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Old 16th April 2012, 10:16 PM   #1
MelB is offline MelB  Canada
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Default LT1083 interesting startup problem

I've never had a linear regulator simply not start into a load. Here are two circuits. Top circuit has a bypass cap (C3) 470uF and the bottom circuit has a bypass cap (C3) of 47uF. The top circuit will NOT start a GM70. It puts out about 2 volts although it is set to put out 19.8V. It will put the 19.8 volts into a 6.8 ohm resistor, but cold the GM70 is only 1.6 ohms. The bottom circuit starts up like a champ. I was just using caps I had lying around. Thought I would share.
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Old 17th April 2012, 12:20 AM   #2
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Quote:
but cold the GM70 is only 1.6 ohms.
Will it start with a 1.6ohm resistor?

I'd guess you're tripping the built-in short circuit protection.
(I'm assuming this reg has it built in,along with thermal,etc protections)
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Old 17th April 2012, 12:53 AM   #3
Arius is offline Arius  United States
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Your issue is with C3. As you stated yourself, it works when C3=47uF and not when C3=470uF. The ADJ cap is not supposed to be any larger than 25uF when R1=120R. With a 120R, your ADJ current is only 100uA.

See the datasheet for more info.
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Old 17th April 2012, 01:27 PM   #4
MelB is offline MelB  Canada
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Ya I read that in the data sheet, but I took that to mean @ 120Hz your Cadj should be at least 25uF. Earlier in the data sheet it says if you use 150uF aluminum on the output it will cover ALL CASES of bypassing the adjustment pin. Should say almost all cases. But starting into a tube filament is probably a special case. You could never run this thing with 1.6 ohms continually at 20V. That would be 12.5 amps. I suppose I got a tad carried away with the 470uF cap, which can only charge as you said with the 100uA adjust current.
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Old 25th April 2015, 11:43 AM   #5
Wolff is offline Wolff  Poland
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I have related problem in this topic:
LM3886 works in a few seconds then it shut off.
Maybe someone from here willl help me...

In addition the Voltage on reference resistor R1 is 0.3V instead of 1,25, and on R2 is like on output ~ 3,5V. If i disconnect LM the Supply recovers to 22V.
I think is something with Overload Recovery?
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Old 25th April 2015, 12:52 PM   #6
MelB is offline MelB  Canada
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From Linear: The circuit design used in the LT1083 family requires the
use of an output capacitor as part of the device frequency
compensation. For all operating conditions, the addition of
a 150μF aluminium electrolytic or a 22μF solid tantalum
on the output will ensure stability. Normally, capacitors
much smaller than this can be used with the LT1083. Many
different types of capacitors with widely varying characteristics
are available. These capacitors differ in capacitor
tolerance (sometimes ranging up to 100%), equivalent
series resistance, and capacitance temperature coeffi cient.
The 150μF or 22μF values given will ensure stability.

If you only have 10uF on the output try 150uF as recommended. It says "normally" smaller caps can be used but I would think 10uF Aluminum would be right at the margin.

Edit: I may have mis-understood. What value do you have on the output?

Last edited by MelB; 25th April 2015 at 12:56 PM. Reason: Ooops
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Old 25th April 2015, 01:48 PM   #7
Wolff is offline Wolff  Poland
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Yes 10uF is on the adjust pin.
On the output I have 470uF at this moment...
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Old 25th April 2015, 03:05 PM   #8
Wolff is offline Wolff  Poland
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And I have after aprox 10cm of wire of course 1000uF + 100nF on LM chip pins.
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Old 25th April 2015, 03:25 PM   #9
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For a filament regulator you don't need ANY capacitor on the ADJ pin as the regulator noise is superfluous and the ripple voltage is so small.

For any other application, the capacitor on the adjust pin reduces regulator noise, but also impairs transient response.

Lastly -- with 470u on the adjust pin you've made a nice slow-start regulator.
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Old 25th April 2015, 04:38 PM   #10
Wolff is offline Wolff  Poland
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Quote:
Originally Posted by jackinnj View Post
For a filament regulator you don't need ANY capacitor on the ADJ pin as the regulator noise is superfluous and the ripple voltage is so small.

For any other application, the capacitor on the adjust pin reduces regulator noise, but also impairs transient response.

Lastly -- with 470u on the adjust pin you've made a nice slow-start regulator.

I removed the 10uF capacitor from the adjust pin and its working!
Nice graph I understand it now. Should I leave it without the capacitor or insert there 1uF - supposly will work also, what do you recomend?

Thanks a lot !
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