Zap a volt regulator w/ no cap??

[eromlignod]

Hi guys:

First of all, forgive me for being a dumb ME meddling in electrical
work.

I have an application in a machine that has a 24-volt power supply. I
have a tilt sensor (analog output) that needs 12 Vdc as its input
supply. Until now this part of the machine used an embedded circuit
that the sensor mounted directly to and that provided its own 12 V.

I converted this part of the machine's control to be read from a PLC
analog input module, so I had to come up with my own 12 V source. What
I did was add an LM2940CT-12 voltage regulator (TO-220 package) to the
circuit right before the analog sensor. This provided my 12V and
worked great for about a day and a half, then my analog sensor fried.
When I measured the voltage from the regulator I found that it was now
the full 24V input...so apparently the regulator failed first and then
cooked the sensor with 24 V.

When I took a look at the spec. sheet on the LM2940, I see that they
recommend putting capacitors from the input and the output to
ground...I didn't do this. Other than this mistake, I can't see
anything else wrong with the circuit. Could the absence of these caps
have caused the regulator to fry? If not, what else could have caused
it? I'd like to know exactly what my problem is before I toast another
sensor (they're $175 a pop).

Thanks for any advice (or chastisement) that you can provide.

Don
Kansas City

 

[John Popelish]

eromlignod wrote:
(snip)

I have an application in a machine that has a 24-volt power supply. I
have a tilt sensor (analog output) that needs 12 Vdc as its input
supply. Until now this part of the machine used an embedded circuit
that the sensor mounted directly to and that provided its own 12 V.

I converted this part of the machine's control to be read from a PLC
analog input module, so I had to come up with my own 12 V source. What
I did was add an LM2940CT-12 voltage regulator (TO-220 package) to the
circuit right before the analog sensor. This provided my 12V and
worked great for about a day and a half, then my analog sensor fried.
When I measured the voltage from the regulator I found that it was now
the full 24V input...so apparently the regulator failed first and then
cooked the sensor with 24 V.

When I took a look at the spec. sheet on the LM2940, I see that they
recommend putting capacitors from the input and the output to
ground...I didn't do this. Other than this mistake, I can't see
anything else wrong with the circuit. Could the absence of these caps
have caused the regulator to fry? If not, what else could have caused
it? I'd like to know exactly what my problem is before I toast another
sensor (they're $175 a pop).

Low drop out regulators (of which the LM2940 is one) are
only stable with certain values of capacitance (and
sometimes only with certain values of series resistance in
that capacitance).
The graph on page 11 shows how tricky this is.
http://info.hobbyengineering.com/specs/NATSEMI-LM2940.pdf

Since you have lots of excess voltage, a follower output
type linear regulator like an LM7812 would be less
problematical. It still works best with a small capacitor
from both input and output to ground, close to the regulator
(say, .1 uF) but the values and series resistance are not at
all critical, compared to the LM2940.
See page 22 of:
http://www.fairchildsemi.com/ds/LM/LM7812.pdf

 

[Tim Wescott]

eromlignod wrote:
(snip)



Low drop out regulators (of which the LM2940 is one) are only stable
with certain values of capacitance (and sometimes only with certain
values of series resistance in that capacitance).
The graph on page 11 shows how tricky this is.
http://info.hobbyengineering.com/specs/NATSEMI-LM2940.pdf

Since you have lots of excess voltage, a follower output type linear
regulator like an LM7812 would be less problematical. It still works
best with a small capacitor from both input and output to ground, close
to the regulator (say, .1 uF) but the values and series resistance are
not at all critical, compared to the LM2940.
See page 22 of:
http://www.fairchildsemi.com/ds/LM/LM7812.pdf

"Works best" is a mild way of saying "leave them off at your peril". I
stopped in the middle of build a circuit with an embedded 78L05
regulator; when powered up without an input cap it oscillated at 80MHz.

