Voltage Regulator - surprise

B

BobP122

Guest
re Voltage Regulator
Hi,
I read the reply by John Popelish (to a 2002 post) on the Power
dissipated by a regulator - which answered my first question.
I'm using a 15V 1A IC regulator with a rechargeable 18Volt supply
battery (20Volts charged).
To my surprise, when the supply voltage drops below 15Volts the
regulator still outputs, (at about 2 Volts below supply) - down to as
low as 5Volts (7Volts supply).
I assumed that the regulator would not output once the supply fell
below (say) 17Volts.
That it continues to output, is not a problem here (it drives a fan
250mA), but the regulator "cooks", fails completely and becomes
unusable.
Two questions:
1. Why does it output to failure - below 15V?
2. As it does becomes unusable (i.e. cooks/fails), what can I do to
have the regulator "cutout" below some supply voltage (say 17volts)?
Thanks in advance
BobP
 
BobP122 wrote:
re Voltage Regulator
Hi,
I read the reply by John Popelish (to a 2002 post) on the Power
dissipated by a regulator - which answered my first question.
I'm using a 15V 1A IC regulator with a rechargeable 18Volt supply
battery (20Volts charged).
To my surprise, when the supply voltage drops below 15Volts the
regulator still outputs, (at about 2 Volts below supply) - down to as
low as 5Volts (7Volts supply).
I assumed that the regulator would not output once the supply fell
below (say) 17Volts.
That it continues to output, is not a problem here (it drives a fan
250mA), but the regulator "cooks", fails completely and becomes
unusable.
Two questions:
1. Why does it output to failure - below 15V?
2. As it does becomes unusable (i.e. cooks/fails), what can I do to
have the regulator "cutout" below some supply voltage (say 17volts)?
Click to expand...
The output stage of the regulator is several transistors
connected as a very high current follower that, when as on
as it can be, drops between 1.5 and 2 volts, depending on
the current. As long as the output voltage is below the
regulated target voltage, this circuit stays fully on.
However, the circuit also includes a temperature sensor that
is supposed to turn the pass circuit off if the chip
temperature exceeds something like 150 C. This is supposed
to prevent rapid destruction of the regulator from
overheating. But if overheated repeatedly, I have seen the
regulators eventually fail.

If your fan draws only 250 mA, you need a heat sink on the
regulator, to keep its temperature down. Another thing that
makes these regulators run hot is if they become unstable
and turn into megahertz oscillators. Do you have 0.1 uF of
ceramic or film capacitance close to the regulator, input to
ground? Do you have any capacitance between the output and
ground?

I don't know any easy way to add a low voltage cut out to
the 78xx series of linear regulators.
--
Regards,

John Popelish
 
On Jan 14, 11:55 am, BobP122 <JohnNico...@sympatico.ca> wrote:


I'm using a 15V 1A IC regulator with a rechargeable 18Volt supply
battery (20Volts charged).
To my surprise, when the supply voltage drops below 15Volts the
regulator still outputs, (at about 2 Volts below supply) - down to as
low as 5Volts (7Volts supply).
...the regulator "cooks", fails completely and becomes
unusable.

1. Why does it output to failure - below 15V?
Click to expand...
A regulator is an amplifier that operates on an output/reference
voltage difference. It saturates HIGH when the output is low
for any reason, including because of inadequate input voltage.
Failure, though, shouldn't happen because of simple input
undervoltage. I'd look at the input/output capacitors, it sounds
like the regulator may be oscillating. The motor might be
a problematic load, too (put a reverse-biased quench diode
on the motor terminals, to be safe).
 
BobP122 wrote:
re Voltage Regulator
Hi,
I read the reply by John Popelish (to a 2002 post) on the Power
dissipated by a regulator - which answered my first question.
I'm using a 15V 1A IC regulator with a rechargeable 18Volt supply
battery (20Volts charged).
To my surprise, when the supply voltage drops below 15Volts the
regulator still outputs, (at about 2 Volts below supply) - down to as
low as 5Volts (7Volts supply).
I assumed that the regulator would not output once the supply fell
below (say) 17Volts.
That it continues to output, is not a problem here (it drives a fan
250mA), but the regulator "cooks", fails completely and becomes
unusable.
Two questions:
1. Why does it output to failure - below 15V?
2. As it does becomes unusable (i.e. cooks/fails), what can I do to
have the regulator "cutout" below some supply voltage (say 17volts)?
Thanks in advance
BobP
Click to expand...
You can add a cutout circuit to the input to the regulator:

+------------------o o---+
| | | |
| S1 |>| |
| __ |
+--o o---+--------+-----+-----+---[7815]----+--> +
| | | | | | |
| [10K] [Rly] [D1] +-[C1]-+-[C2]-+
|+ | | | |
B | +-----+ |
A P | k |
T 10K O<----[TL431] |
T T | |
| | | |
+---------+--------+------------------+----------> Gnd

S1 (momentary) turns the circuit on. When the battery drops
below the set point established by the pot, the relay drops
out and the circuit draws no power, even if the battery
voltage rises (because current is no longer being drawn
from it). The circuit will protect both your battery (from
being discharged too far) and the 7815. Although the TL431
can sink up to 100 mA, it is best to use a low current
relay, such as Allelectronics CAT# RLY-622, which draws
about 16.3 mA. That is a 12 volt relay with a 740 ohm coil,
so you would add a 330 ohm resistor in series. The low current
relay reduces the current the circuit draws from the battery.
You can add a high efficiency LED between the 10K and the pot
as an on indicator, if you want.

Ed
 
On Mon, 14 Jan 2008 15:13:48 -0500, John Popelish <jpopelish@rica.net>
wrote:

That it continues to output, is not a problem here (it drives a fan
250mA), but the regulator "cooks", fails completely and becomes
unusable.
Two questions:
1. Why does it output to failure - below 15V?
2. As it does becomes unusable (i.e. cooks/fails), what can I do to
Click to expand...

Three terminal regulators would cook if the battery is connected to
the output and the input falls below the battery voltage - it is in
the application notes - they recommend an isolation diode on the
output or diode from the output to feed voltage to the input (which is
reversed biased during normal operation and keeps the regulator happy
when the input voltage is missing).

Having a battery on the output and no input is just like connecting
the regulator backwards in the circuit. A large cap on the output can
kill a regulator when power is removed.

The diode makes a cutout circuit unnecessary unless you are concerned
about the small power (quiescent current) the regulator wastes when
sitting idle
--
 
On Jan 15, 9:50 am, default <defa...@defaulter.net> wrote:
On Mon, 14 Jan 2008 15:13:48 -0500, John Popelish <jpopel...@rica.net
wrote:

That it continues to output, is not a problem here (it drives a fan
250mA), but the regulator "cooks", fails completely and becomes
unusable.
Two questions:
1. Why does it output to failure - below 15V?
2. As it does becomes unusable (i.e. cooks/fails), what can I do to

Three terminal regulators would cook if the battery is connected to
the output and the input falls below the battery voltage - it is in
the application notes - they recommend an isolation diode on the
output or diode from the output to feed voltage to the input (which is
reversed biased during normal operation and keeps the regulator happy
when the input voltage is missing).

Having a battery on the output and no input is just like connecting
the regulator backwards in the circuit. A large cap on the output can
kill a regulator when power is removed.

The diode makes a cutout circuit unnecessary unless you are concerned
about the small power (quiescent current) the regulator wastes when
sitting idle
--
Click to expand...
Hi again,
To eshjr
I can see where a relay configured to 'cutout' the battery would avoid
problems. Thanks.

To default
The battery is connected to the VR only. There aren't any direct
connections from the battery to the load.
Nor would I know how to. But now due to a curious nature - I'll be
looking into that! So thanks a lot - I think.

A big Thank you to all JP, eshjr, & default- much appreciated.
The fan is rated 6 to 15 volts. The rechargeable battery power supply
was 18Volts. My 'too simple' remedy was to use a 15V Regulator. Once
again I learned how true the adage "a liitle learning is a dangerous
thing".
After your explanation, I replaced the Voltage regulator with a 33ohm
5W resistor in series with the load (DC Fan 60ohms). I tried a 5.1Volt
5Watt Zener (in series with the Fan),but without any calculations it
just felt warmer to the touch. I'm hoping that this is a safer
solution. . .?
I'm still curious to know why the Voltage Regulator 'cooks' when the
input voltage drops to (around) 7Volts and the regulator's voltage to
way below regulation - (around) 5Volts. Shouldn't there be much less
heat generated at this stage - or at least not enough to heat the VR
to failure?
I did have (recommended) caps at the input and the output. I'm
guessing that at the Fan's dropout Voltage (5Volts), a 'switching' on-
off oscillation occurred, and the momentary 'on current' was much
higher than 250mAmps, enough to burn the VR.
Any explanation would be apprecated,
BobP
 
