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Discussion Groups | Piclist | 2 x 7805 VS 7805-7905
A discussion group for the PICMicro microcontroller. Also called the Microchip PIC, this list is dedicated to the use and abuse of this fine, simple, microcontroller. Close to topic posts are welcome, ie. general electronics.
2 x 7805 VS 7805-7905 - Trikoupis Nikolaos - May 7 12:58:31 2006
Hi to everyone!!
Although it is not really a pic issue I am going to describe what
happened to me. If anyone knows why I would be really grateful to
hear about.
During one of my projects I used 2 x 7805 for dual +-5v supply.
Each one of them had was supplied by a 9V battery(that means 2
seperated batteries).
I connected the output of the second 7805 to the ground of the first
one.
As you can imagine I had both +5 and -5 supply. The circuit was
working with, let say, resistors,etc.
But when I used an op amp the supply circuit didn't work.
I had ground +5V and instead of -5V there was something like -1 V.
A lot of current was drawn through the second 7805. Like a short-
circuit was achieved.
Why do we really need the 7905 negative regulator?
What is happening when using 2x7805 with some circuitry?
I can't find an answer out there. Hope some guy knows the answer.
to unsubscribe, go to http://www.yahoogroups.com and follow the instructions
(You need to be a member of piclist -- send a blank email to piclist-subscribe@yahoogroups.com )
Although it is not really a pic issue I am going to describe what
happened to me. If anyone knows why I would be really grateful to
hear about.
During one of my projects I used 2 x 7805 for dual +-5v supply.
Each one of them had was supplied by a 9V battery(that means 2
seperated batteries).
I connected the output of the second 7805 to the ground of the first
one.
As you can imagine I had both +5 and -5 supply. The circuit was
working with, let say, resistors,etc.
But when I used an op amp the supply circuit didn't work.
I had ground +5V and instead of -5V there was something like -1 V.
A lot of current was drawn through the second 7805. Like a short-
circuit was achieved.
Why do we really need the 7905 negative regulator?
What is happening when using 2x7805 with some circuitry?
I can't find an answer out there. Hope some guy knows the answer.
to unsubscribe, go to http://www.yahoogroups.com and follow the instructions
(You need to be a member of piclist -- send a blank email to piclist-subscribe@yahoogroups.com )
Re: 2 x 7805 VS 7805-7905 - rtstofer - May 7 13:54:56 2006
--- In p...@yahoogroups.com, "Trikoupis Nikolaos" wrote:
>
> Hi to everyone!!
>=20
> Although it is not really a pic issue I am going to describe what=20
> happened to me. If anyone knows why I would be really grateful to=20
> hear about.
>=20
> During one of my projects I used 2 x 7805 for dual +-5v supply.
>=20
> Each one of them had was supplied by a 9V battery(that means 2=20
> seperated batteries).
>=20
> I connected the output of the second 7805 to the ground of the first=20
> one.
>=20=20
> As you can imagine I had both +5 and -5 supply. The circuit was=20
> working with, let say, resistors,etc.=20
>=20
> But when I used an op amp the supply circuit didn't work.
>=20
> I had ground +5V and instead of -5V there was something like -1 V.=20
> A lot of current was drawn through the second 7805. Like a short-
> circuit was achieved.
>=20
> Why do we really need the 7905 negative regulator?
Because you can't connect a 7805 to produce a negative output relative
to ground.
> What is happening when using 2x7805 with some circuitry?
The regulator controls the output voltage relative to 'ground' but the
series pass transsistor that controls current flow is connected
between input and output. There is no ability to flow large currents
out of the ground pin. And, if that transistor conducts between input
and ground as you have it connected, it will try to short the battery.
Bottom line: if two 7805s could be connected as you hoped, they
wouldn't have bothered to design the 7905.
>=20
> I can't find an answer out there. Hope some guy knows the answer.
>
Richard
=20
to unsubscribe, go to http://www.yahoogroups.com and follow the instruction=
s=20
=20
(You need to be a member of piclist -- send a blank email to piclist-subscribe@yahoogroups.com )
>
> Hi to everyone!!
>=20
> Although it is not really a pic issue I am going to describe what=20
> happened to me. If anyone knows why I would be really grateful to=20
> hear about.
>=20
> During one of my projects I used 2 x 7805 for dual +-5v supply.
>=20
> Each one of them had was supplied by a 9V battery(that means 2=20
> seperated batteries).
>=20
> I connected the output of the second 7805 to the ground of the first=20
> one.
>=20=20
> As you can imagine I had both +5 and -5 supply. The circuit was=20
> working with, let say, resistors,etc.=20
>=20
> But when I used an op amp the supply circuit didn't work.
>=20
> I had ground +5V and instead of -5V there was something like -1 V.=20
> A lot of current was drawn through the second 7805. Like a short-
> circuit was achieved.
>=20
> Why do we really need the 7905 negative regulator?
Because you can't connect a 7805 to produce a negative output relative
to ground.
> What is happening when using 2x7805 with some circuitry?