You should also check on power dissipation. These regulators are
supposed to have thermal shutdowns, but that doesn't mean they work
right, and you don't want a shutdown event anyway. Check the input
current to your 12V module; you'll be dissipating that times 12V (24V -
12V), and more if your 24V supply goes higher.

Make sure you have enough heat sinking to keep the regulator cool. The
best rule of thumb I know is to run it for a while then put your thumb
on it. If you pull back and say "ouch" then the heat sink isn't
dissipating enough.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google? See http://cfaj.freeshell.org/google/

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[PeteS]

"Works best" is a mild way of saying "leave them off at your peril". I
stopped in the middle of build a circuit with an embedded 78L05
regulator; when powered up without an input cap it oscillated at 80MHz.

You should also check on power dissipation. These regulators are
supposed to have thermal shutdowns, but that doesn't mean they work
right, and you don't want a shutdown event anyway. Check the input
current to your 12V module; you'll be dissipating that times 12V (24V -
12V), and more if your 24V supply goes higher.

Make sure you have enough heat sinking to keep the regulator cool. The
best rule of thumb I know is to run it for a while then put your thumb
on it. If you pull back and say "ouch" then the heat sink isn't
dissipating enough.

Tim's caution about heat sinking should be heeded; I have worked with
some industrial sensor modules (OPTO-22) and they can draw significant
amounts of current. In your case, that would give a regulator power
dissipation equal to the sensor (12V x load current) and even a TO-220
without a heat sink can't handle too much if it's within a box (so
ambient is pretty high).

Do you know your load current?

Cheers

PeteS

 

[Eeyore]

Hi guys:

First of all, forgive me for being a dumb ME meddling in electrical
work.

I have an application in a machine that has a 24-volt power supply. I
have a tilt sensor (analog output) that needs 12 Vdc as its input
supply. Until now this part of the machine used an embedded circuit
that the sensor mounted directly to and that provided its own 12 V.

I converted this part of the machine's control to be read from a PLC
analog input module, so I had to come up with my own 12 V source. What
I did was add an LM2940CT-12 voltage regulator (TO-220 package) to the
circuit right before the analog sensor. This provided my 12V and
worked great for about a day and a half, then my analog sensor fried.
When I measured the voltage from the regulator I found that it was now
the full 24V input...so apparently the regulator failed first and then
cooked the sensor with 24 V.

When I took a look at the spec. sheet on the LM2940, I see that they
recommend putting capacitors from the input and the output to
ground...I didn't do this. Other than this mistake, I can't see
anything else wrong with the circuit. Could the absence of these caps
have caused the regulator to fry? If not, what else could have caused
it? I'd like to know exactly what my problem is before I toast another
sensor (they're $175 a pop).

Thanks for any advice (or chastisement) that you can provide.

The capacitors are rather important so you should fit them.

Other than that I assume the current drawn @ 12V isn't greater than say 100mA.
I'm thinking of the thermal limitation of the regulator here without a heatsink.

Graham

 

[Rich Grise]

"Works best" is a mild way of saying "leave them off at your peril". I
stopped in the middle of build a circuit with an embedded 78L05
regulator; when powered up without an input cap it oscillated at 80MHz.

I've always used the caps, and never had a regulator problem. But, being
obsessive/compulsive about capacitation, I soldered the caps right to
the regulator leads right where the leads neck down.

Cheers!
Rich

 

[Tim Auton]

eromlignod wrote: [snip]

When I took a look at the spec. sheet on the LM2940, I see that they
recommend putting capacitors from the input and the output to
ground...I didn't do this. Other than this mistake, I can't see
anything else wrong with the circuit. Could the absence of these caps
have caused the regulator to fry? If not, what else could have caused
it? I'd like to know exactly what my problem is before I toast another
sensor (they're $175 a pop).

Thanks for any advice (or chastisement) that you can provide.

The capacitors are rather important so you should fit them.
Absolutely.

