Voltage Controlled Voltage Amplifier/ Mr. Al

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simf14
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Voltage Controlled Voltage Amplifier/ Mr. Al

Post by simf14 »

Mr. Al you recently spent an tremendous effort helping me get this circuit working exceptionally. I have a problem now in actual usage which maybe you could offer a few suggestions. The circuit powers a 12V, 30W, max 3A hydraulic solenoid. The 338K and large heatsink gets very "hot" (too hot to touch, burning smell) in only a few minutes of usage. I checked and only 1.5A at 5V is being outputted. Any suggestions would as always be so appreciated. I am sorry if this is not the appropriate way to contact you also. http://upload2.postimage.org/74276/photo_hosting.html
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Chris Smith
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Post by Chris Smith »

Regulators can boil water and still function.

Their heat rating is quite high.

[150 C / 302 F and they do have thermal shut downs]

Do you have heat sinks on the reg?
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Post by simf14 »

Hi Chris- nice big one, this one really does boil...it is so far below it's rated cap (-3X) I can't believe it should get that hot, I haven't be daring enough to see if it hits thermal shutdown but I believe it would.
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Chris Smith
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Post by Chris Smith »

You do have a huge heat sink on it.

If so, and you have a thermometer, then try and watch the value directly at the reg plastic.

Also a drop of water on top can tell a story. [match stick and drop]

It is incredible but in some circumstances they will dissipate a lot of heat, and they can burn.

I don’t like to do things that way especially if it has a heat sink, but I would start there first?

I have run a few things with a digital temp reading on top, and it ran real hot but way below the specs.

It makes me a little queasy but ...?

My idea of normal is 50 to 75c or 120 to 170 F but it can go higher.
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Post by simf14 »

A little note is if I change the circuit input voltage from 12V to 9V, the 338 doesn't even get warm...Probably takes most of the work away from the 338(?)
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MrAl
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Post by MrAl »

Hi again simf,

Yes, dropping the voltage reduces the power dissipation which allows
the LM device to run cooler.

Here are the results of a complete thermal study on the LM338 and related
LM138 and in two packages, the TO-220 and the TO-3, and operated
at your input voltages of 10 volts to 12 volts and your output load
resistance calculated to be 8/3 ohms.....................................


The following assumes a heat sink with a rating of 2 deg C/W
(which is a fairly large heat sink some 4 inches across
with enough fins...check your heat sink specs)
and decent thermal paste and the device is mounted properly
with enough pressure by torquing the mounting screw(s).

Temperature for both packages (deg C) at
a given Vout and load resistance of 2.667 ohms...

Vin=12 volts constant
-----------------------------------
Vout=1.00, TO220= 74, TO3= 62
Vout=1.50, TO220= 85, TO3= 67
Vout=2.00, TO220= 95, TO3= 72
Vout=2.50, TO220=103, TO3= 76
Vout=3.00, TO220=110, TO3= 80
Vout=3.50, TO220=116, TO3= 83
Vout=4.00, TO220=122, TO3= 86
Vout=4.50, TO220=125, TO3= 87
Vout=5.00, TO220=128, TO3= 89
Vout=5.50, TO220=130, TO3= 90
Vout=6.00, TO220=131, TO3= 90
Vout=6.50, TO220=130, TO3= 90
Vout=7.00, TO220=128, TO3= 89
Vout=7.50, TO220=125, TO3= 87
Vout=8.00, TO220=122, TO3= 86

TO220 max=131, TO3 max=90

The LM338 device max T is 125, so
this TO220 max temperature (131) exceeds
the rating of the LM338 device.
The TO3 package is within spec because
it's max temperature (90) is less than
125 deg C.

Note also that the LM138 device is spec'd
up to 150 deg C, so that device should work
here too, but the package/device that does not
work here is the LM338 device in TO220 package
because there will be at least one operating
point where it's temperature rating is exceeded.

Possibilities:
1. Switch to a higher rated package, the TO-3 package.
2. Switch to a higher rated device, the LM138.
3. Both of the above.
4. Add a fan for forced air cooling.
5. Lower the input voltage.

Since solutions 1 through 4 require purchasing a new product, lets look
at solution #5 .