The regulator controls the output voltage relative to 'ground' but the
series pass transsistor that controls current flow is connected
between input and output. There is no ability to flow large currents
out of the ground pin. And, if that transistor conducts between input
and ground as you have it connected, it will try to short the battery.
Bottom line: if two 7805s could be connected as you hoped, they
wouldn't have bothered to design the 7905.
>=20
> I can't find an answer out there. Hope some guy knows the answer.
>
Richard
=20
to unsubscribe, go to http://www.yahoogroups.com and follow the instruction=
s=20
=20
(You need to be a member of piclist -- send a blank email to piclist-subscribe@yahoogroups.com )
Re: Re: 2 x 7805 VS 7805-7905 - Phil Seakins - May 7 22:11:48 2006
rtstofer, the 7805 provides a regulated 5V output across its two pins=20
regardless of which one is grounded. Whether the internal series=20
regulator transistor is placed behind its +ve pin or -ve pin is=20
irrelevant as it is all a simple series circuit.
The 7905 was designed to complement the 7805 in a more conventional=20
circuit arrangement with a common ground rail (not necessarily earthed).
While Trikoupis's circuit configuration is unusual the theory behind=20
it is sound. It should work.
Trikoupis, there is something else happening in your circuit that you=20
have overlooked. I would suggest that you recheck the wiring. You=20
need to take care that the +ve pin of the -ve battery is connected=20
nowhere but to the regulator input pin. You are using correct value=20
bypass capacitors individually around each regulator, aren't you?
At 03:54 AM 8/05/2006, rtstofer wrote:
>--- In p...@yahoogroups.com, "Trikoupis Nikolaos" wrote=
:
>>
>> Hi to everyone!!
>>
>> Although it is not really a pic issue I am going to describe what
>> happened to me. If anyone knows why I would be really grateful to
>> hear about.
>>
>> During one of my projects I used 2 x 7805 for dual +-5v supply.
>>
>> Each one of them had was supplied by a 9V battery(that means 2
>> seperated batteries).
>>
>> I connected the output of the second 7805 to the ground of the first
>> one.
>>
>> As you can imagine I had both +5 and -5 supply. The circuit was
>> working with, let say, resistors,etc.
>>
>> But when I used an op amp the supply circuit didn't work.
>>
>> I had ground +5V and instead of -5V there was something like -1 V.
>> A lot of current was drawn through the second 7805. Like a short-
>> circuit was achieved.
>>
>> Why do we really need the 7905 negative regulator?
>
>Because you can't connect a 7805 to produce a negative output relative
>to ground.
>
>> What is happening when using 2x7805 with some circuitry?
>
>The regulator controls the output voltage relative to 'ground' but the
>series pass transsistor that controls current flow is connected
>between input and output. There is no ability to flow large currents
>out of the ground pin. And, if that transistor conducts between input
>and ground as you have it connected, it will try to short the battery.
>
>Bottom line: if two 7805s could be connected as you hoped, they
>wouldn't have bothered to design the 7905.
>
>>
>> I can't find an answer out there. Hope some guy knows the answer.
>>
>
>Richard
>
>
>
>
>
>
>
>
>
>to unsubscribe, go to http://www.yahoogroups.com and follow the instructi=
ons
>
=20
(You need to be a member of piclist -- send a blank email to piclist-subscribe@yahoogroups.com )
regardless of which one is grounded. Whether the internal series=20
regulator transistor is placed behind its +ve pin or -ve pin is=20
irrelevant as it is all a simple series circuit.
The 7905 was designed to complement the 7805 in a more conventional=20
circuit arrangement with a common ground rail (not necessarily earthed).
While Trikoupis's circuit configuration is unusual the theory behind=20
it is sound. It should work.
Trikoupis, there is something else happening in your circuit that you=20
have overlooked. I would suggest that you recheck the wiring. You=20
need to take care that the +ve pin of the -ve battery is connected=20
nowhere but to the regulator input pin. You are using correct value=20
bypass capacitors individually around each regulator, aren't you?
At 03:54 AM 8/05/2006, rtstofer wrote:
>--- In p...@yahoogroups.com, "Trikoupis Nikolaos"
:
>>
>> Hi to everyone!!
>>
>> Although it is not really a pic issue I am going to describe what
>> happened to me. If anyone knows why I would be really grateful to
>> hear about.
>>
>> During one of my projects I used 2 x 7805 for dual +-5v supply.
>>
>> Each one of them had was supplied by a 9V battery(that means 2
>> seperated batteries).
>>
>> I connected the output of the second 7805 to the ground of the first
>> one.
>>
>> As you can imagine I had both +5 and -5 supply. The circuit was
>> working with, let say, resistors,etc.
>>
>> But when I used an op amp the supply circuit didn't work.
>>
>> I had ground +5V and instead of -5V there was something like -1 V.
>> A lot of current was drawn through the second 7805. Like a short-
>> circuit was achieved.
>>
>> Why do we really need the 7905 negative regulator?
>
>Because you can't connect a 7805 to produce a negative output relative
>to ground.
