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'[PIC]: [PDB] PIClist Dev Project - Pwr Consumption'
2002\09\14@114759 by Jim

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  "And a heatsink [for the 7805] "

NOT to beat a dead horse, but what are the power
requirements so far and what is the expected
power budget for any peripherals?

What I am getting at is, the 78xx series (this includes
the 7805) *also* contains internal overload and thermal
protection (at least the data from National Semi shows
this - is this specific *only* to the Natinal Semi
parts?) and with mounting to the board and to any ground
plane on the board this *may* prove as adequate heat sinking
for the PIC on the PBK in normal operation without the
added complexity and noise of a switcher.

I have used the TERN TD-40 (which pulls over 300 mA total with
most of that consumed by a big, fat, 40 MHz 80186 5V processor)
with a 10 to 11 VDC unreg from a wall-wart and used a small
piece of AL sheet (about 3 cm by 6 cm) together with attachment
of the 7805 to the PC board -

- and the 7805 will gets warm but not excessively nor to the
point where it shuts down.

How much does a 20 MHz PIC draw?

RF Jim

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2002\09\14@130557 by Peter L. Peres

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On Sat, 14 Sep 2002, Jim wrote:

>   "And a heatsink [for the 7805] "
>
>NOT to beat a dead horse, but what are the power
>requirements so far and what is the expected
>power budget for any peripherals?

The horse is alive.

He said ~300mA @ 5V from nominally 12V wall wart (15-16V peak unloaded).
The usual calculations apply to come up with a terrific efficiency of 30%
or so. The switcher will improve this to 75%. We are looking at lowering
the dissipation in the regulator from 2-3W to under 1W, which is probably
ok for the regulator choosen, assuming an external diode is used (so the
reg will burn ~= 0.5W at correct design).

Peter

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2002\09\14@133108 by Jim

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Question:

Why is such a high-output wall-wart required?

Is this to support certain programming operations
with certain PICs?

If not, why not choose something more suitable with
lower voltage?

I would guess that most folks here understand that
the unreg output, as you cited, from a "12V wall-wart"
can be 15 - 16 V with light load. I have good
operation of the TERN TD-40 board (where the 40 MHz
80186 CMOS CPU runs alsmost hot to the touch) using
unregulated wall-warts labled "9 VDC" and a 7805
regulator. There *are* times I wish this were a
switcher yielding more efficiency (some of the newer
products do offer switchers as an option).

RF Jim

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2002\09\14@134748 by Alan B. Pearce

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>NOT to beat a dead horse, but what are the power
>requirements so far and what is the expected
>power budget for any peripherals?

OK, here are the "power dissipating blocks" as I see the circuit.

1. Always present. 16F877(A) or 18F452 iPIC using 3.6MHz crystal.
Miscellaneous 74HC(T) chips. MAX202 and FTDI FT232 USB I/F. Some LED
indicators (about 4, of which probably only 2 on at any one time). Some low
power op amp and other linear circuit used by (2) or (3).

2. Often present. 16F877(A) or 18F452 Target 1 PIC, frequency controlled by
crystal, resonator or RC depending on user requirement. Outputs include 2 x
10 led bargraph (one run by LM3914, one digital drive), 1 x 4 digit 7
segment display with digits multiplexed.

3. Sometimes present. 16F877A or 18F452 target 2 PIC. Clock rate
possibilities as per (2). Minimal peripherals, but additional unknown chips
possibly fitted to breadboard area by end user.

Admittedly a 7805 may well handle these requirements, and I think that it
will be a case of trying a 7805 to see if it will handle the situation.
However I was planning to include the tracking for a switchmode on the
prototype board as an alternative at this stage.

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2002\09\14@135012 by Alan B. Pearce

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>The horse is alive.

Great, cos at the moment I'm riding it :))

>He said ~300mA @ 5V from nominally 12V wall wart (15-16V peak unloaded).

Umm, I did not quite say that, but that I was looking at a switcher that
could supply this much, and a "wet finger in the air" guess that it would
have enough current capability, without actually sitting down and
calculating it out. The query at that stage was for something with a bit
more input margin than the chip I was looking at.

However having said that, I think ~300mA would probably do it.

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2002\09\14@164608 by Dwayne Reid

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At 06:46 PM 9/14/02 +0100, Alan B. Pearce wrote:

>OK, here are the "power dissipating blocks" as I see the circuit.
>
>1. Some LED indicators (about 4, of which probably only 2 on at any one
>time).
>
>2. Often present.  2 x 10 led bargraph (one run by LM3914, one digital
>drive), 1 x 4 digit 7
>segment display with digits multiplexed.

As I see it, the power-hungry devices are the LEDs.  All the other stuff
should consume less than 30 mA or so - that includes all 3 PICs, the
op-amps, a LCD display.  Heck - even a 78L05 would probably work just fine
if it weren't for the LEDs.

1) the LM3914 can operate from the un-regulated supply.  It will get hot if
operating in bar mode but will work quite nicely in dot mode.

2) How many of the other LEDs can be made to operate from the un-regulated
supply?  2003 or 2803 type o/c buffer chips are cheap - compare their cost
to the cost of the switch mode supply.

