Suitable Power Supply Circuit...

Hi there,

I'm looking for a suitable power supply circuit to supply 5v & 3v DC Voltages from possibly a 9.6v supply...

Basically I have an 8-way AA power pack that I plan on powering from Duracell Plus AA Batteries, but I am unable to find any information on them...

A quick search on line shows that Alkaline batteries range from 1700 mAh to 3000 mAh

At the moment I am using a LM7805 to power the circuit but once I am running from a battery supply I will need to use as little power as possible...

I know zener diodes and potential divider circuits are power hungry, so any suggestions would be greatly appreciated...

I have no power requirments as of yet for the circuit, so just looking for suggestions at the moment...

Hi,
Im not a fan of switchers, but in that case its a good power saver solution to select some step-down regulators..
If the current are not so much,& these is not a product (only for private using); I would overthink the selection of (higher power) selected Zener diodes, but in serie with your electronics on the 9.6V.
Its not very professional and is relative inexact, but you can select individual diodes for your needed5 & 3,3V!
Because it will be in serie with your electronic; its dissipation is minimal.
Of course, the switcher/step-down regulator (i.e. national.com; simple switcher_or some better products) is clearly better & exacter!
K.

The information as provided is vague.

First thing to remember is the voltage drop: to regulate from 9.6V to 5V, you are having a drop of 4.6V which gives you approx power loss of 4.6V*(amount of current being used by 5V). So the goal is to minimize the amount of current flow and decrease the voltage drop. Try changing your spec from input voltage less than 9.6V but above the needed output voltage. The PIC chip can be put into sleep mode to conserve energy.

Try using a LDO regulator instead of using a L7805 which has higher internal voltage drop.

If your spec is fixed input voltage of 9.6V, then look at DC-DC buck converter to make it more efficient.

Sorry, its wrong.
An LDO isnt the solution if a system has a voltage delta what "fixed" is to i.e. between 3.6 & 4.6V! :-(
If not a switcher or serial Zener will be used; its only a "low own current" type analog regulator is OK_ but you can forget generell an LDO...
K.

Hawk I don't think that the current draw is be more than 100ma max I reckon about 50ma by looking at what your running. I have some SMT chips that only 15-100Ua(microamps) from the supply.
**broken link removed** but the output is 50ma. You could alo consider a MC4063 step/up regulator and this one
**broken link removed**
and this one
**broken link removed**.
I've seen a simple curuit that drops 5V to 3.3V not sure on the current draw I will look and check but you could put a transistor in line so the 3V supply is only on when needed.

You could use any of the above then use 4 x 1.5V in series then add another 4 in series then parallel them this will double the capacity of them

I think you need a step down buck converter. It's a single chip that uses switching to do the DC-DC conversion. Typically they're like 80%+ efficient. They will have some external componts you'll need, such as an inductor, diode, and several capacitors.

A good example of a buck converter that would be useful for your application is:

https://focus.ti.com/lit/ds/symlink/tps54231.pdf

There is a complete design example, and step-by-step guide for picking the external components. Additionally, TI has software you can use to generate the design for you.

Matt

I prefer to use the MC34063 here is a little tool to work out the parts and values required. Yes DC to DC conveters work better and draw less current

**broken link removed**

Karesz
If you read through my comments, you will see I had mentioned "Try changing your spec from input voltage less than 9.6V". The gyst of the argument was to lower the input voltage to a small level so an LDO could be used. In other other words, a small input - output differential voltage. This way the loss in energy would be small.

russ said:

Karesz
If you read through my comments, you will see I had mentioned "Try changing your spec from input voltage less than 9.6V". The gyst of the argument was to lower the input voltage to a small level so an LDO could be used. In other other words, a small input - output differential voltage. This way the loss in energy would be small.

Click to expand...

Hi russ,
Dont worry I did understand you, but your system thinking isnt OK_in my opinion.
Why is a LDO to use if the input voltage diff is at 4-4.5V!?
If you would select a 12V /9V or even 6V relay, driven direct from battery, than your supply curents are minim, as Wizpic assumed too.
In taht case is a small SMD, low self consuming (eventually double output step down switcher if you both voltages, as 3.3 & 5V needs) is to use_but as you will; its your project!
Also, these was the one way, if your input is i.e. 9V.