>
>> What is happening when using 2x7805 with some circuitry?
>
>The regulator controls the output voltage relative to 'ground' but the
>series pass transsistor that controls current flow is connected
>between input and output. There is no ability to flow large currents
>out of the ground pin. And, if that transistor conducts between input
>and ground as you have it connected, it will try to short the battery.
>
>Bottom line: if two 7805s could be connected as you hoped, they
>wouldn't have bothered to design the 7905.
>
>>
>> I can't find an answer out there. Hope some guy knows the answer.
>>
>
>Richard
>
>
>
>
>
>
>
>
>
>to unsubscribe, go to http://www.yahoogroups.com and follow the instructi=
ons
>
=20
(You need to be a member of piclist -- send a blank email to piclist-subscribe@yahoogroups.com )
Re: Re: 2 x 7805 VS 7805-7905 - "Roy E. Burrage" - May 7 22:54:44 2006
He also needs to be sure that the 2 supplies are "stacked", i.e. the
lower 7805 will have its ground pin at the negative rail and its output
driving the circuit common. The upper 7805 will have its ground pin at
the circuit return and its output driving the positive supply rail. The
circuit used to be available in a National Semiconductor application
note, perhaps the 78xx part datasheet, if you need to see a schematic.
Be sure that total supply current, both positive and negative, does not
exceed the rating of the 7805 as all circuit current will flow through
the lower 7805. Heat sinking will probably be necessary.
This circuit works quite well and we have used it many times here to
keep from having to maintain stock of the 2 parts as well as the 78xx
parts being less expensive and more readily available than the 79xx parts.
REB
Phil Seakins wrote:
>rtstofer, the 7805 provides a regulated 5V output across its two pins
>regardless of which one is grounded. Whether the internal series
>regulator transistor is placed behind its +ve pin or -ve pin is
>irrelevant as it is all a simple series circuit.
>
>The 7905 was designed to complement the 7805 in a more conventional
>circuit arrangement with a common ground rail (not necessarily earthed).
>
>While Trikoupis's circuit configuration is unusual the theory behind
>it is sound. It should work.
>
>Trikoupis, there is something else happening in your circuit that you
>have overlooked. I would suggest that you recheck the wiring. You
>need to take care that the +ve pin of the -ve battery is connected
>nowhere but to the regulator input pin. You are using correct value
>bypass capacitors individually around each regulator, aren't you?
>At 03:54 AM 8/05/2006, rtstofer wrote:
> >--- In p...@yahoogroups.com, "Trikoupis Nikolaos" wrote:
> >>
> >> Hi to everyone!!
> >>
> >> Although it is not really a pic issue I am going to describe what
> >> happened to me. If anyone knows why I would be really grateful to
> >> hear about.
> >>
> >> During one of my projects I used 2 x 7805 for dual +-5v supply.
> >>
> >> Each one of them had was supplied by a 9V battery(that means 2
> >> seperated batteries).
> >>
> >> I connected the output of the second 7805 to the ground of the first
> >> one.
> >>
> >> As you can imagine I had both +5 and -5 supply. The circuit was
> >> working with, let say, resistors,etc.
> >>
> >> But when I used an op amp the supply circuit didn't work.
> >>
> >> I had ground +5V and instead of -5V there was something like -1 V.
> >> A lot of current was drawn through the second 7805. Like a short-
> >> circuit was achieved.
> >>
> >> Why do we really need the 7905 negative regulator?
> >
> >Because you can't connect a 7805 to produce a negative output relative
> >to ground.
> >
> >> What is happening when using 2x7805 with some circuitry?
> >
> >The regulator controls the output voltage relative to 'ground' but the
> >series pass transsistor that controls current flow is connected
> >between input and output. There is no ability to flow large currents
> >out of the ground pin. And, if that transistor conducts between input
> >and ground as you have it connected, it will try to short the battery.
> >
> >Bottom line: if two 7805s could be connected as you hoped, they
> >wouldn't have bothered to design the 7905.
> >
> >>
> >> I can't find an answer out there. Hope some guy knows the answer.
> >>
> >
> >Richard
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >to unsubscribe, go to http://www.yahoogroups.com and follow the instructions
> >Yahoo! Groups Links
> >
> >
> >
> >to unsubscribe, go to http://www.yahoogroups.com and follow the instructions
>Yahoo! Groups Links
>
>
(You need to be a member of piclist -- send a blank email to piclist-subscribe@yahoogroups.com )
lower 7805 will have its ground pin at the negative rail and its output
driving the circuit common. The upper 7805 will have its ground pin at
the circuit return and its output driving the positive supply rail. The
circuit used to be available in a National Semiconductor application
note, perhaps the 78xx part datasheet, if you need to see a schematic.
Be sure that total supply current, both positive and negative, does not
exceed the rating of the 7805 as all circuit current will flow through
the lower 7805. Heat sinking will probably be necessary.
This circuit works quite well and we have used it many times here to
keep from having to maintain stock of the 2 parts as well as the 78xx
parts being less expensive and more readily available than the 79xx parts.
