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Question regarding 78xx series devices

Could some electronics guru please clarify thhe following ? For the 78xx series voltage regulators, does the ground pin have to be ground(zero) always ? To be more specific, I have successfully used the following setup: Step down(9 - 0 -9) xfrmr -> bridge rectifier -> positive output of bridge rectifier connected to pin 1 of a 7805 and the negative output of the bridgerectifier connected to the ground pin of the 7805. I get 4.99 volts (filtered with

8 capacitors) at the output pin of the 7805. No problems at all. If I try the same setup with a 9 - 0 - 9 output transformer, and the positive and negative ouput pins of the bridge rectifier connected respectively to the positive and ground pins of a 7815, I get zero volts output. I was wondering if my set up is incorrect, i.e., does the ground pin always need to be connected to 0 Volts ? All hints/suggestions are welcome. Thanks in advance.
Reply to
dakupoto
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I'm not entirely sure what you're saying here. You mean that you have identical schematics, except that one has a 7805 and the other has a 7815?

If so, and if your input voltage is high enough, you should be getting

15V from the 15V version.

If you're not, and you bought your 7815s off of eBay or other dodgy supplier, I'd suspect the chip. If you bought them from DigiKey or some other reliable place, double check that they're not 7915s.

Beyond that -- I dunno.

--

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

Sounds like you're measuring between the output pin and 'ground', instead of between the output and ground pins of the regulator. If so, your meter isn't connected across the actual power supply output--it's measuring an open circuit.

There's nothing too special about a connection to the Earth, but current does have to travel in closed loops.

Cheers

Phil Hobbs

Reply to
Phil Hobbs

Yes it has to be connected to 0 volts, in the last case your

0 volts is the center tap of the transformer, not the (-)side of the bridge. In a Setup like that, it is customary to use a 7905 for the (-) side of the bridge and 7805 for the (+) side of the bridge (inputs to the regs). Both polarities need to connect to 0 volts, the center tap of the transformer.

It's the convenience of a bridge that makes it an atractive use, instead of using 4 diodes; two for each polarity in the case of a split phase.

Jamie

Reply to
M Philbrook

I don't quite understand your setup, but the ground pin of a 78xx series regulator can be tied to any potential that can SINK the ground pin current (read the data sheet).

That "any potential" becomes the reference point... the "output" will be the regulator voltage PLUS the reference. ...Jim Thompson

--
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Reply to
Jim Thompson

exactly. Whether the 78's reference point is 0v according to input or output circuits is of no import. So you can easily use a 7905 to get +5 - as long as the input isn't already grounded.

NT

Reply to
tabbypurr

I often stack regulators. An LM1117 can make 1.2 volts for an FPGA core, and you can stack a second one on top to make 2.5 volts for i/o. No resistors.

Reply to
John Larkin

Works great as long as the load on the lower regulator never goes away even in a fault condition. A resistor to ground from the lower regulator, sized to take the worst-case ground current of the upper one, is comforting in cases like that.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Sure, but an FPGA core will gobble tons of milliamps without complaining, even before it's clocked.

Reply to
John Larkin

Or you can have zener type overvoltage protection to handle the current, if you can handle the voltage shift in such circumstances.

NT

Reply to
tabbypurr

Assuming it's actually connected. At 10 cents per kWH, a resistor is cheap insurance. ;)

Cheers

Phil Hobbs

Reply to
Phil Hobbs

p insurance. ;)

not so cheap on batteries. In any case your R has to pass the whole max upp er load i minus minimum lower load i - all the time. IOW the stacked reg ci rcuit is fine when that scenario doesn't happen, but wooden blades & waste can get overinvolved if/when it does, if you don't have the relevant protec tion.

Zenering gives a regulation hit, but no power waste. Sometimes it can be ac ceptable.

NT

Reply to
tabbypurr

I fully agree with and understand what you say, and precisely why I am confused. For the case of the 7805, the negative output of the bridge rectifier is the reference point, and when an identical configuration is applied to the 7815, the output is 0 Volts.

Reply to
dakupoto

Please check Jim Thompson's post. The ground pin has to be connected to any node that can sink current. That is precisely why the 7805, with its center pin(GND) connected to the bridge rectifier negative terminal works fine, producing a steady 4.99 Volts at its output pin. An identical configuration with the 7815 does not work. Very odd.

Reply to
dakupoto 

--
If you'd just post a schematic... 

JF
Reply to
John Fields

Dead 7815? Or is input

Reply to
Jim Thompson

+1, it would ave a lot of guess work.

GH

Reply to
George Herold

No it doesn't, it just has to prevent an unloaded lower supply from being dragged upwards by the bias current of the upper regulator. Think

5 mA or thereabouts for a 7805. Some LDOs dump a lot more current out their ground pins when they get near their dropout limit, due to the PNP pass element starting to saturate.

If that extra 5 mA were really limiting battery life, then one wouldn't be using an LDO there in the first place.

A shunt regulator such as a TLV431 hung on the lower regulator would be another approach--it wouldn't draw any current unless the voltage got too high.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

I can only think we're talking about 2 different circuits

yes, the zener approach

NT

Reply to
tabbypurr

_________ Vin | |

0----*---| 7805 |--------------0 +10V | |_________| | | | | | *------------* | _________ | | | | | *---| 7805 |------*-*-----0 +5V |_________| | | R | R < 1k GND R | GND

Arguably one should put a resistor across the top regulator as well, just to make sure its minimum load current spec is met. 78xx parts work okay at zero load, not every regulator does, I think.

There are other problems with this circuit as well, e.g. the tendency to blow up the top regulator if you short its output.

What this country needs is a good 1.2V zener. ;)

The other nice thing about the TLV431 is that you can put a resistor in series with its anode to keep it from fighting the regulator too hard and blowing up, without hurting the accuracy much.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

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