Hello,
I want to power my circuit with four 1,5V batteries.
I don't find a 5V voltage regulator that would do the job unless it is fed with at least 10V.
What would do the trick best?
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Hello,
I want to power my circuit with four 1,5V batteries.
I don't find a 5V voltage regulator that would do the job unless it is fed with at least 10V.
What would do the trick best?
You have a few options that you could use. It all depends on what you are trying to do with your application. Is this something you want to produce or is this something you are working with at home?
You could use a Zener diode rated at 5 volts. A zener is used to clamp the voltage.
You could make a voltage divider and by selecting the correct resistor values get an output voltage of 5V.
You could use a diode in series with your batteries, not a bad idea as it will give you polarity protection. This should drop the input voltage down by 0.7V.
You could purchase a low dropout voltage regulator. They are alittle more costly than standard voltage regulator. You can find them labeled as LDO's. Take a look on digikey.com and see what you can find.
Hi,
Use the voltage regulator LP2950.
75 μA quiescent current
100 mA output current
http://cache.national.com/ds/LP/LP2950.pdf
The LP2950 and LP2951 are micropower voltage regulators
with very low quiescent current (75 μA typ.) and very low
dropout voltage (typ. 40 mV at light loads and 380 mV at
100 mA). They are ideally suited for use in battery-powered
systems. Furthermore, the quiescent current of the LP2950/
LP2951 increases only slightly in dropout, prolonging battery
life.
Distrelec = 640722 (CHF 1.02)
Best regards,
Luciano
Thanks a lot!
I'll have a look at those LDOs I didn't know about.
The problem with the divider and the diodes is that they will follow the batteries voltage drop.
I'll have a look at DISTRELEC.
Thanks again.
Dear flotulopex,
LDOs are a good choice as suggested by Luciano. Depending on your application if you are using a nanowatt PIC then your PIC may work within the range of 2V-5.5V or 2V-6.0 volts. If you are using CMOS logic as well then this should not be a problem. Also for alkaline batteries the cell voltage is 1.5V, but if you are using a rechargeable solution with Ni-Cd or Ni-Mh then your cell voltages are 1.2 volts per cell. So let us know more about your app.
I have made a kind of timer - more or less something like the ones for the chess players but for up to 5 players. My mother just drives me crazy when we play Rami and she wins but she takes almost triple time for reflection....
So to the project, there's a PIC, an LCD display, 5 LEDs, a piezo and a few external pull-up resistors.
On my breadboard, I had no problem to get the timer work properly. I have a lab power supply so no trouble here.
When I made the prototype, my thought went first to the power supply. Due to the boxe size contraints, I choose to go with four UM4-AAA batteries (I didn't want to go with accus and a charger).
After some tests, I had to modify quite a few things such as:
- external 10k pull-ups (the internal ones look to be to weak because both buttons just switch from time to time even if nobody presses them)
- addition of 100nF cap between Vdd and Vss (I never needed before) and another one between LCD and Vss (same)
Actually, the circuit consumes around 50mA when the display is ON and one LED is activated (normal operation). Strangely, when I start the timer, the current drops to 20mA. I'm still invastigating this.
I also used another display for the final version than the one I use on my breadboard and this display is extremely sensible to power fluctuation. If I have 4,8V then the display will dimm a lot and if the power is over 5,4V, the display will just go black (the contrast setting is very sensible).
I think the LDO is effectively the best solution in my case.
Grazie Luciano ;-)
Hi,
LDO would remove all headaches perhaps. By the time you get it consider this.
1. Use a simple 5.1V Zener shunt regulator for your PIC and LCD.
Both of them are not that current hungry. (Beware not the backlight)
2. Power the LEDs, Buzzer (and the LCD backlight if you are using) from the unregulated supply. (Batt +ve)
3. For lighting up LEDs or Buzzer invert your logic and drive the cathodes while the anodes are connected to Batt +ve through necessary limiting resistors.
Hi,
Incorrect battery insertion is a common occurrence.
See this link:
FET Supplies Low-Voltage Reverse-Polarity Protection
http://www.elecdesign.com/Articles/I...ArticleID=9945
* * *
Also possible:
http://img6.picsplace.to/img6/23/thumbs/fuse.PNG
(Click to enlarge the picture).
Best regards,
Luciano
Sougata,
Why would it be better to "invert the logic" ?
Is it because the PIC can "sink" more current than "source" or is there any other reason?
well i probably missed something but i don't see the advantage of it. Sure Sougata will explain why.
Usually you should initialize your hardware Somewhere at the Top of your code in a way that there's no current need. Let's say your LEDs are connected on PORTB between i/o and GND, you'll use....
PORTB=0
TRISB=0
For your Piezo... use a capacitor in serie. Directly it's just Killer. It may damage your PIC one day or another. 0.1 uF or higher is good.
ANY piezo sounder have there resonant frequency, if you select the right one, 0.1-0.47 uF is enough.
Do a seah for G-Spot on this forum... yeah no kidding!