REB
Phil Seakins wrote:
>rtstofer, the 7805 provides a regulated 5V output across its two pins
>regardless of which one is grounded. Whether the internal series
>regulator transistor is placed behind its +ve pin or -ve pin is
>irrelevant as it is all a simple series circuit.
>
>The 7905 was designed to complement the 7805 in a more conventional
>circuit arrangement with a common ground rail (not necessarily earthed).
>
>While Trikoupis's circuit configuration is unusual the theory behind
>it is sound. It should work.
>
>Trikoupis, there is something else happening in your circuit that you
>have overlooked. I would suggest that you recheck the wiring. You
>need to take care that the +ve pin of the -ve battery is connected
>nowhere but to the regulator input pin. You are using correct value
>bypass capacitors individually around each regulator, aren't you?
>At 03:54 AM 8/05/2006, rtstofer wrote:
> >--- In p...@yahoogroups.com, "Trikoupis Nikolaos"
> >>
> >> Hi to everyone!!
> >>
> >> Although it is not really a pic issue I am going to describe what
> >> happened to me. If anyone knows why I would be really grateful to
> >> hear about.
> >>
> >> During one of my projects I used 2 x 7805 for dual +-5v supply.
> >>
> >> Each one of them had was supplied by a 9V battery(that means 2
> >> seperated batteries).
> >>
> >> I connected the output of the second 7805 to the ground of the first
> >> one.
> >>
> >> As you can imagine I had both +5 and -5 supply. The circuit was
> >> working with, let say, resistors,etc.
> >>
> >> But when I used an op amp the supply circuit didn't work.
> >>
> >> I had ground +5V and instead of -5V there was something like -1 V.
> >> A lot of current was drawn through the second 7805. Like a short-
> >> circuit was achieved.
> >>
> >> Why do we really need the 7905 negative regulator?
> >
> >Because you can't connect a 7805 to produce a negative output relative
> >to ground.
> >
> >> What is happening when using 2x7805 with some circuitry?
> >
> >The regulator controls the output voltage relative to 'ground' but the
> >series pass transsistor that controls current flow is connected
> >between input and output. There is no ability to flow large currents
> >out of the ground pin. And, if that transistor conducts between input
> >and ground as you have it connected, it will try to short the battery.
> >
> >Bottom line: if two 7805s could be connected as you hoped, they
> >wouldn't have bothered to design the 7905.
> >
> >>
> >> I can't find an answer out there. Hope some guy knows the answer.
> >>
> >
> >Richard
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >to unsubscribe, go to http://www.yahoogroups.com and follow the instructions
> >Yahoo! Groups Links
> >
> >
> >
> >to unsubscribe, go to http://www.yahoogroups.com and follow the instructions
>Yahoo! Groups Links
>
>
(You need to be a member of piclist -- send a blank email to piclist-subscribe@yahoogroups.com )
Re: Re: 2 x 7805 VS 7805-7905 - Neil - May 8 1:59:15 2006
Are the tabs heat-sunk together (and therefore shorted)?
-Neil.
On Sunday 07 May 2006 21:11, Phil Seakins wrote:
> rtstofer, the 7805 provides a regulated 5V output across its two pins
> regardless of which one is grounded. Whether the internal series
> regulator transistor is placed behind its +ve pin or -ve pin is
> irrelevant as it is all a simple series circuit.
>
> The 7905 was designed to complement the 7805 in a more conventional
> circuit arrangement with a common ground rail (not necessarily earthed).
>
> While Trikoupis's circuit configuration is unusual the theory behind
> it is sound. It should work.
>
> Trikoupis, there is something else happening in your circuit that you
> have overlooked. I would suggest that you recheck the wiring. You
> need to take care that the +ve pin of the -ve battery is connected
> nowhere but to the regulator input pin. You are using correct value
> bypass capacitors individually around each regulator, aren't you?
>
> At 03:54 AM 8/05/2006, rtstofer wrote:
> >--- In p...@yahoogroups.com, "Trikoupis Nikolaos" wro=
te:
> >> Hi to everyone!!
> >>
> >> Although it is not really a pic issue I am going to describe what
> >> happened to me. If anyone knows why I would be really grateful to
> >> hear about.
> >>
> >> During one of my projects I used 2 x 7805 for dual +-5v supply.
> >>
> >> Each one of them had was supplied by a 9V battery(that means 2
> >> seperated batteries).
> >>
> >> I connected the output of the second 7805 to the ground of the first
> >> one.
> >>
> >> As you can imagine I had both +5 and -5 supply. The circuit was
> >> working with, let say, resistors,etc.
> >>
> >> But when I used an op amp the supply circuit didn't work.
> >>
> >> I had ground +5V and instead of -5V there was something like -1 V.
> >> A lot of current was drawn through the second 7805. Like a short-
> >> circuit was achieved.
> >>
> >> Why do we really need the 7905 negative regulator?
> >
> >Because you can't connect a 7805 to produce a negative output relative
> >to ground.