On Jan 15, 11:11 am, BobP122 <JohnNico...@sympatico.ca> wrote:

I did have (recommended) caps at the input and the output. I'm
guessing that at the Fan's dropout Voltage (5Volts), a 'switching' on-
off oscillation occurred, and the momentary 'on current' was much
higher than 250mAmps, enough to burn the VR.
Click to expand...
Alternately, consider that the regulator might have shutdown
due to overtemperature, thus going 'open circuit' with
the (inductive) motor at 5V, 200 mA. The motor winding
would, as the current dropped, go to -45V and the resultant
in/out voltage transient can kill the regulator. Thirty
volts input-to-output is allowed, 52 volts isn't.

I've had regulators feeding short circuits for, literally, years
and when the short was lifted, the regulator worked fine.
I'm skeptical of the overcurrent hypothesis.
 
BobP122 wrote:
On Jan 15, 9:50 am, default <defa...@defaulter.net> wrote:

On Mon, 14 Jan 2008 15:13:48 -0500, John Popelish <jpopel...@rica.net
wrote:


That it continues to output, is not a problem here (it drives a fan

250mA), but the regulator "cooks", fails completely and becomes
unusable.
Two questions:
1. Why does it output to failure - below 15V?
2. As it does becomes unusable (i.e. cooks/fails), what can I do to

Three terminal regulators would cook if the battery is connected to
the output and the input falls below the battery voltage - it is in
the application notes - they recommend an isolation diode on the
output or diode from the output to feed voltage to the input (which is
reversed biased during normal operation and keeps the regulator happy
when the input voltage is missing).

Having a battery on the output and no input is just like connecting
the regulator backwards in the circuit. A large cap on the output can
kill a regulator when power is removed.

The diode makes a cutout circuit unnecessary unless you are concerned
about the small power (quiescent current) the regulator wastes when
sitting idle
--


Hi again,
To eshjr
I can see where a relay configured to 'cutout' the battery would avoid
problems. Thanks.

To default
The battery is connected to the VR only. There aren't any direct
connections from the battery to the load.
Nor would I know how to. But now due to a curious nature - I'll be
looking into that! So thanks a lot - I think.

A big Thank you to all JP, eshjr, & default- much appreciated.
The fan is rated 6 to 15 volts. The rechargeable battery power supply
was 18Volts. My 'too simple' remedy was to use a 15V Regulator. Once
again I learned how true the adage "a liitle learning is a dangerous
thing".
After your explanation, I replaced the Voltage regulator with a 33ohm
5W resistor in series with the load (DC Fan 60ohms). I tried a 5.1Volt
5Watt Zener (in series with the Fan),but without any calculations it
just felt warmer to the touch. I'm hoping that this is a safer
solution. . .?
I'm still curious to know why the Voltage Regulator 'cooks' when the
input voltage drops to (around) 7Volts and the regulator's voltage to
way below regulation - (around) 5Volts. Shouldn't there be much less
heat generated at this stage - or at least not enough to heat the VR
to failure?
Click to expand...
The thermal overtemperature cutout for these regulators compares a
reference voltage derived from an internal 6.3V zener diode to the
cut-in Vbe of an internal temperature sensing transistor positioned to
shunt the base drive away from the output transistors at a die
temperature of 175oC. At your low input voltage of 7V or so, the zener
bias current is too low which makes the zener reference voltage low, and
this in effect shifts the resulting thermal cut-out threshold to a
higher die temperature causing the 'VR' to fail. The output transistor
SOA protection circuit will also be inoperative, leaving the current
limiting protection as the only safety feature in place. You also risk
damaging your battery by running it down to one third of its full charge
cell voltage. You need to include a battery voltage monitor which
latches the whole battery load off when it drops below a threshold of
say 12V under load. Because this is a battery powered system, you would
be better off using CMOS logic and a MOSFET series pass element for the
monitoring and control.

I did have (recommended) caps at the input and the output. I'm
guessing that at the Fan's dropout Voltage (5Volts), a 'switching' on-
off oscillation occurred, and the momentary 'on current' was much
higher than 250mAmps, enough to burn the VR.
Click to expand...
Not plausible...
 
