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Topic: LM78L05 and LM79L05 supply arduino (Read 2162 times) previous topic - next topic

Yaro

I need to adjust voltage from two 9v batteries in such a way to have an +5v and -5v output.
This is the scheme:

But I have doubts. I've checked pins on datasheet and they are the same on the sketch, but what is the right way of the piece? seen from above the component is like this: D
I've connected it, with the flat face front on me, like in the sketch. But when I measure the voltage between the two outputs O I note 10v, it's right? a kind of sum between +5v and -5v? What is the explanation?
Instead the voltage measured between O of LM78L05 and the gnd of LM79L05 is +5 v.

If it is right so I can power arduino mini (3v3) with the two outputs of the voltage regulators or do I need to use the output of LM78L05 and the gnd of LM79L05.

Ty!

Datasheets links:
http://www.sparkfun.com/datasheets/Components/LM7805.pdf
http://www.hep.upenn.edu/SNO/daq/parts/lm7915.pdf

Grumpy_Mike

First off you need capacitors from input to ground on both regulators.

Quote
But when I measure the voltage between the two outputs O I note 10v, it's right?

Yes that is right. One probe is at -5V the other is at +5V so the potential difference between the two probes is 10V.

Quote
do I need to use the output of LM78L05 and the gnd of LM79L05.

Yes.

Small 9V batteries are rubbish for powering arduinos, they don't last long.

Nick Gammon

You don't need -5V for the mini. A single LM7805 should do the trick.

However you might be better off with two or three x 1.5V batteries. Then you don't need the voltage regulator and they will last longer. Especially if you read up on using sleep mode.

http://www.gammon.com.au/power
Please post technical questions on the forum, not by personal message. Thanks!

More info:
http://www.gammon.com.au/electronics

dc42

You don't need the -5V supply at all. The INA122 will work down to 2.2V total supply voltage and its common mode input voltage includes ground. So ditch the 79L05 and connect pin 4 of the INA122 to ground instead. See Fig 5 on the datasheet.
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.

Yaro


First off you need capacitors from input to ground on both regulators.

Please, can you explain why i need capacitors to input? 1.0mF will be ok?

I've noticed that if i power arduino with 5v i have 5v on vcc, but i need to power an 23k256 too, that work with 3v3. How i can lower voltage to 3v3? I've read that a zener and a resistance can provide a stable supply. But how I can apply that system and what type of zener and resistance I need if i use arduino mini pro with power supply from 78L05?

oric_dan

#5
Dec 04, 2012, 03:39 am Last Edit: Dec 04, 2012, 07:43 pm by oric_dan(333) Reason: 1
Quote
Please, can you explain why i need capacitors to input?


Yes, read the datasheets for the parts. They tell you what is proper.

Quote
How i can lower voltage to 3v3?


First off, many but not all standard-size Arduino boards have a 3.3V v.reg on board.
You can add your own on other boards, but the zener idea may work.

Secondly, if you have an Arduino board that runs at 3.3V rather than 5V, you can
probably power it off a 5V regulator ok. However, you'll be running a bit close on loading
with powering from a 78L05, which can deliver only 100-mA. Would be better to use
a 7805 v.reg, ie 1 Amp device.

Thirdly, if you have mixed systems of 5V and 3.3V, you will need level-shifters to interface
the 23K256 to 5V I/O pins on an Arduino.

[Edit: I meant to say you can probably supply 5V to the Vin pin of a 3.3V board to power
the 3.3V v.reg on the board].

Grumpy_Mike

Quote
Please, can you explain why i need capacitors to input

To lower the impedance of the voltage source to prevent oscillation of the regulator.
Quote
1.0mF will be ok?

For those parts you linked to then yes. But only if the part is actually made by the same manufacturer who wrote the data sheet.

Quote
but i need to power an 23k256 too, that work with 3v3. How i can lower voltage to 3v3?

You use a voltage regulator designed to give you a 3V3 output.
That part is an SPI device so you will have to lower the drive voltage from the arduino. What I have used for this is a 74LS07 open collector buffer. You get 6 in on IC package, the inputs come from the arduino and the outputs to the SRAM with a 1K8 or so pull up resistor to the SRAM's 3V3 supply. The output of this chip can be fed straight into the arduino.

dc42


That part is an SPI device so you will have to lower the drive voltage from the arduino. What I have used for this is a 74LS07 open collector buffer. You get 6 in on IC package, the inputs come from the arduino and the outputs to the SRAM with a 1K8 or so pull up resistor to the SRAM's 3V3 supply. The output of this chip can be fed straight into the arduino.


Mike, I'd be interested to know why you are suggesting using a 74LS07 instead of the simpler option of a voltage divider?
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.

Grumpy_Mike

Quote
why you are suggesting using a 74LS07 instead of the simpler option of a voltage divider?

Voltage dividers can often slow down the rise time of the edge given the capacitance of the input. On the Ada wave shield their version 1 had voltage dividers but they changed to level translators for version 2 because of the trouble the users were getting with dividers. With a buffer ( which I already had so that was a plus factor for that part ) the falling edge could be very fast and the rising edge controllable by the size of the pull up resistor.
I used this arrangement on my SpoonDuino, worked first time.  :)

dc42


Quote
why you are suggesting using a 74LS07 instead of the simpler option of a voltage divider?