> >
> >> What is happening when using 2x7805 with some circuitry?
> >
> >The regulator controls the output voltage relative to 'ground' but the
> >series pass transsistor that controls current flow is connected
> >between input and output. There is no ability to flow large currents
> >out of the ground pin. And, if that transistor conducts between input
> >and ground as you have it connected, it will try to short the battery.
> >
> >Bottom line: if two 7805s could be connected as you hoped, they
> >wouldn't have bothered to design the 7905.
> >
> >> I can't find an answer out there. Hope some guy knows the answer.
> >
> >Richard
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >to unsubscribe, go to http://www.yahoogroups.com and follow the
> > instructions
=20
(You need to be a member of piclist -- send a blank email to piclist-subscribe@yahoogroups.com )
-Neil.
On Sunday 07 May 2006 21:11, Phil Seakins wrote:
> rtstofer, the 7805 provides a regulated 5V output across its two pins
> regardless of which one is grounded. Whether the internal series
> regulator transistor is placed behind its +ve pin or -ve pin is
> irrelevant as it is all a simple series circuit.
>
> The 7905 was designed to complement the 7805 in a more conventional
> circuit arrangement with a common ground rail (not necessarily earthed).
>
> While Trikoupis's circuit configuration is unusual the theory behind
> it is sound. It should work.
>
> Trikoupis, there is something else happening in your circuit that you
> have overlooked. I would suggest that you recheck the wiring. You
> need to take care that the +ve pin of the -ve battery is connected
> nowhere but to the regulator input pin. You are using correct value
> bypass capacitors individually around each regulator, aren't you?
>
> At 03:54 AM 8/05/2006, rtstofer wrote:
> >--- In p...@yahoogroups.com, "Trikoupis Nikolaos"
te:
> >> Hi to everyone!!
> >>
> >> Although it is not really a pic issue I am going to describe what
> >> happened to me. If anyone knows why I would be really grateful to
> >> hear about.
> >>
> >> During one of my projects I used 2 x 7805 for dual +-5v supply.
> >>
> >> Each one of them had was supplied by a 9V battery(that means 2
> >> seperated batteries).
> >>
> >> I connected the output of the second 7805 to the ground of the first
> >> one.
> >>
> >> As you can imagine I had both +5 and -5 supply. The circuit was
> >> working with, let say, resistors,etc.
> >>
> >> But when I used an op amp the supply circuit didn't work.
> >>
> >> I had ground +5V and instead of -5V there was something like -1 V.
> >> A lot of current was drawn through the second 7805. Like a short-
> >> circuit was achieved.
> >>
> >> Why do we really need the 7905 negative regulator?
> >
> >Because you can't connect a 7805 to produce a negative output relative
> >to ground.
> >
> >> What is happening when using 2x7805 with some circuitry?
> >
> >The regulator controls the output voltage relative to 'ground' but the
> >series pass transsistor that controls current flow is connected
> >between input and output. There is no ability to flow large currents
> >out of the ground pin. And, if that transistor conducts between input
> >and ground as you have it connected, it will try to short the battery.
> >
> >Bottom line: if two 7805s could be connected as you hoped, they
> >wouldn't have bothered to design the 7905.
> >
> >> I can't find an answer out there. Hope some guy knows the answer.
> >
> >Richard
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >to unsubscribe, go to http://www.yahoogroups.com and follow the
> > instructions
=20
(You need to be a member of piclist -- send a blank email to piclist-subscribe@yahoogroups.com )
Re: Re: 2 x 7805 VS 7805-7905 - Nick Trik - May 8 8:19:20 2006
The wiring must be ok.
I will tell you exactly what happened.
Using the 2 x 7805 as I mentioned (with the appropriate capacitors), the circuit had +5 and -5 volt without any load.
When entering a resistor across +5 to -5 or any possibly combination (+5,ground,-5) everything was working fine.
The problem occured when I inserted an opa4227 which needs positive and negative supply and has no ground pin.
At the -V pin I had something -1 V instead of -5V. A lot of current was drawn through the second battery.
To check if the opa4227 was ok I removed the second 7805 and used directly -9V to the negative supply of the opa4227.
The circuit was working fine.
Unfortunately I don't have a simulation file for the opa4227 to test it.
Any conclusion?
Nikolaos Trikoupis
Neil έγραψε:
Are the tabs heat-sunk together (and therefore shorted)?
-Neil.
On Sunday 07 May 2006 21:11, Phil Seakins wrote:
> rtstofer, the 7805 provides a regulated 5V output across its two pins
> regardless of which one is grounded. Whether the internal series
> regulator transistor is placed behind its +ve pin or -ve pin is
> irrelevant as it is all a simple series circuit.
>
> The 7905 was designed to complement the 7805 in a more conventional
> circuit arrangement with a common ground rail (not necessarily earthed).
>
> While Trikoupis's circuit configuration is unusual the theory behind
> it is sound. It should work.