On Jan 16, 6:51 am, Fred Bloggs <nos...@nospam.com> wrote:
BobP122 wrote:
On Jan 15, 9:50 am, default <defa...@defaulter.net> wrote:

On Mon, 14 Jan 2008 15:13:48 -0500, John Popelish <jpopel...@rica.net
wrote:

That it continues to output, is not a problem here (it drives a fan

250mA), but the regulator "cooks", fails completely and becomes
unusable.
Two questions:
1. Why does it output to failure - below 15V?
2. As it does becomes unusable (i.e. cooks/fails), what can I do to

Three terminal regulators would cook if the battery is connected to
the output and the input falls below the battery voltage - it is in
the application notes - they recommend an isolation diode on the
output or diode from the output to feed voltage to the input (which is
reversed biased during normal operation and keeps the regulator happy
when the input voltage is missing).

Having a battery on the output and no input is just like connecting
the regulator backwards in the circuit. A large cap on the output can
kill a regulator when power is removed.

The diode makes a cutout circuit unnecessary unless you are concerned
about the small power (quiescent current) the regulator wastes when
sitting idle
--

Hi again,
To eshjr
I can see where a relay configured to 'cutout' the battery would avoid
problems. Thanks.

To default
The battery is connected to the VR only. There aren't any direct
connections from the battery to the load.
Nor would I know how to. But now due to a curious nature - I'll be
looking into that! So thanks a lot - I think.

A big Thank you to all JP, eshjr, & default- much appreciated.
The fan is rated 6 to 15 volts. The rechargeable battery power supply
was 18Volts. My 'too simple' remedy was to use a 15V Regulator. Once
again I learned how true the adage "a liitle learning is a dangerous
thing".
After your explanation, I replaced the Voltage regulator with a 33ohm
5W resistor in series with the load (DC Fan 60ohms). I tried a 5.1Volt
5Watt Zener (in series with the Fan),but without any calculations it
just felt warmer to the touch. I'm hoping that this is a safer
solution. . .?
I'm still curious to know why the Voltage Regulator 'cooks' when the
input voltage drops to (around) 7Volts and the regulator's voltage to
way below regulation - (around) 5Volts. Shouldn't there be much less
heat generated at this stage - or at least not enough to heat the VR
to failure?

The thermal overtemperature cutout for these regulators compares a
reference voltage derived from an internal 6.3V zener diode to the
cut-in Vbe of an internal temperature sensing transistor positioned to
shunt the base drive away from the output transistors at a die
temperature of 175oC. At your low input voltage of 7V or so, the zener
bias current is too low which makes the zener reference voltage low, and
this in effect shifts the resulting thermal cut-out threshold to a
higher die temperature causing the 'VR' to fail. The output transistor
SOA protection circuit will also be inoperative, leaving the current
limiting protection as the only safety feature in place. You also risk
damaging your battery by running it down to one third of its full charge
cell voltage. You need to include a battery voltage monitor which
latches the whole battery load off when it drops below a threshold of
say 12V under load. Because this is a battery powered system, you would
be better off using CMOS logic and a MOSFET series pass element for the
monitoring and control.

I did have (recommended) caps at the input and the output. I'm
guessing that at the Fan's dropout Voltage (5Volts), a 'switching' on-
off oscillation occurred, and the momentary 'on current' was much
higher than 250mAmps, enough to burn the VR.

Not plausible...- Hide quoted text -

- Show quoted text -
Click to expand...
BobP
Wow! and Thanks! Your explanation may be just a 'little too clever for
me', but there was a flash of Iight here and there, enough for me to
at least "believe" if not "undestand". By the way just in passing, who
did create the universe? It wasn't you or - was it? Please say it
wasn't.
Now I have a real value for 'deep cycling' batteries. That alone is
great information.
Another Question: The circuit for all this came from an old mercury
'heat' type thermostat.
The contacts are open to a set 'low' temperature (e.g. 70oF), at which
point the contacts close.
The contacts stay closed as long as the temp is below 70oF and stay
open as long as the temp is above 70oF.
Problem was, I needed the Thermostat to operate as 'cooling' or air
conditioning type. As such - it would have to turn on a small fan when
the contacts were 'open' and turn off the fan when the contacts were
'closed' - the exact opposite to its normal usage.
A kind of Solution: To do this, I used a PNP (Darlington) transistor
configured as a switch.
One Thermostat contact is tied to B+.
The Other Thermostat contact is tied to the base of the Transistor and
Ground (B-) through a resistor (10k).
With the contacts 'open', the base of the transistor base is grounded
through the resistor. i.e. Transistor on.
With the contacts 'closed', the base of the transistor base is
positive - i.e. Transistor off.
The circuit works - but is somewhat inelegant.. For a start - it draws
power even with the transistor off. i.e thru th 10k resistor - a mA or
two.
Its kept awake to find a better configuration (using the same 'heat'
type thermostat). Any suggestions?
BobP
P.S. I could have and maybe should have just 'recyled' the thermostat
- but then again, look at all I've learned. (There was a previous
"lesson" on discarded lithium-ion batteries. Enough just to say I'm
using
Nickel-Cadmium now.)
 