Other than that I assume the current drawn @ 12V isn't greater than say 100mA.
I'm thinking of the thermal limitation of the regulator here without a heatsink.

The LM2940 has thermal overload protection, so given the symptoms
described (overvoltage rather than shutdown) I doubt that was the
problem in this instance.

Shutdown isn't desirable though, so the thermals do need checking.
eromlignod: If your sensor does draw more than around 100mA and/or the
ambient temperature is much higher than room temperature you can let us
know and we can show you how to calculate the required heatsink (or
check your calculations if you've already figured that out).


Tim

 

[David L. Jones]

Hi guys:

First of all, forgive me for being a dumb ME meddling in electrical
work.

I have an application in a machine that has a 24-volt power supply. I
have a tilt sensor (analog output) that needs 12 Vdc as its input
supply. Until now this part of the machine used an embedded circuit
that the sensor mounted directly to and that provided its own 12 V.

I converted this part of the machine's control to be read from a PLC
analog input module, so I had to come up with my own 12 V source. What
I did was add an LM2940CT-12 voltage regulator (TO-220 package) to the
circuit right before the analog sensor. This provided my 12V and
worked great for about a day and a half, then my analog sensor fried.
When I measured the voltage from the regulator I found that it was now
the full 24V input...so apparently the regulator failed first and then
cooked the sensor with 24 V.

When I took a look at the spec. sheet on the LM2940, I see that they
recommend putting capacitors from the input and the output to
ground...I didn't do this. Other than this mistake, I can't see
anything else wrong with the circuit. Could the absence of these caps
have caused the regulator to fry?

Yes.
All regulators should have input and output caps, but "Low dropout"
regulators like the LM2940 in particular are very touchy and require an
output capacitance of a certain type and value to stop it oscillating,
and hence stop the vital smoke from escaping the case.

Unless you have a very low input voltage (<14V or so) then don't use
the LM2940, use a bog standard LM7812 instead, they are much more
stable. Yes, use an input and output cap on the LM7812 as well, value
is not critical, 0.1uF should be fine.

Dave

 

[eromlignod]

Thanks for the help so far, guys.

The National spec. sheet (and my chip is a National) mentions reverse
polarity protection and short circuit protection, but not overtemp
protection, so I'm not sure if it was heat or not that made it fail.

The sensor is all that the output drives and it's a very low
load--maybe a few mA. The regulator is mounted perpendicular on a PC
board with no heat sink and the ambient is normal indoor room
temperature (maybe 75 F). It isn't enclosed in a box and gets good
ventillation. I remember touching the regulator several times to see
if it was red hot. It was hot, but not soldering-iron hot, like I've
seen some power transistors get before they blow. I had experienced
some low voltage problems on the 24 V supply due to a separate problem,
but no lower than about 22 V or so. The regulator failure did occur
during this low voltage period though...I'm not sure if it's related or
not.

I have ordered an LM7812 and the appropriate capacitors to go with it
to try out (as well as a new sensor). Do you think I'll be all right
without a heat sink?

Don
Kansas City

 

[Baron]

Thanks for the help so far, guys.

The National spec. sheet (and my chip is a National) mentions reverse
polarity protection and short circuit protection, but not overtemp
protection, so I'm not sure if it was heat or not that made it fail.

The sensor is all that the output drives and it's a very low
load--maybe a few mA. The regulator is mounted perpendicular on a PC
board with no heat sink and the ambient is normal indoor room
temperature (maybe 75 F). It isn't enclosed in a box and gets good
ventillation. I remember touching the regulator several times to see
if it was red hot. It was hot, but not soldering-iron hot, like I've
seen some power transistors get before they blow. I had experienced
some low voltage problems on the 24 V supply due to a separate
problem,
but no lower than about 22 V or so. The regulator failure did occur
during this low voltage period though...I'm not sure if it's related
or not.

I have ordered an LM7812 and the appropriate capacitors to go with it
to try out (as well as a new sensor). Do you think I'll be all right
without a heat sink?