The problem with command like SOUND, PWM, FREQOUT is that you never know if the PIN is set to HIGH or LOW at the end of the cycle. In case your PIEZO is between I/O and GND AND the SOUND command leave the I/o to HIGH, your Piezo will s u c k juice from your supply 'till you disable it the right way...
SOUND Piezo, ......
LOW PIEZO
Hi All,
If you are using a simple shunt regulator to power your PIC and the LCD (not the backlight) then while driving outputs high would require your shunt regulator to use a lower value resistor. That means to keep the zener under conduction with all the loads. When the loads are less the zener will act as the load. Keeping in mind that it is battery operated it results in higher quiscent current. I recommended using the LEDs,Buzzer and the backlight to be driven from the unregulated supply. This is only possible if you are using an inverted logic. And setting PORTB = 0 or PORTB = $FF takes the same amount of labor and code. Also the PIC and the LCD would be a fair enough steady load to dimension your resistor-zener shunt regulator for maximum supply efficiency. I hope I have made myself clear or am I missing something??? And using a cap with buzzer is always a good practice. To find out the root of the problem I suggested the experiment to find out exactly whats going wrong and where. Even if the problem is solved with software I would strongly recommend to use a cap as Steve mentioned. The buzzer is a killer.
Hello Sougata,
I have a similar requirement, but without the LCD. Same 6V operation and a Piezo buzzer.
Its basically a code lock operating on 6V. A couple of flashing LED, Sounder (piezo) and a Solenoid.
I have a design before me which does not use even a series diode to drop the voltage, So I presume , the PIC can take that, till the Alkaline battery drops down to safer level on its own. Seeing this , I feel a series diode would be good.
I used a GP PNP transistor with zener at its base to regulate the voltage. Don't know if that's the right way.
Any help ?
Check the new TPS60002(xxx1 for 3.3 volts) from TI. It will take 2.5 to 5.5 volts input, and put out a regulated 5VDC up to 800 mA. It will allow you to wring every drop of juice out of four Alkalines. It's a tiny 3x3 package, but works like a champ.
A shunt-based approach is possibly THE worst avenue you could take for a battery powered circuit. Zener diodes require at least 5mA of current at all time in order to maintain satisfactory regulation. As you can see, this is a highly undesirable attribute, particularly for batteries. Go with the suggested low dropout regulator - clearly your best option.
IMHO - I feel a series diode dropping the voltage by 0.7V would be the most ideal. Still I would try those LDO Ron has suggested.
I also plan to use a TC54 voltage detector circuit to measure Low voltage on the PIC.
Thanks RON for that reference.
Thanks to all
Just the series diode alone will be inadequate. Four fresh new cells, series connected, with no load will be around 1.7 - 1.8 V ea. So 4 x say 1.75V = 7V. As the batteries begin to discharge to about 85% capacity remaining, then you'll have your 1.5V p/cell. You need more than just a diode unfortunately.
Hi,Quote:
Originally Posted by charudatt
In order to ensure long battery life, your circuit must use almost no current in standby.
The PIC has a wide operating voltage range and has a sleep mode for power saving.
To drive the LEDs, piezo and solenoid, make the I/Os sink current. Drive the
solenoid with a PNP transistor. If you use four 1.5V cells in series, just connect the
PIC after the third cell (4.5V). The rest of the circuit (solenoid) will use four cells. (6V).
A diode in series on the battery minus terminal will be the simplest form of battery-reversal
protection.
Best regards,
Luciano
Hi,
As I am about to build a circuit powered from 4 x 1.5v cells I just looked at the maximum ratings for the 18F2525 and it says it can take a max of 7.5 volts! So running a 2525 from 4 x 1.5v cells should be ok. Check the datasheet for the PIC you are using.
But personally I would put a diode in series with the +ive supply. It’s surprising some of the kit I have seen where the user has 'overcome' the plastic mouldings preventing incorrect battery fitting!!!
Bob
Hi Bob,Quote:
Originally Posted by BobP
The voltage 7.5V is not the maximum operating voltage of the chip.
The 7.5V visible in the "Absolute Maximum Ratings" page 323, is
the voltage where you can go without physical damage to the chip.
The PIC18F2525 will operate reliable only when the voltage is between
4.2V and 5.5V.
This note is visible on page 323:
NOTICE: Stresses above those listed under “Absolute Maximum Ratings” may cause permanent
damage to the device. This is a stress rating only and functional operation of the device
at those or any other conditions above those indicated in the operation listings of this
specification is not implied. Exposure to maximum rating conditions for extended periods
may affect device reliability.
See the datasheet page 326, parameter No. D001 for the operating voltage.
Datasheet:
http://ww1.microchip.com/downloads/e...Doc/39626b.pdf
Best regards,
Luciano
Many thanks Luciano, I thought it seemed a bit high but as usual I get interupted with phones etc. before reading it properly.
Thanks,
Bob
If possible, add another call and use 7805. but after an hour of use the voltahge will fall below the I/O differental of 2.5V.then the problem comes.
As many devices work upto 3.3V, why not consider using 3 cells at 4.5V?