Lowering the input voltage (as you have found out) lowers the
total package dissipation for any output voltage. The lower
limit is set by the drop out spec for the regulator chip.
For this chip, the dropout is 2 volts. This means that in order
to make sure you can get the full 8 volts output, you need to
supply at least 10 volts input. If you use a tracking pre-regulator
you can get around this, but i am assuming you dont want that
additional complexity.

Ok, so lets look at the temperature rise with only 10v input now...

Vin=10 volts constant
-----------------------------------
Vout=1.00, TO220= 70, TO3= 60
Vout=1.50, TO220= 78, TO3= 64
Vout=2.00, TO220= 86, TO3= 68
Vout=2.50, TO220= 92, TO3= 71
Vout=3.00, TO220= 97, TO3= 73
Vout=3.50, TO220=101, TO3= 75
Vout=4.00, TO220=104, TO3= 77
Vout=4.50, TO220=105, TO3= 77
Vout=5.00, TO220=106, TO3= 78
Vout=5.50, TO220=105, TO3= 77
Vout=6.00, TO220=104, TO3= 77
Vout=6.50, TO220=101, TO3= 75
Vout=7.00, TO220= 97, TO3= 73
Vout=7.50, TO220= 92, TO3= 71
Vout=8.00, TO220= 86, TO3= 68

TO220 max=106, TO3 max=78

As we can see from the above, now the max is 106 which is under
the 125 deg C spec.

Note actual temperatures will vary a bit with the heat sink specs.
Also, note that if the voltage is lowered to 10 volts, it must
be fairly constant and not subject to dip with the higher current
of 3 amps. If for some reason the supply must be 12 volts, you
can add one or two diodes on the input side (perhaps with a little
additional capacitance after the diodes) to drop some of the input
voltage. For example, if you start with 12v and subtract two diode
drops of 0.8 volts each, you'll be down to 10.4 volts on the input.
The diodes have to be rated for at least 3 amps but prefer higher
rated devices for better cooling like 5 amp devices, and they
can be the typical rectifier type diodes. You might try one first,
then a second one if the temperature doesnt drop enough.

As a final note, when the device overheats due to operation out of
it's comfort zone the internal circuitry starts to cut back the
output current. This is probably why you saw only 5 volts on
the output with the 12v input supply noting that the temperature for
a 5v output is only 3 degrees away from being at the max for this setup.
Once the thermal issues have been adequately managed, this problem
should permanently go away.
LEDs vs Bulbs, LEDs are winning.
simf14
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Post by simf14 »

Dear Mr. Al

I am using a smaller heat sink for the 338K TO-3 that is diamond shape with slotted vertical fins and about 2in long 1in wide and 2 in high so it would not take too much space on the board and it was the most common one I found. I will pick up a digital thermometer tonight but I know it is running at least twice the theoretical temp of 86 (maybe three times):

Temperature (deg C) at
at Vout and load resistance of 2.667 ohms...

Vin=12 volts constant
-----------------------------------
Vout=8.00, TO3= 86

I have built 7 boards and they are all running the same so I don't think I have a short or mis-build. I believe that maybe the load is much higher or something else is up. Ultimately switching to a common 9V power supply will work and still get enough max output for the application. I will also look for a 9V 3-5A regulated power supply today. I would still like to find out why the 12V actual temps are so much higher than the theoretical and will have some hard data tonight once I hit Radio Shack : )
Bye the way, this is the first electronic project I ever made...dual layer, woohoo! Hope there isn't too may mistakes!!! (grounding bolts to be added on this one)
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Post by Robert Reed »

Sim
just one look at your heat sink and I can tell it is too small for your application. When you get your thermometer readings and I would assume you will be measuring case temperatures, you must factor in junction to case thermal resistance by working backwards from your readings to the actual junction temperature. Use your readings along with the manu's spec sheet to do this. Most regulators will come in a 1 or 2 or 3 prefix. These generally differ only in the rated junction temperature that that they will perform satisfactorily in. The one prefix is usually Mil-Spec and will have a max. junction temp. of 150 degrees C. Your chip is a consumer grade with a max. junction of 125 degrees C. you want to leave a comfortable margin here of 25 degrees, so that the max. junction temperature of your device will not exeed 100 degees C at its junction. This will mean that your measured case temperature will show a much cooler temperature than what the actual junction temperature actually is. But by working backwards with the spec sheet, you will obtain a fairly accurate number for that junction temperature.
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Post by philba »

Al's posting said this but I thought it would be good to have a simple rule of thumb to go along with it.