Voltage dividers can often slow down the rise time of the edge given the capacitance of the input. On the Ada wave shield their version 1 had voltage dividers but they changed to level translators for version 2 because of the trouble the users were getting with dividers. With a buffer ( which I already had so that was a plus factor for that part ) the falling edge could be very fast and the rising edge controllable by the size of the pull up resistor.


Thanks, but I'm still puzzled. Both the Arduino and the 74LS07 have a maximum output current of 40mA (it's under "recommended operating conditions" for the 74LS07, and is the highest current for which Vol is specified). So I don't see that using the 'LS07 improves the rise time.

Example:

1. 74LS07 with 82 ohm pullup resistor to +3.3v. 40mA sink current, right at the top of the recommended operating conditions.

2. Voltage divider with 100 ohms to ground and 51 ohms to the Ardino pin. Current sourced by the Arduino is a little under 33mA. The equivalent circuit is 3.3V fed through (51 || 100) ohms = 33 ohms.

So it appears to me that the voltage divider should give a rise time less than half that of the 'LS07, if pin input capacitance is the dominant factor. OTOH the 'LS07 will probably give a faster fall time.

I guess that if you wanted to level shift lots of signals, using t he voltage divider approach you would have to design for a lower current draw to stay within the 200mA chip limit, then the 'LS07 might be better.
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.

Grumpy_Mike

Quote
So I don't see that using the 'LS07 improves the rise time.

True, but then I never said it would. It is the fall time that controls the switch on of a ~CE signal and that is what I think causes problems.

Papa G


You don't need the -5V supply at all. The INA122 will work down to 2.2V total supply voltage and its common mode input voltage includes ground. So ditch the 79L05 and connect pin 4 of the INA122 to ground instead. See Fig 5 on the datasheet.


Consider the situation where the unbalanced state of the bridge, Vunb, is <= 0.1V/G, where G is the gain of the instrumentation amp. While the common mode input voltage includes ground, the output voltage of the instrumentation amplifier when operated from a single supply does not.

The OP's original plan takes that into consideration. If eliminating the negative supply is important, the output could be biased to a value > 0.1V by applying a reference voltage to pin 5 (Ref). A 2.5V reference might make sense in this application.

oric_dan

Quote
Voltage dividers can often slow down the rise time of the edge given the capacitance of the input. On the Ada wave shield their version 1 had voltage dividers but they changed to level translators for version 2


Sounds surprising. Do you know what values the Rs were in the v.divider? After all, a
2.4K:4.7K looking at a high 40 pF gives a 60 nsec time-constant, which sounds pretty quick.
And the R values could be reduced.

Do the typical SPI device loads have excessively high Cin?

oric_dan

To answer my own question, the schematics for both versions of the wave shield are
found here, click on downloads.

http://www.adafruit.com/products/94

The earlier version used 4.7K:10K v.dividers. These are interfacing to an SD Card, and not
to a typical little CMOS chip. So, the SD Card apparently has more stringent requirements
for driving. I don't know the schematic for an SD card.

Yaro


Quote
Please, can you explain why i need capacitors to input

To lower the impedance of the voltage source to prevent oscillation of the regulator.
Quote
1.0mF will be ok?

For those parts you linked to then yes. But only if the part is actually made by the same manufacturer who wrote the data sheet.

So i have find these:
7805
http://www.datasheetcatalog.org/datasheet/stmicroelectronics/2143.pdf
7905
http://www.datasheetcatalog.org/datasheet/stmicroelectronics/2149.pdf

7805 : 0.33 mF and 0.1 mF
7905 : 2.2 mF and 1.0 mF
I found all capacitor in tantalum but not the 1.0 mF
With what would be better to replace it? I was advised that ceramic is good too. And electrolytic is too slow.
But in these datasheet:
http://www.hep.upenn.edu/SNO/daq/parts/lm7915.pdf (it's not the same producer)
Say this: "²Required for stability. For value given, capacitor must be
solid tantalum. 25 mF aluminum electrolytic may be substituted.
Values given may be increased without limit."
So it's better 1.0 mF ceramic or 25 mF electolytic?



Quote
How i can lower voltage to 3v3?


First off, many but not all standard-size Arduino boards have a 3.3V v.reg on board.
You can add your own on other boards, but the zener idea may work.

Secondly, if you have an Arduino board that runs at 3.3V rather than 5V, you can
probably power it off a 5V regulator ok. However, you'll be running a bit close on loading
with powering from a 78L05, which can deliver only 100-mA. Would be better to use
a 7805 v.reg, ie 1 Amp device.

Thirdly, if you have mixed systems of 5V and 3.3V, you will need level-shifters to interface
the 23K256 to 5V I/O pins on an Arduino.

[Edit: I meant to say you can probably supply 5V to the Vin pin of a 3.3V board to power
the 3.3V v.reg on the board].


I've changed to 1 Amp devices, and i'll use raw pin on arduino pro mini (3v3 ver.). It'll be ok?

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