Don
Kansas City

Based on this and previous posts, its quite likely that the LM2940
simply took off ( Self Oscillation ). As others have pointed out, the
capacitors are essential. They also need to be as close as practicable
to the regulator and if there is more than a few inches of wire on the
output, another one near the load.

 

[David L. Jones]

Thanks for the help so far, guys.

The National spec. sheet (and my chip is a National) mentions reverse
polarity protection and short circuit protection, but not overtemp
protection, so I'm not sure if it was heat or not that made it fail.

The sensor is all that the output drives and it's a very low
load--maybe a few mA. The regulator is mounted perpendicular on a PC
board with no heat sink and the ambient is normal indoor room
temperature (maybe 75 F). It isn't enclosed in a box and gets good
ventillation. I remember touching the regulator several times to see
if it was red hot. It was hot, but not soldering-iron hot, like I've
seen some power transistors get before they blow. I had experienced
some low voltage problems on the 24 V supply due to a separate problem,
but no lower than about 22 V or so. The regulator failure did occur
during this low voltage period though...I'm not sure if it's related or
not.

I have ordered an LM7812 and the appropriate capacitors to go with it
to try out (as well as a new sensor). Do you think I'll be all right
without a heat sink?

If it's too hot to keep your finger on the metal tab then add a
heatsink.

From memory the TO-220 package is about 50degC/W, so if you are

dissipating say 0.5W or less you'll be fine without a heatsink. For a
24V input and 12V output that's an output current of about 40mA

Dave

 

[Tim Auton]

Thanks for the help so far, guys.

The National spec. sheet (and my chip is a National) mentions reverse
polarity protection and short circuit protection, but not overtemp
protection, so I'm not sure if it was heat or not that made it fail.

The Feb 2006 National datasheet I have here mentions thermal overload
protection, buried in the text at the start. The last sentence in the
paragraph at the top of the second column of the first page:

"Familiar regulator features such as short circuit and
thermal overload protection are also provided."

The sensor is all that the output drives and it's a very low
load--maybe a few mA. The regulator is mounted perpendicular on a PC
board with no heat sink and the ambient is normal indoor room
temperature (maybe 75 F). It isn't enclosed in a box and gets good
ventillation. I remember touching the regulator several times to see
if it was red hot. It was hot, but not soldering-iron hot, like I've
seen some power transistors get before they blow. I had experienced
some low voltage problems on the 24 V supply due to a separate problem,
but no lower than about 22 V or so. The regulator failure did occur
during this low voltage period though...I'm not sure if it's related or
not.

Anything from around 13V to 26V would be OK for the LM2940, so your 10%
dip in input voltage won't have been a problem. It's only rated for
brief pulses of over 26V input though - could your supply have gone much
higher than the nominal 24V, as well as lower? Anyway, that question is
academic now you're using the LM7812, which will be OK with anything
from around 15V to 35V input.

I have ordered an LM7812 and the appropriate capacitors to go with it
to try out (as well as a new sensor). Do you think I'll be all right
without a heat sink?

The LM7812 in a TO-220 has similar thermal specs to the LM2940 and will
be dissipating the same power, so if the LM2940 didn't burn your finger
the LM7812 should be fine.

If you want to do the calculations, you'll need to find out how much
current the sensor draws. Multiply that by the voltage drop across the
regulator (input - output, 12V in this case) to get the power the
regulator has to dissipate. Multiply that by the 'junction to ambient'
or 'junction to air' thermal resistance (65 C/W according to the
datasheet I have to hand). That'll give you the temperature rise, so add
worst-case ambient ( hot summer day... ) to get the junction
temperature. Specified absolute maximum junction temperature is 150 C,
so you want to be well short of that to allow for accumulated dust,
higher than normal input voltage etc. Anything less than, say, 110 C
would be conservative.