BobP122 wrote:
On Jan 16, 6:51 am, Fred Bloggs <nos...@nospam.com> wrote:

BobP122 wrote:

On Jan 15, 9:50 am, default <defa...@defaulter.net> wrote:

On Mon, 14 Jan 2008 15:13:48 -0500, John Popelish <jpopel...@rica.net
wrote:

That it continues to output, is not a problem here (it drives a fan

250mA), but the regulator "cooks", fails completely and becomes
unusable.
Two questions:
1. Why does it output to failure - below 15V?
2. As it does becomes unusable (i.e. cooks/fails), what can I do to

Three terminal regulators would cook if the battery is connected to
the output and the input falls below the battery voltage - it is in
the application notes - they recommend an isolation diode on the
output or diode from the output to feed voltage to the input (which is
reversed biased during normal operation and keeps the regulator happy
when the input voltage is missing).

Having a battery on the output and no input is just like connecting
the regulator backwards in the circuit. A large cap on the output can
kill a regulator when power is removed.

The diode makes a cutout circuit unnecessary unless you are concerned
about the small power (quiescent current) the regulator wastes when
sitting idle
--

Hi again,
To eshjr
I can see where a relay configured to 'cutout' the battery would avoid
problems. Thanks.

To default
The battery is connected to the VR only. There aren't any direct
connections from the battery to the load.
Nor would I know how to. But now due to a curious nature - I'll be
looking into that! So thanks a lot - I think.

A big Thank you to all JP, eshjr, & default- much appreciated.
The fan is rated 6 to 15 volts. The rechargeable battery power supply
was 18Volts. My 'too simple' remedy was to use a 15V Regulator. Once
again I learned how true the adage "a liitle learning is a dangerous
thing".
After your explanation, I replaced the Voltage regulator with a 33ohm
5W resistor in series with the load (DC Fan 60ohms). I tried a 5.1Volt
5Watt Zener (in series with the Fan),but without any calculations it
just felt warmer to the touch. I'm hoping that this is a safer
solution. . .?
I'm still curious to know why the Voltage Regulator 'cooks' when the
input voltage drops to (around) 7Volts and the regulator's voltage to
way below regulation - (around) 5Volts. Shouldn't there be much less
heat generated at this stage - or at least not enough to heat the VR
to failure?

The thermal overtemperature cutout for these regulators compares a
reference voltage derived from an internal 6.3V zener diode to the
cut-in Vbe of an internal temperature sensing transistor positioned to
shunt the base drive away from the output transistors at a die
temperature of 175oC. At your low input voltage of 7V or so, the zener
bias current is too low which makes the zener reference voltage low, and
this in effect shifts the resulting thermal cut-out threshold to a
higher die temperature causing the 'VR' to fail. The output transistor
SOA protection circuit will also be inoperative, leaving the current
limiting protection as the only safety feature in place. You also risk
damaging your battery by running it down to one third of its full charge
cell voltage. You need to include a battery voltage monitor which
latches the whole battery load off when it drops below a threshold of
say 12V under load. Because this is a battery powered system, you would
be better off using CMOS logic and a MOSFET series pass element for the
monitoring and control.


I did have (recommended) caps at the input and the output. I'm
guessing that at the Fan's dropout Voltage (5Volts), a 'switching' on-
off oscillation occurred, and the momentary 'on current' was much
higher than 250mAmps, enough to burn the VR.