>
> Trikoupis, there is something else happening in your circuit that you
> have overlooked. I would suggest that you recheck the wiring. You
> need to take care that the +ve pin of the -ve battery is connected
> nowhere but to the regulator input pin. You are using correct value
> bypass capacitors individually around each regulator, aren't you?
>
> At 03:54 AM 8/05/2006, rtstofer wrote:
> >--- In p...@yahoogroups.com, "Trikoupis Nikolaos" wrote:
> >> Hi to everyone!!
> >>
> >> Although it is not really a pic issue I am going to describe what
> >> happened to me. If anyone knows why I would be really grateful to
> >> hear about.
> >>
> >> During one of my projects I used 2 x 7805 for dual +-5v supply.
> >>
> >> Each one of them had was supplied by a 9V battery(that means 2
> >> seperated batteries).
> >>
> >> I connected the output of the second 7805 to the ground of the first
> >> one.
> >>
> >> As you can imagine I had both +5 and -5 supply. The circuit was
> >> working with, let say, resistors,etc.
> >>
> >> But when I used an op amp the supply circuit didn't work.
> >>
> >> I had ground +5V and instead of -5V there was something like -1 V.
> >> A lot of current was drawn through the second 7805. Like a short-
> >> circuit was achieved.
> >>
> >> Why do we really need the 7905 negative regulator?
> >
> >Because you can't connect a 7805 to produce a negative output relative
> >to ground.
> >
> >> What is happening when using 2x7805 with some circuitry?
> >
> >The regulator controls the output voltage relative to 'ground' but the
> >series pass transsistor that controls current flow is connected
> >between input and output. There is no ability to flow large currents
> >out of the ground pin. And, if that transistor conducts between input
> >and ground as you have it connected, it will try to short the battery.
> >
> >Bottom line: if two 7805s could be connected as you hoped, they
> >wouldn't have bothered to design the 7905.
> >
> >> I can't find an answer out there. Hope some guy knows the answer.
> >
> >Richard
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >to unsubscribe, go to http://www.yahoogroups.com and follow the
> > instructions Yahoo! Groups Links
> to unsubscribe, go to http://www.yahoogroups.com and follow the
> instructions Yahoo! Groups Links
>
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I will tell you exactly what happened.
Using the 2 x 7805 as I mentioned (with the appropriate capacitors), the circuit had +5 and -5 volt without any load.
When entering a resistor across +5 to -5 or any possibly combination (+5,ground,-5) everything was working fine.
The problem occured when I inserted an opa4227 which needs positive and negative supply and has no ground pin.
At the -V pin I had something -1 V instead of -5V. A lot of current was drawn through the second battery.
To check if the opa4227 was ok I removed the second 7805 and used directly -9V to the negative supply of the opa4227.
The circuit was working fine.
Unfortunately I don't have a simulation file for the opa4227 to test it.
Any conclusion?
Nikolaos Trikoupis
Neil
Are the tabs heat-sunk together (and therefore shorted)?
-Neil.
On Sunday 07 May 2006 21:11, Phil Seakins wrote:
> rtstofer, the 7805 provides a regulated 5V output across its two pins
> regardless of which one is grounded. Whether the internal series
> regulator transistor is placed behind its +ve pin or -ve pin is
> irrelevant as it is all a simple series circuit.
>
> The 7905 was designed to complement the 7805 in a more conventional
> circuit arrangement with a common ground rail (not necessarily earthed).
>
> While Trikoupis's circuit configuration is unusual the theory behind
> it is sound. It should work.
>
> Trikoupis, there is something else happening in your circuit that you
> have overlooked. I would suggest that you recheck the wiring. You
> need to take care that the +ve pin of the -ve battery is connected
> nowhere but to the regulator input pin. You are using correct value
> bypass capacitors individually around each regulator, aren't you?
>
> At 03:54 AM 8/05/2006, rtstofer wrote:
> >--- In p...@yahoogroups.com, "Trikoupis Nikolaos"
> >> Hi to everyone!!
> >>
> >> Although it is not really a pic issue I am going to describe what
> >> happened to me. If anyone knows why I would be really grateful to
> >> hear about.
> >>
> >> During one of my projects I used 2 x 7805 for dual +-5v supply.
> >>
> >> Each one of them had was supplied by a 9V battery(that means 2
> >> seperated batteries).
> >>
> >> I connected the output of the second 7805 to the ground of the first
> >> one.
> >>
> >> As you can imagine I had both +5 and -5 supply. The circuit was
> >> working with, let say, resistors,etc.
> >>
> >> But when I used an op amp the supply circuit didn't work.
> >>
> >> I had ground +5V and instead of -5V there was something like -1 V.
> >> A lot of current was drawn through the second 7805. Like a short-
> >> circuit was achieved.
> >>
> >> Why do we really need the 7905 negative regulator?
> >
> >Because you can't connect a 7805 to produce a negative output relative
> >to ground.
> >
> >> What is happening when using 2x7805 with some circuitry?
> >
> >The regulator controls the output voltage relative to 'ground' but the
> >series pass transsistor that controls current flow is connected
> >between input and output. There is no ability to flow large currents
> >out of the ground pin. And, if that transistor conducts between input
> >and ground as you have it connected, it will try to short the battery.