Other way; if your battery stack has only 2/4 pices: for power saving is a linear reulator never the best philosophy, even if its a super LDO type!
I say it because a battery circuit must have, we say 30% or maybe up to 50% voltage reserve, while a battery delivers for you not a stabile voltage!
Up to the time if your battery voltage isnt so low that an LDO is exactly supplyed with the minimum possible input voltage: you have the said (& not needed + plus(!) waste dissipation), & that practically in the full lifetime your battery...
I hate basically switchers, but for power saving their are the best components!
Good progress!
K.

Karsz
You are fixated on 9V input voltage. The original email was vague. Depending on the battery arrangement, one can arrange the batteries so that the input to output differential is small. The original email does not mention a specific source supply, hence my mentioning "the posting was vague". Besides the LDO, I did suggest using SMPS.

No one said an LDO circuit was going to be more efficient than an SMPS circuit. In fact the SMPS circuit was suggested in my first posting, purely for efficiency purposes. SMPS are more expensive. Depending on one's end goal cost vs efficiency, even a linear regulator can be effective. It is how one justifies using the component in the circuit.

Before claiming someone is wrong, I think due diligence should be practiced. I do not know the totality of the design of the original poster. Hence, I provided a few solution. From the solution, the original poster needs to see what fits his needs.

OK,
Sorry; I was really more fixed on 9V as others...
I asked you some earlyer; that is 3.3V or 5V or both to same time needed?
I think the relay is direct from incoming 5V to supply...
Only one porbleme is real, pls do understand me: the batteries has an input voltage that is strong variant.
I think you dont wish to us batteries only in a small i.e. 10% voltage range? these is the problem for LDOs, or better said -for analog regulators...
In battery equipment is a switcher surly more power saver_pls calculate it for efficiences by both systems, but with extreme parameters too!
Input supply voltage range(Min/max) & output power min/max situations and even all four situations for an typical LDO & a Switcher...
You will be surprised!
K.

Karesz
Just like I mentioned, original poster was vague in his email. I had provided 2 solution which was also vague in nature.
From your last email, you are supposing that you know "Input supply voltage range(Min/max) & output power min/max situations and even all four situations for an typical LDO & a Switcher... " and making an analysis and providing a solution at the same time.
Just like I mentioned, depending on the scenario, the designer will have to justify a solution based on the specification on hand. The specification and the design analysis will drive the end circuit. Where I am failing to come across is, you or I do not know the specification of the design nor do we know the end application.
Do not get me wrong. Your points are well taken and the questions as posed by you is also legit but you fail in your premise whereby you are making up input voltage range and output power range. The original poster was very vague in his posting and never provided any specification.
Whatever numbers I choose is only to explain the circuit on hand and in no way constitutes the final solution to the original posting. They are only possible solution. At the end of the day, the designer will have to look at all the solution, do his due diligence and provide the best answer.
Regards

Hi russ,
OK, Im agree with your aspects_ we have to waite_maybe "hawk" will bring us the quen`s greetings and tell us more of hes spec`s!
Regards!
K.

Sorry things are a bit vague...

I'm still trying to work out how to get things done and write the required code, as the whole unit is so far running from a bench powersupply throught a L7805 whilst I get the basic circuit up and running...

Basically this is what I am trying to do...



But at the moment I have a few teething problems with the code, which I hope to resolve this evening...

The main area that I am now struggling with is the power supply side of things... just not sure how to even approach the subject been awhile since I designed anything but with the time constraints forced upon me by the University I have no option but to race headlong into the code and getting the basic design up and running as fast as possible...

Can anyone please specify what they actually require from me and i'll do what I can to quickly supply them with the data... basically I am so stressed over the code that I cannot focus on the power side of things...

So any help will be greatly appreciated...

Been reading the suggestions so far but it's quickly becomming aparent that you require a few more bits of information to assist me to the best of your knowledge...

Ok here goes...

I require the circuit to supply 5v & 3v to the circuit attached in the previous thread for a period of at least 10 weeks... as the unit will be placed in a field in all sorts of weather from August to November... so it will be exposed to rain, perhaps the occasional frost it may even dip below zero on the odd occasion...

As for the current rating, I still need to work that out...

I am using: -

1off - 16F684
2off - DS18B20
2off - FM25V02

16F684 - Low-Power Features:-
Standby Current: - 50 nA @ 2.0V, typical
Operating Current: - 11μA @ 32 kHz, 2.0V
typical - 220μA @ 4 MHz, 2.0V,
Watchdog Timer Current: - 1μA @ 2.0V, typical

FM25V02 - Low Voltage, Low Power
• Low Voltage Operation 2.0V – 3.6V
• 90 µA Standby Current (typ.)
• 5 µA Sleep Mode Current (typ.)

