Use of a LM317 with 5 volt dc 1.5 amp power supply

7/7/2009 4:36:17 PMBrian

I now have determined that my power supply is a linear type with a heavy transformer and filters to smooth out dc. Can someone help me determine how to include the LM317 for output of 2.2 volt and up to 1 amp to power my battery radio? Input voltage will be 5 dc.

7/7/2009 6:59:17 PMEdd

Sir Brian. . .

Found specs on the unit. . .ehhh . . .?

If it is putting out 5vdc at the required amperage or greater, there is a fault in wanting to use a LM317 in trying to acquiire that 2+ V filament supply.

Its called "head room" . . . seems that IC wants a cushion of MORE than that small margin of voltage difference between your 5VDC voltage input and that 2-2.2 VDC output.

Welcome to the world of solid state design and its commited parameters and paradigms.

For just such situations, they make what is called a low drop out regulator. . .the LMS8117-ADJ or the LM1084-ADJ which are drop in, direct replacements for the ultra familiar LM-317, but with their capabilities being within 1.5VDC of the Input - Output voltage. . .and with that '1084 even being rated at a hefty 5 amps of current capability, should a design require that.. . .a 1 amp spec on the other unit or for the standard LM317.

BUT and that's HEAV-EE on the BUT, How come I think that unit of yours might be what is referred to as an "open frame" power supply unit, with its power transformer, heat sink and PCB circuitry visible. In it, you may already HAVE all that you will be needing OF which, I am expecting that design as already being a fully regulated supply with its control electronics and a visible TO-3 type of pass transistor being mounted on the heat sink.
(Or some dedicated house numberred special IC ) ALONG with a pot some where ? on the unit, that is used to trim in the voltage output from the unit to approx~ +/- 10 % of that 5.00 vdc value.

Fill me in, if an examination does not confirm those elements being present.

That may be in the form of either a larger knobbed or end slotted short shaft pot, or probably in the form of a small 10 turn trim pot mounted on the edge of the PCB.

IF such is the case, it should be possible to measure the total resistance of the adjustment pot from its end to end and then figure on using a small fixed resistor ~25% of that value and then you lift an end wire from the pots 3 terminals and insert that resistor in series.

That will shift the adjustment parameters fed back to the voltage sampling of the regulator control circuitry and skew the degree of total adjustment range. That should then let you vary further from the original manner of adjustment values, to then be able to trim in the units regulated output level to your desired 2---2.2 v . . . .under load.

So now, you say, WHICH end terminal ? . . . .experiment. . . initially, while dumping into a 44 pilot lamp for a load.

If the voltage units Voltage out ends up being from 5-6.5 VDC, you guessed the wrong terminal end.

I expect the unit to be having a raw DC output from the rectifier /electrolytic filter bank in the range of 10 VDC on down to a minimal 7.5 VDC initially going into the regulator circuitry.

Thasssitt. . . . . questions ?

73's de Edd

7/7/2009 7:29:39 PMLewis L

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:Sir Brian. . .
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:Found specs on the unit. . .ehhh . . .?
:
:
:If it is putting out 5vdc at the required amperage or greater, there is a fault in wanting to use a LM317 in trying to acquiire that 2+ V filament supply.
:
: Its called "head room" . . . seems that IC wants a cushion of MORE than that small margin of voltage difference between your 5VDC voltage input and that 2-2.2 VDC output.
:
:
:Welcome to the world of solid state design and its commited parameters and paradigms.
:
:
:For just such situations, they make what is called a low drop out regulator. . .the LMS8117-ADJ or the LM1084-ADJ which are drop in, direct replacements for the ultra familiar LM-317, but with their capabilities being within 1.5VDC of the Input - Output voltage. . .and with that '1084 even being rated at a hefty 5 amps of current capability, should a design require that.. . .a 1 amp spec on the other unit or for the standard LM317.
:
:
:BUT and that's HEAV-EE on the BUT, How come I think that unit of yours might be what is referred to as an "open frame" power supply unit, with its power transformer, heat sink and PCB circuitry visible. :In it, you may already HAVE all that you will be needing OF which, I am expecting that design as already being a fully regulated supply with its control electronics and a visible TO-3 type of pass transistor being mounted on the heat sink.
:(Or some dedicated house numberred special IC ) ALONG with a pot some where ? on the unit, that is used to trim in the voltage output from the unit to approx~ +/- 10 % of that 5.00 vdc value.
:
:Fill me in, if an examination does not confirm those elements being present.
:
:
:That may be in the form of either a larger knobbed or end slotted short shaft pot, or probably in the form of a small 10 turn trim pot mounted on the edge of the PCB.
:
:
:IF such is the case, it should be possible to measure the total resistance of the adjustment pot from its end to end and then figure on using a small fixed resistor ~25% of that value and then you lift an end wire from the pots 3 terminals and insert that resistor in series.
:
:That will shift the adjustment parameters fed back to the voltage sampling of the regulator control circuitry and skew the degree of total adjustment range. That should then let you vary further from the original manner of adjustment values, to then be able to trim in the units regulated output level to your desired 2---2.2 v . . . .under load.
:
:
:So now, you say, WHICH end terminal ? . . . .experiment. . . initially, while dumping into a 44 pilot lamp for a load.
:
:If the voltage units Voltage out ends up being from 5-6.5 VDC, you guessed the wrong terminal end.
:
:I expect the unit to be having a raw DC output from the rectifier /electrolytic filter bank in the range of 10 VDC on down to a minimal 7.5 VDC initially going into the regulator circuitry.
:
:
:
:
:Thasssitt. . . . . questions ?
:
:
:
:
:
:73's de Edd
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:
:
:

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:
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Brian:
You also might could use a NPN transistor with 3 or 4 diodes in series forward biased in the base to ground, fed from the emitter with a 1k resistor.
Lewis

7/7/2009 7:30:19 PMLewis L

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:
:
:
:

:
:
:
:
:
:
:
:
:
:Sir Brian. . .
:
:
:Found specs on the unit. . .ehhh . . .?
:
:
:If it is putting out 5vdc at the required amperage or greater, there is a fault in wanting to use a LM317 in trying to acquiire that 2+ V filament supply.
:
: Its called "head room" . . . seems that IC wants a cushion of MORE than that small margin of voltage difference between your 5VDC voltage input and that 2-2.2 VDC output.
:
:
:Welcome to the world of solid state design and its commited parameters and paradigms.
:
:
:For just such situations, they make what is called a low drop out regulator. . .the LMS8117-ADJ or the LM1084-ADJ which are drop in, direct replacements for the ultra familiar LM-317, but with their capabilities being within 1.5VDC of the Input - Output voltage. . .and with that '1084 even being rated at a hefty 5 amps of current capability, should a design require that.. . .a 1 amp spec on the other unit or for the standard LM317.
:
:
:BUT and that's HEAV-EE on the BUT, How come I think that unit of yours might be what is referred to as an "open frame" power supply unit, with its power transformer, heat sink and PCB circuitry visible. :In it, you may already HAVE all that you will be needing OF which, I am expecting that design as already being a fully regulated supply with its control electronics and a visible TO-3 type of pass transistor being mounted on the heat sink.
:(Or some dedicated house numberred special IC ) ALONG with a pot some where ? on the unit, that is used to trim in the voltage output from the unit to approx~ +/- 10 % of that 5.00 vdc value.
:
:Fill me in, if an examination does not confirm those elements being present.
:
:
:That may be in the form of either a larger knobbed or end slotted short shaft pot, or probably in the form of a small 10 turn trim pot mounted on the edge of the PCB.
:
:
:IF such is the case, it should be possible to measure the total resistance of the adjustment pot from its end to end and then figure on using a small fixed resistor ~25% of that value and then you lift an end wire from the pots 3 terminals and insert that resistor in series.
:
:That will shift the adjustment parameters fed back to the voltage sampling of the regulator control circuitry and skew the degree of total adjustment range. That should then let you vary further from the original manner of adjustment values, to then be able to trim in the units regulated output level to your desired 2---2.2 v . . . .under load.
:
:
:So now, you say, WHICH end terminal ? . . . .experiment. . . initially, while dumping into a 44 pilot lamp for a load.
:
:If the voltage units Voltage out ends up being from 5-6.5 VDC, you guessed the wrong terminal end.
:
:I expect the unit to be having a raw DC output from the rectifier /electrolytic filter bank in the range of 10 VDC on down to a minimal 7.5 VDC initially going into the regulator circuitry.
:
:
:
:
:Thasssitt. . . . . questions ?
:
:
:
:
:
:73's de Edd
:
:
:

:
:
:
:

:
:
:
:
:
Brian:
You also might could use a NPN transistor with 3 or 4 diodes in series forward biased in the base to ground, fed from the emitter with a 1k resistor.
Lewis