On Wed, 01 Sep 2004 11:36:29 -0500, John Fields
jfields@austininstruments.com> wrote:
From last time:
PHEW... That's it for this installment. Next one will be: "Since
You
Have A Comparator Left Over What'll It Take To Turn It into A
Flasher?"
Stay tuned...
AND NOW, SPORTS FANS...
+-----+------+---------------------------------------+-----+
| | | |R3 |
| [634K] [316K] +-[634K]-+ R2 [2.4K] |
| | | | | +-[100K]-+ | |
+|9V +------|-----+--|+\ | R1 | | [LED] |
[BAT] | | U1A | >--+---[100K]--+--|+\ | |K |
| | 1.2V-+--------|-/ U1B| >--+---+ |
| | | +--|-/ | [LOAD]
| [78.7K][LM385] | R4 | |
| | | +-----------+----[1M]----+ |
| | | | R5 | |
| | | +----+-[100k]--[IN4148>]-+ |
| | | | | CR1 |
| | | C1[1ľF][10M]R6 |
| | | | | |
+-----+------+--------------+----+-------------------------+
That's the basic scheme, with component values chosen just to get us
off the ground and the plan being to provide something like a 10%
duty
cycle for the LED flash, with an on time of about 100ms and an off
time of about 900ms.
So, where do we start? With U1A.
With the battery voltage above about 6V, the output of U1A will
always
be high. If we can assure that when that's the case the + input of
U1B
will be higher than the - input, the output of U1B will also always
be
high when the output of U1A is high, and the LED will be off.
Let's assume, for some reason, that that's not the case and that we
have a transient condition where the output of U1B is driven low,
but
that we have the circuit connected to a fresh battery with a 9 volt
output. If that's true, then the output of U1A will be high, and
since R1 and R2 are equal resistances the voltage on U1B+ will be
half
the battery voltage; 4.5V.
Now, if the voltage on U1B- is higher than that, the output of U1B
will stay low and the LED will stay lit, but only for as long as it
will take for C1 to discharge to <4.5V through R4, R5, and CR1.
When
that happens, the output of U1B will go high, the LED will turn off,
U1B+ will go to 9V, and C1 will start charging through R4. Because
of
the offset voltage differences between U1B+ and - and the slight
differences in the output structures of U1A and B, it's conceivable
that when C1 charged up close to the supply voltage U1B- might go
enough higher than U1B+ to drive the output of U1B low, so R6 is in
there to prevent that from happening since it forms a voltage
divider
with R4 and is chosen to make sure that U1B- can never go higher
than
U1B+ when the outputs of U1A and B are both high. More on that
later.
Assume now that the battery voltage has fallen enough to cause the
output of U1A to go low. When that happens, U1B+ will go to about
3V,
making it less positive than U1B-. That will force the output of
U1B
low, which will turn on the LED and force C1 to start discharging
through R4, R5, and CR1 until U1B goes slightly less positive than
U1B+. When that happens, the output of U1B will go positive, the
LED
will go out, and C1 will start charging through R4 until U1B- goes
slightly more positive than U1B+, starting the cycle anew.
Notice that there is more resistance in the charge path than there
is
in the discharge path, which means that the capacitor will discharge
more quickly than it will charge. We can use this to tailor the
duty
cycle of the LED to be whatever we want, in this case about 10% with
a
total period of about 1 second.
But how do we choose the values of R and C to get us what we want?
Looking at R1 and R2 just at the threshold of switching, we can see
that when the output of U1A is high the circuit looks like this:
6V (U1A OUT)
|
[R1]
|
+---> 6V (U1B+)
|
[R2]
|
6V (U1B OUT)
Then, when U1A OUT goes low, forcing U1B OUT low:
0V (U1A OUT)
|
[R1]
|
+---> 0V (U1B+)
|
[R2]
|
0V (U1B OUT)
Then, when C1 discharges to 0V:
0V (U1A OUT)
|
[R1]
|
+---> 3V (U1B+)
|
[R2]
|
6V (U1B OUT)
Then, when it charges to 3V:
0V (U1A OUT)
|
[R1]
|
+---> 0V (U1B+)
|
[R2]
|
0V (U1B OUT)
Thereafter, U1B will continue to oscillate as the voltage on U1B+
switches from 3V to 0V, so our job becomes one of making C1 charge
from 0V to 3V in 100ms and making it discharge from 3V to 0V in
900ms.
But, we have a problem in that there's no guarantee that, even with
both comparator outputs low, U1B- will go lower than U1B+ when the
cap
discharges.
So we'll add a bias resistor (R7) to make sure that when both
comparator outputs are low U1B+ will be positive enough to allow the
cap to discharge below that voltage and switch the output high.