Not plausible...- Hide quoted text -

- Show quoted text -


BobP
Wow! and Thanks! Your explanation may be just a 'little too clever for
me', but there was a flash of Iight here and there, enough for me to
at least "believe" if not "undestand". By the way just in passing, who
did create the universe? It wasn't you or - was it? Please say it
wasn't.
Now I have a real value for 'deep cycling' batteries. That alone is
great information.
Another Question: The circuit for all this came from an old mercury
'heat' type thermostat.
The contacts are open to a set 'low' temperature (e.g. 70oF), at which
point the contacts close.
The contacts stay closed as long as the temp is below 70oF and stay
open as long as the temp is above 70oF.
Problem was, I needed the Thermostat to operate as 'cooling' or air
conditioning type. As such - it would have to turn on a small fan when
the contacts were 'open' and turn off the fan when the contacts were
'closed' - the exact opposite to its normal usage.
A kind of Solution: To do this, I used a PNP (Darlington) transistor
configured as a switch.
One Thermostat contact is tied to B+.
The Other Thermostat contact is tied to the base of the Transistor and
Ground (B-) through a resistor (10k).
With the contacts 'open', the base of the transistor base is grounded
through the resistor. i.e. Transistor on.
With the contacts 'closed', the base of the transistor base is
positive - i.e. Transistor off.
The circuit works - but is somewhat inelegant.. For a start - it draws
power even with the transistor off. i.e thru th 10k resistor - a mA or
two.
Its kept awake to find a better configuration (using the same 'heat'
type thermostat). Any suggestions?
BobP
P.S. I could have and maybe should have just 'recyled' the thermostat
- but then again, look at all I've learned. (There was a previous
"lesson" on discarded lithium-ion batteries. Enough just to say I'm
using
Nickel-Cadmium now.)
Click to expand...
Well, the "elegant solution" that occurs to me is to use
a regulated wall wart that provides the proper voltage and
sufficient current for your relay circuit to start with. That
eliminates the need for a 7815 regulator and battery and
protection circuit for both of those. It also makes the 1 or 2
mA quiescent current unimportant. You can then make it
"more elegant" (at the cost of more current) by doing this:

/
+ 12---+---o o----+---[1.5K]---+
| | |
+---[Rly]---+ |
|
Gnd ----------------------------+

It uses a MOSFET relay (Mouser #653-G3VM-61A1)
for low current draw (< 7 mA). The relay output
is open when the t-stat is closed, and vice versa.
It is rated for 500 mA continuous load current.

Ed

Ed
 
On Jan 18, 2:36 am, ehsjr <eh...@bellatlantic.net> wrote:
BobP122 wrote:
On Jan 16, 6:51 am, Fred Bloggs <nos...@nospam.com> wrote:

BobP122 wrote:

On Jan 15, 9:50 am, default <defa...@defaulter.net> wrote:

On Mon, 14 Jan 2008 15:13:48 -0500, John Popelish <jpopel...@rica.net
wrote:

That it continues to output, is not a problem here (it drives a fan

250mA), but the regulator "cooks", fails completely and becomes
unusable.
Two questions:
1. Why does it output to failure - below 15V?
2. As it does becomes unusable (i.e. cooks/fails), what can I do to

Three terminal regulators would cook if the battery is connected to
the output and the input falls below the battery voltage - it is in
the application notes - they recommend an isolation diode on the
output or diode from the output to feed voltage to the input (which is
reversed biased during normal operation and keeps the regulator happy
when the input voltage is missing).

Having a battery on the output and no input is just like connecting
the regulator backwards in the circuit. A large cap on the output can
kill a regulator when power is removed.

The diode makes a cutout circuit unnecessary unless you are concerned
about the small power (quiescent current) the regulator wastes when
sitting idle
--

Hi again,
To eshjr
I can see where a relay configured to 'cutout' the battery would avoid
problems. Thanks.

To default
The battery is connected to the VR only. There aren't any direct
connections from the battery to the load.
Nor would I know how to. But now due to a curious nature - I'll be
looking into that! So thanks a lot - I think.

A big Thank you to all JP, eshjr, & default- much appreciated.
The fan is rated 6 to 15 volts. The rechargeable battery power supply
was 18Volts. My 'too simple' remedy was to use a 15V Regulator. Once
again I learned how true the adage "a liitle learning is a dangerous
thing".
After your explanation, I replaced the Voltage regulator with a 33ohm
5W resistor in series with the load (DC Fan 60ohms). I tried a 5.1Volt
5Watt Zener (in series with the Fan),but without any calculations it
just felt warmer to the touch. I'm hoping that this is a safer
solution. . .?
I'm still curious to know why the Voltage Regulator 'cooks' when the
input voltage drops to (around) 7Volts and the regulator's voltage to
way below regulation - (around) 5Volts. Shouldn't there be much less
heat generated at this stage - or at least not enough to heat the VR
to failure?

