dual power supply -15 and 15 help
grasshopper
Posts: 438
I am using the L7815 and the L7915 with an input DC power supply of 47Volts .56Amps trying to make a -15 and +15 volt dual supply.
data sheet l7915 www.st.com/stonline/products/literature/ds/2149/l7905c.pdf
data sheet l7815 www.st.com/stonline/books/pdf/docs/2144.pdf
Schematic is attached. I basically used the same schematic listed in the data sheet of the L7819 figure 5.
My problem is that the voltages I am measuring is not -15 and +15 its more around +15 and -7Volts. My initial thoughts is that my DC power supply does not have enough current. However I am running no load on the regulators as of yet. Any ideas as to why this is not working correctly?
* Also i use the diodes listed in the data sheet. My picture did not list the diodes but they are in place. *
Thanks.
data sheet l7915 www.st.com/stonline/products/literature/ds/2149/l7905c.pdf
data sheet l7815 www.st.com/stonline/books/pdf/docs/2144.pdf
Schematic is attached. I basically used the same schematic listed in the data sheet of the L7819 figure 5.
My problem is that the voltages I am measuring is not -15 and +15 its more around +15 and -7Volts. My initial thoughts is that my DC power supply does not have enough current. However I am running no load on the regulators as of yet. Any ideas as to why this is not working correctly?
* Also i use the diodes listed in the data sheet. My picture did not list the diodes but they are in place. *
Thanks.
664 x 501 - 28K
Comments
If your power source is just· +/- 47 volts between 2 points, it isn't going to work.
If you use a centertapped·transformer and bridge rectifier, the DC voltage will be 1.4 times the rms ac voltage, so a 25v ct transformer will get you the +/- 15vdc you're looking for.
· Darlene
Post Edited (LilDi) : 10/7/2008 3:25:18 PM GMT
Rick
Hum...
Thanks for the reply's. I will post my results when or if I ever get this to work.
That would be because you are not drawing any current. What you designed in the battery + resistor schematic is a voltage divider, which works fine for things that require tiny amounts of current and have high impedances. If you are running some sort of SPICE simulation then you better be providing a load, otherwise you're going to end up with an unrealistic situation.
Why do you need +-15V anyways? I'm guessing it's for some sort of design that uses op amps. Perhaps you could elaborate more on what you are doing since there may be easier solutions (such as using rail-to-rail opamps).
As far as the voltage divider drawing I used. I was told in school that common was just a relative point in the circuit, or, not always the negative terminal on the power supply. I mean Ground is earth ground and common it a point on a circuit that all other measurements are relative to. Perhaps I am wrong...
·
In this case though you have a floating ground or floating reference point.· If one regulator is required to deliver more current than the other regulator, then your ground reference will migrate.· You need to anchor this point somehow so that it is fixed.· One method is to use a split voltage supply... you don't need a transformer to do this ,·just a center tap on your DC·supply.· A resistor divider might work to help stabilize the ground reference, but you’re still going to see a migration depending on the·current demands of the regulators.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
The ground pin on those regulators is indeed a point of reference, however, about 5 mA current flows out of that ground pin and has to go somewhere. It usually sinks into the ground reference point. On the data sheets, that current is referred to either as the quiescent current or as the ground pin current, and it is the current that it takes to operate the regulator itself, apart from the current that is actually delivered to the load. It will change with load and with temperature and is certainly not the same for the 7815 and the 7915. In your circuit without a load, you are asking the 7815 and 7915 to supply one another's quiescent currents, but it won't balance and one or the other or both will go weird.
The problem will not be improved when you attach a load to the circuit. Suppose the positive half of the circuit needs to supply 100mA while the negative half supplies 10mA. Where will that current go? Think about it. It has to get back to the battery, but the only path back to the battery is through either the other regulator or through the quiescent current path, and that simply doesn't add up. You are right to think of this as a resistive circuit, but the currents have to flow in loops, per Kirchoff's laws.
It is possible to make a supply splitter with a high power op-amp, but the op-amp has to be able to absorb all the current at the virtual ground.
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Tracy Allen
www.emesystems.com
Sorry I am so hard headed but this is type of clarity that I need. It makes me think and better, conclude that I was a bit off. Thanks everyone for the help. I guess I need to get a better power supply and forgo the current idea.
Also I must admit that I forgot about quiescent currents - needing some refreshing.
Firstly, you don't need 47 volts, but you do need two supplies of roughly 17.5 to 20 which are in series and provide a ground via their middle connection. Going much higher in voltage only creates the need to dump a lot of excess voltage as heat. Why bother?
Secondly, if you follow that schematic properly, I don't think you need to worry about current imbalances or much of the theory mentioned above. Your schematic is doomed to failure because your power supply is wrong.
_____+20
|
--
----
--
----
|_______ Common Ground
|
--
----
--
----
|____ -20 Volts
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It's sunny and warm here. It is always sunny and warm here.... (unless a typhoon blows through).
Tropically, G. Herzog [noparse][[/noparse] 黃鶴 ] in Taiwan
Post Edited (Kramer) : 10/8/2008 1:55:22 PM GMT
Anyway, here is a link with some information that maybe be of help:
www.tangentsoft.net/elec/vgrounds.html
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- Rick