Yo! Marc:
Friday September 06 1996 15:10, Marc Chabot wrote to Bill Cheek:
MC> I use and old Sega 9V 1A on my PRO43 with 6 AA Rat-Shack HiCapNiCad
MC> in it. How much ma does the internal PRO43 charger gives to the
MC> batteries? (When the scanner is off and same question when it's in
MC> use while recharging)
Your questions really can't be answered without your taking measurements and
doing some calculations.
It isn't a fixed rate because the internal charging circuit is as simple as
it gets. Here is the effective (simplified) circuit:
22 ohms choke
(+)____/\/\/\____ CCCC___
|
|
Recharger ----- 6-cell
Input --- NiCd Full Charge
----- Pack 8.40-8.64v
---
| Full Discharge
diode choke | 6.0 volts
(-)_______|/______CCCC___|
|\
Here are the issues:
1. The full recharge level of a NiCd cell is 1.44v. Any more than that
and the cell could be damaged. Therefore, 1.40v per cell is considered
a safe recharge maximum.
2. The full discharge point of a NiCd cell is 1.0v. Any less than that
and the cell could become damaged due to a variety of conditions not
discussed here.
3. The maximum recharge rate is one-third of the mA/H rating of the cell.
This is the "fast charge" rate, known as C/3. Higher rates could damage
the cell.
4. The safe trickle charge rate is one-tenth the mA/H rating of the cell,
and is known as the C/10 rate. A NiCd cell can be recharged at this
rate indefinitely without damage.
5. Ideally, the terminal recharge voltage to the above 6-cell NiCd pack
will be 8.40 to 8.60 volts, let's say 8.50v for example, then to get
that supplied to the cells, you have to take into account voltage drops
across the two chokes, the 22-ohm resistor and the series diode. The
two chokes are probably 1-ohm each, and added to the resistor, make
24-ohms of series resistance. The silicon diode drops a continuous
0.6v, regardless of current, so we have this to consider:
A. Hi-Cap NiCd's are 850-mA/H rated, so a trickle charge (C/10) will
be 85-mA.
B. 85-mA flowing through 24 ohms (E = I*R) gives a drop of:
Edrop = (.085)(24)
= 2.04 volts
C. The series diode drops 0.6v.
D. Total drops are 2.04v + 0.6v = 2.64 volts
E. If the desired terminal voltage to the pack is 8.50v, then 2.64v
must be added to that to calculate the desired Recharger voltage:
8.50 + 2.64 = 11.14 volts.
Knowing all the above, the first step is to recharge the pack for 18-hrs,
and then quickly remove the pack and measure the terminal voltage. This
will tell you whether the pack is being under charged or over charged. See
(5) above.
Next, you will have to plug the charger back in and measure the charger's
eed
voltage while it is charging the pack. Compare that to the results in (5.E)
above. The rest should be intuitive, but get back to me if you have a snag.
Do NOT measure the recharger's output when it is not connected to the pack.
The value will be meaningless.
Bill Cheek | Internet: bcheek@cts.com | Compu$erve: 74107,1176
Windows 95 Juggernaut Team | Microsoft MVP
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