The thermal overtemperature cutout for these regulators compares a
reference voltage derived from an internal 6.3V zener diode to the
cut-in Vbe of an internal temperature sensing transistor positioned to
shunt the base drive away from the output transistors at a die
temperature of 175oC. At your low input voltage of 7V or so, the zener
bias current is too low which makes the zener reference voltage low, and
this in effect shifts the resulting thermal cut-out threshold to a
higher die temperature causing the 'VR' to fail. The output transistor
SOA protection circuit will also be inoperative, leaving the current
limiting protection as the only safety feature in place. You also risk
damaging your battery by running it down to one third of its full charge
cell voltage. You need to include a battery voltage monitor which
latches the whole battery load off when it drops below a threshold of
say 12V under load. Because this is a battery powered system, you would
be better off using CMOS logic and a MOSFET series pass element for the
monitoring and control.

I did have (recommended) caps at the input and the output. I'm
guessing that at the Fan's dropout Voltage (5Volts), a 'switching' on-
off oscillation occurred, and the momentary 'on current' was much
higher than 250mAmps, enough to burn the VR.

Not plausible...- Hide quoted text -

- Show quoted text -

BobP
Wow! and Thanks! Your explanation may be just a 'little too clever for
me', but there was a flash of Iight here and there, enough for me to
at least "believe" if not "undestand". By the way just in passing, who
did create the universe? It wasn't you or - was it? Please say it
wasn't.
Now I have a real value for 'deep cycling' batteries. That alone is
great information.
Another Question: The circuit for all this came from an old mercury
'heat' type thermostat.
The contacts are open to a set 'low' temperature (e.g. 70oF), at which
point the contacts close.
The contacts stay closed as long as the temp is below 70oF and stay
open as long as the temp is above 70oF.
Problem was, I needed the Thermostat to operate as 'cooling' or air
conditioning type. As such - it would have to turn on a small fan when
the contacts were 'open' and turn off the fan when the contacts were
'closed' - the exact opposite to its normal usage.
A kind of Solution: To do this, I used a PNP (Darlington) transistor
configured as a switch.
One Thermostat contact is tied to B+.
The Other Thermostat contact is tied to the base of the Transistor and
Ground (B-) through a resistor (10k).
With the contacts 'open', the base of the transistor base is grounded
through the resistor. i.e. Transistor on.
With the contacts 'closed', the base of the transistor base is
positive - i.e. Transistor off.
The circuit works - but is somewhat inelegant.. For a start - it draws
power even with the transistor off. i.e thru th 10k resistor - a mA or
two.
Its kept awake to find a better configuration (using the same 'heat'
type thermostat). Any suggestions?
BobP
P.S. I could have and maybe should have just 'recyled' the thermostat
- but then again, look at all I've learned. (There was a previous
"lesson" on discarded lithium-ion batteries. Enough just to say I'm
using
Nickel-Cadmium now.)

Well, the "elegant solution" that occurs to me is to use
a regulated wall wart that provides the proper voltage and
sufficient current for your relay circuit to start with. That
eliminates the need for a 7815 regulator and battery and
protection circuit for both of those. It also makes the 1 or 2
mA quiescent current unimportant. You can then make it
"more elegant" (at the cost of more current) by doing this:

/
+ 12---+---o o----+---[1.5K]---+
| | |
+---[Rly]---+ |
|
Gnd ----------------------------+

It uses a MOSFET relay (Mouser #653-G3VM-61A1)
for low current draw (< 7 mA). The relay output
is open when the t-stat is closed, and vice versa.
It is rated for 500 mA continuous load current.

Ed

Ed- Hide quoted text -

- Show quoted text -
Click to expand...
BobP
Thanks ehsjr. The location makes a "wall wart" impractical.
I'm in Montreal ( = cold). I have a heating radiator in a closed
cabinet under a sink (not my design). Needless to say, it gets quite
warm in the cabinet maybe 15oF higher than the room temperature. The
fan is supposed to blow out the warmer air into the room. Ergo the
thermostat. I know there are (inexpensive) air conditioning
thermostats that would solve the design problem. But . . .
BobP
P.S. Had to google 'wall wart' just to be sure. Very colorful
description.
 
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