&catalogId=10001&ddkey=http:StoreCatalogDrillDownView




And at two for $5 I see in drug stores they are a very uneconomical method of powering even a minimum arduino based project. They do look cute though.  ;)

Cue2

Thanks CrossRoads for the insightful read about V.reg damage. Thanks everyone for the advice regarding the battery and daisy chaining regulators too. It is a standalone Atmega328p project as some of you may have guessed. I currently use a power jack but the the thought of using a 9V battery crossed my mind since I have some spare rechargable 9Vs but I will look into the better suited alternative batteries.

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