> >
> >Bottom line: if two 7805s could be connected as you hoped, they
> >wouldn't have bothered to design the 7905.
> >
> >> I can't find an answer out there. Hope some guy knows the answer.
> >
> >Richard
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >to unsubscribe, go to http://www.yahoogroups.com and follow the
> > instructions Yahoo! Groups Links
> to unsubscribe, go to http://www.yahoogroups.com and follow the
> instructions Yahoo! Groups Links
>
to unsubscribe, go to http://www.yahoogroups.com and follow the instructions
SPONSORED LINKS
Microcontroller Microprocessor Microchip pic Microcontrollers Picmicro 8051 microprocessor
---------------------------------
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---------------------------------
---------------------------------
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Re: 2 x 7805 VS 7805-7905 - john - May 8 15:47:58 2006
Why are you using 7805's with 9v battery's? You could just connect on=20
battery straight to the op-amp (with a series resistor and a zener=20
for protection). Then use a resistor divider network to get the=20
ground point. What output swing do you need for the op-amp? The=20
opa4227 isn't a rail to rail op-amp so you wouldn't get a 10V swing=20
output even with the 7805's.
--- In p...@yahoogroups.com, Nick Trik wrote:
>
> The wiring must be ok.
>=20=20=20=20
> I will tell you exactly what happened.
>=20=20=20=20
> Using the 2 x 7805 as I mentioned (with the appropriate=20
capacitors), the circuit had +5 and -5 volt without any load.
>=20=20=20=20
> When entering a resistor across +5 to -5 or any possibly=20
combination (+5,ground,-5) everything was working fine.
>=20=20=20=20
> The problem occured when I inserted an opa4227 which needs=20
positive and negative supply and has no ground pin.
>=20=20=20=20
> At the -V pin I had something -1 V instead of -5V. A lot of=20
current was drawn through the second battery.
>=20=20=20=20
> To check if the opa4227 was ok I removed the second 7805 and used=20
directly -9V to the negative supply of the opa4227.=20
>=20=20=20=20
> The circuit was working fine.
>=20=20=20=20
> Unfortunately I don't have a simulation file for the opa4227 to=20
test it.
>=20=20=20=20
> Any conclusion?
>=20=20=20=20
> Nikolaos Trikoupis
>=20
> Neil =DD=E3=F1=E1=F8=E5:
> Are the tabs heat-sunk together (and therefore shorted)?
> -Neil.
>=20
>=20
> On Sunday 07 May 2006 21:11, Phil Seakins wrote:
> > rtstofer, the 7805 provides a regulated 5V output across its two=20
pins
> > regardless of which one is grounded. Whether the internal series
> > regulator transistor is placed behind its +ve pin or -ve pin is
> > irrelevant as it is all a simple series circuit.
> >
> > The 7905 was designed to complement the 7805 in a more=20
conventional
> > circuit arrangement with a common ground rail (not necessarily=20
earthed).
> >
> > While Trikoupis's circuit configuration is unusual the theory=20
behind
> > it is sound. It should work.
> >
> > Trikoupis, there is something else happening in your circuit that=20
you
> > have overlooked. I would suggest that you recheck the wiring. You
> > need to take care that the +ve pin of the -ve battery is connected
> > nowhere but to the regulator input pin. You are using correct=20
value
> > bypass capacitors individually around each regulator, aren't you?
> >
> > At 03:54 AM 8/05/2006, rtstofer wrote:
> > >--- In p...@yahoogroups.com, "Trikoupis Nikolaos"=20
wrote:
> > >> Hi to everyone!!
> > >>
> > >> Although it is not really a pic issue I am going to describe=20
what
> > >> happened to me. If anyone knows why I would be really=20
grateful to
> > >> hear about.
> > >>
> > >> During one of my projects I used 2 x 7805 for dual +-5v=20
supply.
> > >>
> > >> Each one of them had was supplied by a 9V battery(that means 2
> > >> seperated batteries).
> > >>
> > >> I connected the output of the second 7805 to the ground of=20
the first
> > >> one.
> > >>
> > >> As you can imagine I had both +5 and -5 supply. The circuit=20
was
> > >> working with, let say, resistors,etc.
> > >>
> > >> But when I used an op amp the supply circuit didn't work.
> > >>
> > >> I had ground +5V and instead of -5V there was something like -
1 V.
> > >> A lot of current was drawn through the second 7805. Like a=20
short-
> > >> circuit was achieved.
> > >>
> > >> Why do we really need the 7905 negative regulator?
> > >
> > >Because you can't connect a 7805 to produce a negative output=20
relative
> > >to ground.
> > >
> > >> What is happening when using 2x7805 with some circuitry?
> > >
> > >The regulator controls the output voltage relative to 'ground'=20
but the
> > >series pass transsistor that controls current flow is connected
> > >between input and output. There is no ability to flow large=20
currents
> > >out of the ground pin. And, if that transistor conducts=20
between input
> > >and ground as you have it connected, it will try to short the=20
battery.
