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I really like Barry's new cover and battery charger jack, nice job. I
am sure they spent some hours figuring that out with the limited
space.
If you have followed some of my earlier posts on connectors and wire
in rig you know that there exists some voltage drop in harness. I was
intimadated at first but determined to eliminate aa much voltage drop
as possible. I was able to remove the connector on PC board by using
solder wick on pins. I then experimented with maximum wire size that
would fit in hole {orginal is 30 gauge and has over 100MV drop from
battery}. If you are carefull you can use 22 gauge, it just barley
fits, I used 22 gauge with .006 thick teflon insulation. I then
installed a Deans Ultra plug and socket, more on this later. I now
have 1.3MV drop under full load from battery to board!.
While experimenting I found the following voltage drops under full
load:
Battery Tray 200MV!
New Connector 35 MV
Wire from connnector to PC board 100MV
Wire from batery to connector 100MV
You can see that if you are using the battery tray and stock wiring.
This is all lost in heat energy. 2amp X .5 volts = 1 watt lost to
voltage drop!!

If you do not want to tackle taking off wire from board, another
member suggested soldering leads to bottom of board bridging the
parallel positive and negative feed with larger wire, 20 gauge will
work for this and you can remove for warranty work.

Deans connectors: The Ultra I used is going to be my new connector
for ARIES and all portable packs. I measured less than 100 MV at 20
amp load, it a tight fit in batery compartment requiring some
trimming plastic case. Their standard 2,3 or 4 pin connectors all
exhibitted 5MV or less under load and fit easily in battery
compartment. The Ultra connector is overkill but I had some from my
R/C flying and standardized on these for the shack.
For ARIES work the Amderson Power Pole Connector is almost as good
and easier to find. I have some pig tails to convert to Power Poles
for ARIES box.

If you do either wire mod and use W4RT's cover with charging jack you
will have the an easily chargable pack and virtually 0 voltage loss.

If there is enough interes I will order a supply of Deans connectors
and offer to group at cost, as I did on the molex connectors.
.73 Tim O'Rourke KG4CHX

--- In FT817@y..., KG4CHX@A... wrote:

While experimenting I found the following voltage drops under full
load:
Battery Tray 200MV!
New Connector 35 MV
Wire from connnector to PC board 100MV
Wire from batery to connector 100MV
You can see that if you are using the battery tray and stock wiring.
This is all lost in heat energy. 2amp X .5 volts = 1 watt lost to
voltage drop!!

Does it really matter? According to K6XX analysis of current drain
during transmit, the current drawn by the power amplifier is largely
unaffected by voltage. As a result, that 1 watt that we could
conceivable recover by replacing the power connectors will just result
in 1 watt more power dissipation in the MOSFETs.

Or am I missing something?
--
Brian N0KZ

Well - yes - I think you are missing something - let's say - since I
don't have the EXACT value - that the radio stops working at 8.5
volts - and you have .4 volt drop between the battery and the load
(the radio) - so when the battery is discharged to 8.9 volts - the
radio quits on TX. Where if you could reduce the voltage drop between
the battery and the load to say .15 volts, then the battery could go
down to 8.65 before the radio sops working. By having less voltage
drop between the source and the load, you are able to use more of the
available battery capacity.


--- In FT817@y..., bbadger@y... wrote:

--- In FT817@y..., KG4CHX@A... wrote:

While experimenting I found the following voltage drops under

full

load:
Battery Tray 200MV!
New Connector 35 MV
Wire from connnector to PC board 100MV
Wire from batery to connector 100MV
You can see that if you are using the battery tray and stock

wiring.

This is all lost in heat energy. 2amp X .5 volts = 1 watt lost to
voltage drop!!

Does it really matter? According to K6XX analysis of current drain
during transmit, the current drawn by the power amplifier is largely
unaffected by voltage. As a result, that 1 watt that we could
conceivable recover by replacing the power connectors will just

result

in 1 watt more power dissipation in the MOSFETs.

Or am I missing something?
--
Brian N0KZ

Which battery technology are we talking about? When NiMH batteries
sag below 1.1 volts per cell at 2 amps draw, they are about 95%
drained (based on the discharge graphs I have on hand), and it drops
off quickly from there, below 1 volt at about 97% of capacity. So
between 8.8 volts and 8.0 volts (where the radio shuts down) you have
about 2% of the usable capacity of the NiMH cells.

So the difference between shut off at 8.5 volts (stock) and 8.0 volts
(after modification) is less than 2% of the capacity of the cell.
Call that maybe 30 seconds of extra transmit time per recharge.

I'm not trying to troll here. Are my numbers right? If they are, I
don't see that there is an argument to be made for replacing the power
connectors and resoldering new leads to the board.

In terms of reliability, a dedicated NiMH pack (like the Maha) is
going to be better than the battery tray, simply because soldered
cells are going to last longer than spring contacts, and because
heating up the NiMH cells from the power dissipation in the battery
tray will unnecessarily shorten their useful lifetime.