DS18B20 - DC Electrical Characteristics
Sink Current IL VI/O = 0.4V 4.0 mA
Standby Current IDDS 750 1000 nA
Active Current IDD VDD = 5V 1 1.5 mA
DQ Input Current IDQ 5 μA

At the moment I have available to me: -

9v - PP3 battery holder
4x - AA battery holder
8x - AA battery holder

It's the eight way battery holder that interests me the most, as this has a suitable area for the PCB...

Not yet had chance to calculate the total voltage it's likely to supply but I beleive it to be around 9.6v depending upon the battery arrangement...

I have been recommend to use D-type batteries as these apparently would be more suitable to my needs...

But at the moment I am unsure which direction to take...

Hi,
I checked the data from Ds; it can be driven between 3--5V & the FM25V02 is 2--3.6V to use, but be care pls; ITS HOLD CAN NOT BE on 5V_hes inputs arent 5V tolerant! :-(
I think the PIC is even with 3.3V to use(refer pls to page133: if you can life with 8MHz it can be 2/2.5V, by 3.3V it works clearly up to 10MHz!) _I can not see why you need the 5V for power saving modus if the used components are existing in 3.3V version?

Only question is for me the ICSP connection, has it not 3V logic levels too?

You are, I think, speaking over a relay too_where is it pls?

Even, I BELIEVE THAT YOU NEED ONLY A 3V3 Supply
Apropos; is the 220K in serie with the XTAL OK?, maybe you dont have these resistor, but I believe-it must be(if needed, see page 24 pls) at 10--22KOhm & ON THE OTHER pole of quartz, PIN=Osc2!!!... You will have eventually to play with C2/C3-can be that i.e. 27pFs are good, or even only at 15-18pF...

Dont forget pls, that C1 must be a 100nF(min. 22-47nF), & same capacitor pls on all ICS, but minimum onto the memory & PIC & ICSP Header/pin1, but it need s a bigger capacitor from i.e. 2.2 or 10uF, a good CERAMIC, on the power input!!
By the calculation of BATTERY CAPACITY be care pls: its desirable values are AT MINUS TEMPERATURES only a portion of your nominal value! :-(

If you have to use D-batteries; are their to be rechargeable?
If not; you will have a Usupply of 10V instead 9.6V, but its detail yet...
These is a clear case for (power saving) Step-down switcher, be care with minus temp. extrems...

You can refer to page 153: PICs Icc is ca 1mA at 3.3V, 10MHz & 3mA at 5V, memory has similar pwr need & the temp sensors too; also I think if you will calculate with so 10mA / 10MHz & 3.3V-it will be real as a systems power need!
If you can apply a week up circuit, needs these current only in all minutes/hours (time to time) & than the actual sleep currents are at 100uA ...
K.

You are spooking over a relay too_where is it pls?

Click to expand...

What relay... there is no requirement for a relay...

Apropos; is the 220K in serie with the XTAL OK?, maybe you dont have these resistor, but I believe-it must be(if needed) at 10--22KOhm & ON THE OTHER pole of quartz, PIN=Osc2!!!... You will have eventually to play with C2/C3-can be that i.e. 27pFs are good, or even only at 15-18pF...

Click to expand...

Who is Apropos?

The parts list is slightly inaccurate especially the BOM it's just a basic one to get something down on paper that I can tweak...

The 32Khz Crystal has been recommended for the WDC to allow me to wake the processor up every hour, the resistor and capacitor are from the datasheet attached...

Only question is for me the ICSP connection, has it not 3V logic levels too?

Click to expand...

Not sure... only ever used 5v to power it... however it depends as that side of things can be powered from the actual programmer...

All batteries are to be off the shelf batteries... nothing special like re-chargable batteries... but I would like to keep an option like the L7805 available on the board so that I can use a bench powersupply to assist with downloading the actual data to save relying upon the actual batteries...

Hi,
Than I remembered wrong with relay...
Even Osc component values are strong from type of Quartz depending., but the serial resistor is to oscpin2 to connect!..
Design pls with bypass diode for connecting bench top supply to internal regulated battery voltage to your electronics.
K.

Hi hawk,
I think the most suitable choice would be to use a switcher. The L4976 would be a very suitable choice as it can regulate supplies upto a current of max 1A, the L4971 upto 1.5A. Furthermore, the package is 8-pin PDIP, which is very convenient for experiments, especially.
For more current: L4960: 2.5A, L4970: 10A, L4977: 7A, L4974: 4A

Hope this helps.
Tahmid.