+-----+------+--------------------------+------------+-----+
| | | |R7 | |
| | | [100K] |R3 |
| [634K] [316K] +-[634K]-+ | R2 [2.4K] |
| | | | | +-[100K]-+ | |
+|9V +------|-----+--|+\ | R1 | | [LED] |
[BAT] | | U1A | >--+-[100K]----+--|+\ | |K |
| | 1.2V-+--------|-/ U1B| >--+---+ |
| | | +--|-/ | [LOAD]
| [78.7K][LM385] | R4 | |
| | | +-----------+----[1M]----+ |
| | | | R5 | |
| | | +----+-[100k]--[IN4148>]-+ |
| | | | | CR1 |
| | | C1[1ľF][10M]R6 |
| | | | | |
+-----+------+--------------+----+-------------------------+
If we choose R7 to be 100kohms, then with both comparator outputs
high
we'll have what amounts to:
6V 6V 6V
|R1 |R7 |R2
[100K] [100K] [100K]
| | |
+----------+---------+--->U1B+
So we'll wind up with the battery voltage on U1B+, which is what we
want, so that's OK.
With U1A OUT low and U1B OUT high (immediately after U1A OUT goes
low,
but before U1B out has a chance to switch) we'll have:
6V 6V
|R7 |R2
[100k] [100K]
| |
+----------+---------+--->U1B+
|R1
[100K]
|
GND
Which reduces to:
6V
|
[50K]
|
+---->4.0V (U1B+)
|R1
[100K]
|
GND
So, since at that moment the cap will be charged up to very nearly
6V,
U1B- will also be at 6V and the comparator's output will go low,
which
is what we want, so that's OK, too.
Now, a microsecond or so later, when U1B OUT goes low we'll have
what
amounts to:
6v
|R7
[100k]
|
+------+------+
| |
[100K] [100K]
| |
GND GND
which reduces to:
6V
|R7
[100K]
|
+---->2.0V (U1B+)
|
[50K]
|
GND
Which is also fine because the voltage on U1B- will still be more
positive than 2.0V and U1B OUT will stay low, lighting the LED and
discharging C1 through R4, R5, and CR1.
When C1 discharges to less than 2V the voltage on U1B- will be less
positive than the voltage on U1B+, which will make U1B OUT go high.
This will make the voltage on U1B+ go to 4V and will turn off the
LED
and start charging the capacitor through R4. Then, when C1 charges
up
to 4V it will cause U1B OUT to go low, starting the cycle anew.
Now, with U1B OUT switching between high and low we'll have (after
the
first discharge from 6V to 2V) C1 oscillating between 2V and 4V, and
what we want is to have it discharge from 4V to 2V in 100
milliseconds
and then charge from 2V to 4V in 900 milliseconds. To do that we'll
need to select something in the right ballpark to work with, and
since
T=RC and we're talking about almost zero load from the comparator
input, a microfarad and a megohm sound about right for a first cut.
Since, when the capacitor is charging, it'll be starting from 2V and
rising to 4V with a 6V supply, that's really only 2 volts out of a 4
volt spread, so if we write
V1
T kRC , where k = ln ----
V2
and V1 is the 4 volt spread and V2 is the two volt excursion we'll
be
making, then
4V
k = ln ---- = ln 2 = 0.693
2V
Then, since we know C is going to be 1ľF, we can rearrange to solve
for R:
T 0.9s
R = --- = -------------- = 1.298 megohms
Ck 1E-6F * 0.693
The closest standard 1% value is 1.3 megohms, so that should be
close
enough!
Now, for the discharge time, we have a diode in there so we have to
subtract about 0.7V from the 4V spread, leaving us with a 3.3V
spread
and a 2V excursion, so this time
3.3
k = ln ----- = ln 1.65 = 0.501
2.0
and
T 0.1s
R = --- = --------------- = 199.6 kohms
Ck 1e-6F * 0.501
The closest standard 1% value for that one is 200k, so that's also
pretty close.
So here's the final schematic:
+-----+------+--------------------------+------------+-----+
| | | |R7 | |
| | | [100K] |R3 |
| [634K] [316K] +-[634K]-+ | R2 [2.4K] |
| | | | | +-[100K]-+ | |
+|9V +------|-----+--|+\ | R1 | | [LED] |
[BAT] | | U1A | >--+-[100K]----+--|+\ | |K |
| | 1.2V-+--------|-/ U1B| >--+---+ |
| | | +--|-/ | [LOAD]
| [78.7K][LM385] | R4 | |
| | | +-----------+---[1.3M]---+ |
| | | | R5 | |
| | | +----+-[200k]--[IN4148>]-+ |
| | | | | CR1 |
| | | C1[1ľF][10M]R6 |
| | | | | |
+-----+------+--------------+----+-------------------------+
One final thing, R6 will increase the charge and discharge times by
about 10%, but for a low battery indicator I don't think that's
going
to cause any heartburn.
Good luck!
--
John Fields
Andy´s ASCII-Circuit v1.25.250804 www.tech-chat.de
I'm guessing that's your answer and not the question...
Thanks Robert,
I bet you say that to all the girls.
Geo,
Yeah. Besides, how didja come up with this "AND Gate" stuff? ;-)
Why did you give us this url ?
No, it doesn't. If anything it is an always gate, the output is always
No. And I can't believe that you are trying to defend that circuit as
Ah-ha! I see what you mean. I thought it referred to the design maximum
This web page is obviously not intended to show examples of faulty
I was not wrong, the web site was!
Hi John,
Owww, you get a howler for that one!! </Harry Potter>