But I don't see voltage drop as being very important, otherwise.
--
Brian N0KZ

--- In FT817@y..., don@h... wrote:

Well - yes - I think you are missing something - let's say - since I
don't have the EXACT value - that the radio stops working at 8.5
volts - and you have .4 volt drop between the battery and the load
(the radio) - so when the battery is discharged to 8.9 volts - the
radio quits on TX. Where if you could reduce the voltage drop

between

the battery and the load to say .15 volts, then the battery could go
down to 8.65 before the radio sops working. By having less voltage
drop between the source and the load, you are able to use more of

the

available battery capacity.


--- In FT817@y..., bbadger@y... wrote:

--- In FT817@y..., KG4CHX@A... wrote:

While experimenting I found the following voltage drops under

full

load:
Battery Tray 200MV!
New Connector 35 MV
Wire from connnector to PC board 100MV
Wire from batery to connector 100MV
You can see that if you are using the battery tray and stock

wiring.

This is all lost in heat energy. 2amp X .5 volts = 1 watt lost

to

voltage drop!!

Does it really matter? According to K6XX analysis of current

drain

during transmit, the current drawn by the power amplifier is

largely

unaffected by voltage. As a result, that 1 watt that we could
conceivable recover by replacing the power connectors will just

result

in 1 watt more power dissipation in the MOSFETs.

Or am I missing something?
--
Brian N0KZ

Well - you very well might be correct - my reply was meant to explain
one reason WHY it could be an issue - the numbers I was using were
for the sake of argument - I don't know what the voltage is that the
radio quits working and I don't have the time to waste studying the
discharge graphs.

But another point is that that power lost in the resistance of the
system has to go somewhere, and it ain't into powering the radio - it
is in heating up the wires and contacts that aren't adequate to do
the job.

Actually, I seldom use the internal tray that I have fused and
stuffed with Radio Shack 1600 mah NiMH AA cells in that it takes so
long to charge them and it is a pain to remove the radio from the
case with the "Plumbers Delight" antenna mount on the rear connector.

So I use the 1600 mah NiMH RC car packs from RS and their 4.5 hour 2-
pack charger with an adapter from the pack connector to the jack on
the rear of the radio - made from 18 gauge wire with a fuse. The pack
velcros to the bottom of the case at the front of the radio and does
double duty in proping up the front of the radio.

I also have a approx 15" square solor panel that charges the packs in
about 6 hours - se ya on Field day QRP Battery power with the natural
power multiplier - portable from the patio table with a Screwdriver
antenna. A ligit location for simulated emergency conditions as I
live about 600 feet from the north/south center of the Hayward fault
where the state OES predicts the next "big one" is most likely to be
centered!

Don
W6ZO




--- In FT817@y..., bbadger@y... wrote:

Which battery technology are we talking about? When NiMH batteries
sag below 1.1 volts per cell at 2 amps draw, they are about 95%
drained (based on the discharge graphs I have on hand), and it drops
off quickly from there, below 1 volt at about 97% of capacity. So
between 8.8 volts and 8.0 volts (where the radio shuts down) you

have

about 2% of the usable capacity of the NiMH cells.

So the difference between shut off at 8.5 volts (stock) and 8.0

volts

(after modification) is less than 2% of the capacity of the cell.
Call that maybe 30 seconds of extra transmit time per recharge.

I'm not trying to troll here. Are my numbers right? If they are, I
don't see that there is an argument to be made for replacing the

power

connectors and resoldering new leads to the board.

In terms of reliability, a dedicated NiMH pack (like the Maha) is
going to be better than the battery tray, simply because soldered
cells are going to last longer than spring contacts, and because
heating up the NiMH cells from the power dissipation in the battery
tray will unnecessarily shorten their useful lifetime.

But I don't see voltage drop as being very important, otherwise.
--
Brian N0KZ

--- In FT817@y..., don@h... wrote:

Well - yes - I think you are missing something - let's say -

since I

don't have the EXACT value - that the radio stops working at 8.5
volts - and you have .4 volt drop between the battery and the

load

(the radio) - so when the battery is discharged to 8.9 volts -

the

radio quits on TX. Where if you could reduce the voltage drop

between

the battery and the load to say .15 volts, then the battery could

go

down to 8.65 before the radio sops working. By having less

voltage

drop between the source and the load, you are able to use more of

the

available battery capacity.


--- In FT817@y..., bbadger@y... wrote:

--- In FT817@y..., KG4CHX@A... wrote:

While experimenting I found the following voltage drops under

full

load:
Battery Tray 200MV!
New Connector 35 MV
Wire from connnector to PC board 100MV
Wire from batery to connector 100MV
You can see that if you are using the battery tray and stock

wiring.

This is all lost in heat energy. 2amp X .5 volts = 1 watt lost

to

voltage drop!!

Does it really matter? According to K6XX analysis of current

drain

during transmit, the current drawn by the power amplifier is

largely

unaffected by voltage. As a result, that 1 watt that we could
conceivable recover by replacing the power connectors will just

result

in 1 watt more power dissipation in the MOSFETs.

Or am I missing something?
--
Brian N0KZ

Re "the next big one"...shouldn't you be moving house rather than playing
wireless? ;o(


David G4YVM