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.
2 x 7805 VS 7805-7905
Started by ●May 7, 2006
Reply by ●May 7, 20062006-05-07
--- In p..., "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
>
> 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
Reply by ●May 7, 20062006-05-07
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..., "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
>
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..., "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
>
Reply by ●May 7, 20062006-05-07
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..., "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
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >Yahoo! Groups Links
> >
> >
> >
> >
>Yahoo! Groups Links
>
>
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..., "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
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >Yahoo! Groups Links
> >
> >
> >
> >
>Yahoo! Groups Links
>
>
Reply by ●May 8, 20062006-05-08
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..., "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
-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..., "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
Reply by ●May 8, 20062006-05-08
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 :
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..., "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..., "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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Reply by ●May 8, 20062006-05-08
Why are you using 7805's with 9v battery's? You could just connect on
battery straight to the op-amp (with a series resistor and a zener
for protection). Then use a resistor divider network to get the
ground point. What output swing do you need for the op-amp? The
opa4227 isn't a rail to rail op-amp so you wouldn't get a 10V swing
output even with the 7805's.
--- In p..., Nick Trik wrote:
>
> 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 :
> 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..., "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
battery straight to the op-amp (with a series resistor and a zener
for protection). Then use a resistor divider network to get the
ground point. What output swing do you need for the op-amp? The
opa4227 isn't a rail to rail op-amp so you wouldn't get a 10V swing
output even with the 7805's.
--- In p..., Nick Trik wrote:
>
> 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 :
> 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..., "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