Tim

 

[[email protected]]

"Works best" is a mild way of saying "leave them off at your peril". I
stopped in the middle of build a circuit with an embedded 78L05
regulator; when powered up without an input cap it oscillated at 80MHz.

You should also check on power dissipation. These regulators are
supposed to have thermal shutdowns, but that doesn't mean they work
right, and you don't want a shutdown event anyway. Check the input
current to your 12V module; you'll be dissipating that times 12V (24V -
12V), and more if your 24V supply goes higher.

Make sure you have enough heat sinking to keep the regulator cool. The
best rule of thumb I know is to run it for a while then put your thumb
on it. If you pull back and say "ouch" then the heat sink isn't
dissipating enough.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Posting from Google? See http://cfaj.freeshell.org/google/

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html

I can't speak for all thermal shutdown schemes, but on the parts I
designed, the shutdown was not very intelligent. That is, it would
sense the die temperature, shut down, then cool due to the shutdown,
power up, and of course get hot again enough to trip the shutdown
circuit. You could make an oscillator using the shutdown. As part of
the characterization of the part, a diode on the chip would be
characterized over temperature, i.e. with the part in an oven and no
load so the die temp equals ambient. Then if you short the output, you
can monitor this diode to see the temperature trip points.

There may be smarter regulators that stay shut down once tripped. It is
a marketing issue since the smart design would need a power on reset
from the input voltage. Not impossible, but POR circuits can be fooled.

 

[[email protected]]

I converted this part of the machine's control to be read from a PLC

Low drop out regulators (of which the LM2940 is one) are
only stable with certain values of capacitance (and
sometimes only with certain values of series resistance in
that capacitance).
The graph on page 11 shows how tricky this is.
http://info.hobbyengineering.com/specs/NATSEMI-LM2940.pdf

Guess it's the 00882206 diagram "Output Capacitor ESR".

Makes me wonder if these LDO really are good. Seems more like liability

 

[John Popelish]

John Popelish wrote: (snip)

Guess it's the 00882206 diagram "Output Capacitor ESR".

Makes me wonder if these LDO really are good. Seems more like liability

I have used them successfully, but I always use a low esr
cap and an external series resistor.

 

[John - KD5YI]

Thanks for the help so far, guys.

The National spec. sheet (and my chip is a National) mentions reverse
polarity protection and short circuit protection, but not overtemp
protection, so I'm not sure if it was heat or not that made it fail.

The sensor is all that the output drives and it's a very low
load--maybe a few mA. The regulator is mounted perpendicular on a PC
board with no heat sink and the ambient is normal indoor room
temperature (maybe 75 F). It isn't enclosed in a box and gets good
ventillation. I remember touching the regulator several times to see
if it was red hot. It was hot, but not soldering-iron hot, like I've
seen some power transistors get before they blow. I had experienced
some low voltage problems on the 24 V supply due to a separate problem,
but no lower than about 22 V or so. The regulator failure did occur
during this low voltage period though...I'm not sure if it's related or
not.

I have ordered an LM7812 and the appropriate capacitors to go with it
to try out (as well as a new sensor). Do you think I'll be all right
without a heat sink?

Don
Kansas City

Since your load is only a few mills, I would suggest using a zener and
resistor to set the 12V. Of course, the zener would not be as accurate in
voltage as the regulator. But, the zener should be inherently safer from a
failure standpoint as they usually fail shorted.

If you insist on using a regulator, you could add some resistance between
the 24V and the regulator input. The capacitors mentioned by others should
be connected directly at the regulator pins. Then put a 13 or 14 V transient
suppressor on the regulator's output. If you have another failure, the clamp
should keep the voltage down to 13 or 14V.

Just a thought. Good luck.

John

 

[David L. Jones]

Guess it's the 00882206 diagram "Output Capacitor ESR".

Makes me wonder if these LDO really are good. Seems more like liability

They are essential when you actually need real LDO operation! e.g.
dropping say 5V to 4.5V with low noise.

Dave