> > >
> > >Bottom line: if two 7805s could be connected as you hoped, they
> > >wouldn't have bothered to design the 7905.
> > >
> > >> I can't find an answer out there. Hope some guy knows the=20
answer.
> > >
> > >Richard
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >to unsubscribe, go to http://www.yahoogroups.com and follow the
> > > instructions
=20
(You need to be a member of piclist -- send a blank email to piclist-subscribe@yahoogroups.com )
battery straight to the op-amp (with a series resistor and a zener=20
for protection). Then use a resistor divider network to get the=20
ground point. What output swing do you need for the op-amp? The=20
opa4227 isn't a rail to rail op-amp so you wouldn't get a 10V swing=20
output even with the 7805's.
--- In p...@yahoogroups.com, Nick Trik
>
> The wiring must be ok.
>=20=20=20=20
> I will tell you exactly what happened.
>=20=20=20=20
> Using the 2 x 7805 as I mentioned (with the appropriate=20
capacitors), the circuit had +5 and -5 volt without any load.
>=20=20=20=20
> When entering a resistor across +5 to -5 or any possibly=20
combination (+5,ground,-5) everything was working fine.
>=20=20=20=20
> The problem occured when I inserted an opa4227 which needs=20
positive and negative supply and has no ground pin.
>=20=20=20=20
> At the -V pin I had something -1 V instead of -5V. A lot of=20
current was drawn through the second battery.
>=20=20=20=20
> To check if the opa4227 was ok I removed the second 7805 and used=20
directly -9V to the negative supply of the opa4227.=20
>=20=20=20=20
> The circuit was working fine.
>=20=20=20=20
> Unfortunately I don't have a simulation file for the opa4227 to=20
test it.
>=20=20=20=20
> Any conclusion?
>=20=20=20=20
> Nikolaos Trikoupis
>=20
> Neil
> Are the tabs heat-sunk together (and therefore shorted)?
> -Neil.
>=20
>=20
> On Sunday 07 May 2006 21:11, Phil Seakins wrote:
> > rtstofer, the 7805 provides a regulated 5V output across its two=20
pins
> > regardless of which one is grounded. Whether the internal series
> > regulator transistor is placed behind its +ve pin or -ve pin is
> > irrelevant as it is all a simple series circuit.
> >
> > The 7905 was designed to complement the 7805 in a more=20
conventional
> > circuit arrangement with a common ground rail (not necessarily=20
earthed).
> >
> > While Trikoupis's circuit configuration is unusual the theory=20
behind
> > it is sound. It should work.
> >
> > Trikoupis, there is something else happening in your circuit that=20
you
> > have overlooked. I would suggest that you recheck the wiring. You
> > need to take care that the +ve pin of the -ve battery is connected
> > nowhere but to the regulator input pin. You are using correct=20
value
> > bypass capacitors individually around each regulator, aren't you?
> >
> > At 03:54 AM 8/05/2006, rtstofer wrote:
> > >--- In p...@yahoogroups.com, "Trikoupis Nikolaos"=20
> > >> Hi to everyone!!
> > >>
> > >> Although it is not really a pic issue I am going to describe=20
what
> > >> happened to me. If anyone knows why I would be really=20
grateful to
> > >> hear about.
> > >>
> > >> During one of my projects I used 2 x 7805 for dual +-5v=20
supply.
> > >>
> > >> Each one of them had was supplied by a 9V battery(that means 2
> > >> seperated batteries).
> > >>
> > >> I connected the output of the second 7805 to the ground of=20
the first
> > >> one.
> > >>
> > >> As you can imagine I had both +5 and -5 supply. The circuit=20
was
> > >> working with, let say, resistors,etc.
> > >>
> > >> But when I used an op amp the supply circuit didn't work.
> > >>
> > >> I had ground +5V and instead of -5V there was something like -
1 V.
> > >> A lot of current was drawn through the second 7805. Like a=20
short-
> > >> circuit was achieved.
> > >>
> > >> Why do we really need the 7905 negative regulator?
> > >
> > >Because you can't connect a 7805 to produce a negative output=20
relative
> > >to ground.
> > >
> > >> What is happening when using 2x7805 with some circuitry?
> > >
> > >The regulator controls the output voltage relative to 'ground'=20
but the
> > >series pass transsistor that controls current flow is connected
> > >between input and output. There is no ability to flow large=20
currents
> > >out of the ground pin. And, if that transistor conducts=20
between input
> > >and ground as you have it connected, it will try to short the=20
battery.
> > >
> > >Bottom line: if two 7805s could be connected as you hoped, they
> > >wouldn't have bothered to design the 7905.
> > >
> > >> I can't find an answer out there. Hope some guy knows the=20
answer.
> > >
> > >Richard
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >to unsubscribe, go to http://www.yahoogroups.com and follow the
> > > instructions
=20
(You need to be a member of piclist -- send a blank email to piclist-subscribe@yahoogroups.com )