W
Winfield Hill
Guest
Ken Smith wrote...
with the help of Chilton and Haynes manuals. The engine compartment is
filled with 20, 30, 40 and 60A distribution fuses for various things, plus
single 175A "Mega fuse" to the generator and "distribution block" (called
a fuseable link on some models). Mine is a removable plastic-encased fuse.
There is also a 15A and an 80A fuse bypassing the Mega fuse. In the case
of these Ford models the starter motor is _not_ fused. With some effort
these big replaceable fuses can be procured as automotive spare parts.
--
Thanks,
- Win
Winfield Hill wrote...In article <cn81ug02too@drn.newsguy.com>,
Winfield Hill <hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:
Ken Smith wrote...
[...]
For the OP I'm adding a bit of extra working things out.
I've added a capacitor to the drawing called "Cbig" and labled some
resistors.
The TIP35 and TIP36 transistors are fairly cheap. I think I'd start
like this:
10 each TIP35
------------------- ---/\/\---
! \ /e R4 !
! ----- !
! TIP36 ! !
+-/\/\/-- -----------/\/\------+
! R2 e \ / R3 !
! ------ !
! ! --------- !
---+--/\/\/\/--+-----! LM317 !---------+-+--
R1 --------- !
! --- Cbig
---
!
GND
The TIP35 and TIP36 are slow. They are extra slow when their collector
voltage drops below about 2V. Along with the usual bypass an LM317 needs,
you also need some added largish capacitor to make up for the fact that it
will take the TIP35,36 stuff time to throttle up and down. The
minimum value of this capacitor I would extimate like this:
Assume that the 150A draw starts suddenly.
Assume that the voltage can droop or spike up by 0.1V.
The TIP35 and 26 have a rise and fall time of about 0.25uS (Check data
sheet to be sure)
The TIP36 has to turn on and then the TIP35 so the total time is 0.5uS.
Rather than having to integrate, we just say the TIP35 and TIP36 stuff
doesn't do anything for 0.5uS and get an over estimate of the minimum.
In 0.5uS, 150A will transfer a charge of 75uC.
The voltage change on the Cbig will be:
V = Q / C
where
Q is the charge
C is the capacitance
We solve for C
C = Q / V = 75uC / 0.1 = 750uF
Now here's the messy bit. We have assumed tha Cbig has no resistance.
All real parts have some resistance. We now need to find the data sheet
on some lets say, 1000uF capacitor with a nice low Equivelent Series
Resistance (ESR) to see how much drop will appear in the ESR. If that
plus the droop in the 1000uF is more than the specification, you need a
larger capacitor.
Finding R1:
If the circuit is lightly loaded, we want the LM317 to pass all of the
current. This means that there should be less than about 0.6V on R1 when
a "small current" is drawn. We will define a small current as 50mA os we
can say
R1 = 0.6V / 0.05A = 12 Ohms
So we make R1 about 12 Ohms.
About R2:
R2 could in most cases be zero. It lowers the gain of the TIP36 part of
things and works to reduce the risk of oscillation. I'd make this a low
inductance resistor of about 0.01 Ohms or a short length of wire.
About R3:
We want about an Amp flowing in the TIP36 before the TIP35s are brought
into things. R3 is something like 0.1 to 1.0 Ohms.
About R4:
This is actually 10 resistors, one resistor in the emitter leg of each
TIP35. When a bipolar transistor heats up, its Vbe drop decreases. If R4
wasn't there, the hottest TIP35 would end up taking on more than its share
of the load, get hotter, draw more, get hotter and etc. This could
destroy one transistor.
Thermal resistance is a lot like electrical resistance. Heat sources are
like current sources. All of the heat must flow out of one transistor
through all of the stuff it goes through to get out to the environment.
In process, one transistor will tend to warm up its brothers. To find the
minimum R4 we start with assuming a current flows in one transistor and
work out how much its temperature rises above the coolest of its brothers.
To make the math easy, just assume 1A flows.
One you have the temperature rise, assume that the individual R4 must add
at least 2.5mV of drop for every amp flowing. It is likely that this will
be such a small value that using 10 wires running from the TIP35s to the
load instead of one big oun will be enough to do it.
A note about layout:
You want Cbig to be as close to the load as practial. Ideally, there
should be 12 wires to the (+) side of the load. The LM317, the TIP36 and
all the TIP35s each have a wire. These wires should be tightly bundled
together.
A note about fuses:
They make special fuses for things like car starter motors. They look
like a thick cable with a wide section in it. Put one of these between
the car battery and the circuit. A car battery can produce nearly
infinite current for a short time.
Yes that's a good point. I learned about this by examining my 1985 Cougar. 7.2V 150A Linear Regulator
. BATT 250A fuse by Winfield Hill
. POS __or breaker
. =====|__|==(O)===+===+===+===+===+===+===+===+===+===+===+===+===+===+===+
. +12.5-14.5V | | | | | | | | | | | | | | |
with the help of Chilton and Haynes manuals. The engine compartment is
filled with 20, 30, 40 and 60A distribution fuses for various things, plus
single 175A "Mega fuse" to the generator and "distribution block" (called
a fuseable link on some models). Mine is a removable plastic-encased fuse.
There is also a 15A and an 80A fuse bypassing the Mega fuse. In the case
of these Ford models the starter motor is _not_ fused. With some effort
these big replaceable fuses can be procured as automotive spare parts.
--
Thanks,
- Win