There are a few LEDs that might need to be driven directly by PIC pins -
such as in item (1) in Alan's message.  Run them at 4 or 5 mA and add that
current to the 5V supply.  They don't need to be blinding - just
visible.  The cheap T1 LEDs I am currently using are nice and bright at 4 mA.

dwayne

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2002\09\15@085904 by Alan B. Pearce

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>There are a few LEDs that might need to be driven directly by PIC pins -
>such as in item (1) in Alan's message.  Run them at 4 or 5 mA and add that
>current to the 5V supply.  They don't need to be blinding - just
>visible.  The cheap T1 LEDs I am currently using are nice and bright at 4
mA.

ha ha. I know this too - I currently have a project where I am running a
bunch of high efficiency LED's direct from pins, with 100 ohm resistors. At
power on I set these to light up, and then extinguish as the initialisation
proceeds through the relevant parts. When you push reset it is quite
blinding :))

I'll be leaving them with 100 ohm resistors though, as they are interrupt
event indicators, so in normal operation they have very short on times, and
the 100 ohm seems to work out about right for this.

>As I see it, the power-hungry devices are the LEDs.  All the other stuff
>should consume less than 30 mA or so - that includes all 3 PICs, the
>op-amps, a LCD display.  Heck - even a 78L05 would probably work just fine
>if it weren't for the LEDs.

Yeah that does seem to be the way of it. I plan to run the LED's from
ULN2003 drivers anyway, so maybe a low power linear is really worth looking
at. Will check it out, and see.

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2002\09\16@091946 by Roman Black

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Jim wrote:

> Why is such a high-output wall-wart required?
>
> Is this to support certain programming operations
> with certain PICs?
>
> If not, why not choose something more suitable with
> lower voltage?


I tend to agree! The 12v wart solution will still
require a resistor/zener for the 12v HVP and will do
the 12v -> 5v conversion for the BULK of the power.
Why convert the bulk??

A switching regulator at 75% efficiency etc is
expensive and *may not* be as efficient as a correctly
chosen 6v wart (putting out 7v etc) and 7805
linear regulator. As the current approaches 300mA(?)
the wart voltage will sag and the 7805 dissipation
will remain low. A cheap cmos gate charge pump etc
can be used for the uA needed at 12v for HVP.

The BIG efficiency losses are the LED resistors!
1600mCd high intensity leds are dirt cheap now and
are very bright even at 1mA. If efficiency is an
issue that would be the place to start, and at the
end of the day when we have to troubleshoot the
"PDB" after the beginner dropped something conductive
on the PSU section i'd prefer if it had a 7805.
-Roman

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2002\09\16@191602 by Peter L. Peres

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On Mon, 16 Sep 2002, Roman Black wrote:

*>I tend to agree! The 12v wart solution will still
*>require a resistor/zener for the 12v HVP and will do
*>the 12v -> 5v conversion for the BULK of the power.
*>Why convert the bulk??
*>
*>A switching regulator at 75% efficiency etc is
*>expensive and *may not* be as efficient as a correctly
*>chosen 6v wart (putting out 7v etc) and 7805
*>linear regulator. As the current approaches 300mA(?)
*>the wart voltage will sag and the 7805 dissipation
*>will remain low. A cheap cmos gate charge pump etc
*>can be used for the uA needed at 12v for HVP.

Unfortunately a bigger case, the need for venting holes, and a radiator
offset the cost of a linear regulator when the power level is 'on the
limit'. There are several unfortunate designs on the market that have this
problem, as you know.

Peter

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2002\09\17@013228 by Roman Black

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Peter L. Peres wrote:
>
> On Mon, 16 Sep 2002, Roman Black wrote:

> *>A switching regulator at 75% efficiency etc is
> *>expensive and *may not* be as efficient as a correctly
> *>chosen 6v wart (putting out 7v etc) and 7805
> *>linear regulator. As the current approaches 300mA(?)
> *>the wart voltage will sag and the 7805 dissipation
> *>will remain low.

> Unfortunately a bigger case, the need for venting holes, and a radiator
> offset the cost of a linear regulator when the power level is 'on the
> limit'.

Huh?? What part of "As the current approaches 300mA
the wart voltage will sag and the 7805 dissipation
will remain low" did you not understand?

I just tested a handful of dirt cheap taiwanese 7805
regs, Vdropout at 300mA is 1.65v, total efficiency of
5.00v/6.65v or 75.18% efficiency. Around the SAME as the
expensive noisy messy switching regulator. Either solution
dissipates about 500 or 600mW. I wish people would do
their homework before shooting their mouth off.

The design probably WON'T require 300mA anyway, more
like 150mA from what i've read so far, so the cost
size and reliability issues of the switching regulator
are not needed.

If you want super efficiency at full power why not a
low-dropout 7805? That could give 90+% with a properly
chosen wall wart.

> There are several unfortunate designs on the market that have this
> problem, as you know.

None I can think of.
-Roman

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2002\09\17@024220 by Wouter van Ooijen

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> If you want super efficiency at full power why not a
> low-dropout 7805? That could give 90+% with a properly
> chosen wall wart.

If the wallwart gives a voltage that accurate you don't need the 7805 at
all. In practice you will - even with a perfectly choosen wallwart -
have to compensate for
- tolerances in the primary voltage (may vary from country to country!)
- tolerances in the tranformation factor
- tolerances in the diode voltage drops
- ripple on the buffer elco

I dunno what the rule of thumb is, but I would guess this asks for a
factor 1.5 between lowest and highest input to the 7805 (in addition to
the drop voltage).

Wouter van Ooijen

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