The reason 9V reduces the heat is because the difference between the regulators output voltage (Vout) and input voltage (Vin) times the current is dissipated as heat. So (Vin-Vout)*Iout is the wattage. for 12V at 1.5A it's 4*1.5 or 6 Watts vs 1*1.5 or 1.5W. 1/4th the heat generated. However, I think the min dropout is higher than 1V so in practive it's probably more than 1.5W.

In general, try to set up Vin to be just a bit higher than the dropout voltage of the linear regulator. This minimizes the heat generation.

On a somewhat different note, do you even need to regulate the voltage to the solenoid? I didn't recall the previous thread but I'd just use a MOSFET and PWM to control the average (unregulated) voltage to the coil. A more sophisticated solution is to use current sensing and a comparator to shut off the MOSFET when the current rises too high - basically how chopper drivers for stepper motors work. Your solenoid can probably handle a wide range of voltages.
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Post by simf14 »

thanks Philba and Robert...will pickup a larger heat sink today and then take some readings- very informative. I am committed to this approch now but I know PWM is used freqently too.
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Post by Edd »

Wow....an initial six "proto" boards comitted and already having been carved in stone.

Looks like that with your potential required dissipation level using that higher raw DC voltage input, and being the utilization a linear regulator design, that the minimal heatsinking might just be the size of adjunct heatsink with its healthy fins that I have shown placed beside your board. Its overall sizings footpring being about the size of your whole board but mounted on the backside. Healthy jumpers completing the I-G-O interconnects. The fins also oriented for optimal natural "thermal upflow" of air...unless further on hands evaluation shows that you might need an additional top heatsink mounted 12 VDC computer / brushless fan for suppplemental forced airflow.


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simf14
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Post by simf14 »

I actually built one protoboard with a breadboard and then a second hardwired board. Tested both under actual conditions but unfortunately not long enough!

Anyhow did the heat test tonight. Added a 3 inch heatsink, the largest I could find, attached it to the LM338 with heatsink compound and solid bolts. With input voltage 12v, output volt 7.87 at 1.34A, ambient temp 85 degrees the temp stabilized at 140 degrees F. measured with a new digital therm after 10 minutes on the top of the case. This is way off the theoreticals. Also it might even be above the shutoff temp. (?) Love to know what I did wrong. :shock:
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Chris Smith
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Post by Chris Smith »

85 is OK, 140 is not, especially if yours is the lower 125 c.

Especially Not being the plastic model, with T03 metal and large heat sink, no.

Two things you can do, one is turn off the power to the trans then check it with the digital probe, the second test is the drop of water.

Some digitals will get interference from the body of trans when they are in operation and then hot, so just pop off the power before the test.

Also water boiling says it all.
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Post by Robert Reed »

Sim
I have not seen your circuit and I am not sure exactly what you want to out put, but I am gathering you want 8.0 volts. You said you read 7.87 volts at your out put and that is only in error of 2% - not bad as this could be further trimmed by the adj. pin components. with the current- voltage levels you gave, you are dissipating 5.5 watts across your juntion.You said your case temperature was 140 degrees F -which is 60 degrees C. I looked up the specs on your device and it shows only a 1 degree junction to case rise per watt of dissapation and I was surprised to see such a low thermal resistance at that point. Now lets throw in approx. 0.5 degree per watt rise for the case to heat sink junction. I will have to assume this value since I don't know your particulars here. Total these two up and we have a 1.5 degree rise per watt of dissipation times 5.5 watts equals slightly over 8 degrees C rise from junction to case. Now since we are working backwards from your case measurment of 60 degrees C, we have to add that 8 degrees to it to arrive at the junction temperature and that gives us about 68 degrees C. This is well within the regulators limit of 125 degrees. So far you look good. What are you looking for beyond this. I hope i got this right as it is bedtime and I am very sleepy and usually prone to mistakes at this point. I will go over this in the morning to confirm,but thats my thinking for now.
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Chris Smith
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Post by Chris Smith »

F is fine
C is not

You said You went over?

Also it might even be above the shutoff temp.

That would be C, not F.
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