Battery charging

Um, how about putting another connector on the input side of the 7805?

Cheers

Phil Hobbs

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Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
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email: hobbs (atsign) electrooptical (period) net
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Hi I do not want to provide more than one connector t the user.

John

Wouldn't the 7803SR be worth considering for this job? Better efficiency from a switcher would extend battery life.

Unclear from your description whether you mean here that the charger could be accidentally plugged into the thing requiring 3.3v regulated. If so that sounds like a recipe for disaster. Ditto if you mean that the

3.3v output is capable of accepting the chargers plug.

No reason why the voltage regulator and its load has to be disconnected when the battery is on charge. Regulators will stand 15v (or even 30v).

Regards, Martin Brown

So you want to put a >12V supply on the output of the 7805 to charge the battery?

If your circuitry can stand 12V, why not just delete the 7805 and be done?

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net

Okay, So use a three contact connector...one and a common for powering your device and a second and common for charging your battery. Same connector, different pins.

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The battery has a connector A on it. The circuit and the battery charger has connector B on them. Connector A is a mate of connector B. The battery is 12 volts goes into 7805 the out put of the

7805( connector A) goes into the circuit that has the voltage regulator that outputs 3.3 volts.

Now , I want the user to plug in the battery into the circuit and when battery gets dischagred he / she can plug the batter into its charger. but 7805 is there now in the middle of the battery and the battery charger. So, how to solve this problem.

Thanks John

Thoroughly confusing. As Confucius say, picture worth thousand words :-) ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |

      Remember: Once you go over the hill, you pick up speed

Since you have volts to spare in both directions how about a diode in series with the regulator output and a diode reverse polarity across the regulator. VA < 5v the series diode passes current to the load and VA >12v the parallel diode passes current from the charger to the battery. The odd resistor may be needed to prevent instability and avoid sharp current surges when the charger is first plugged in.

That is if I have understood what you want to do.

Regards, Martin Brown

This is absolutely easy, I'm just not sure why you would do it. Just use diodes to isolate and bypass the circuit sections see:

note U1 is supposed to be a 7805 but I didn't have one.

Hi,

Please check the following link

or

The thing is that I got only two wires coming out of the 7805 and I want to power up the circuit with it and than also want to charge the

12 volt battery throught the same two wires.Offcourse the circuit or the charger, only one will be connected at one time. They will not be connected together at the same time. But I need to avoid 7805. Can a four diode bridge work, if yes than how?

Thanks John

I can only find an "equivalent circuit" for a 7805, but I'd guess the following will work...

Diode, anode to battery, cathode to _input_ of 7805

Diode, anode to _output_ of 7805, cathode to battery

I don't foresee any sneak paths that could damage the 7805, but you'll have to try it to make sure ;-)

Caution, don't exceed 20V in the charging direction. ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |

      Remember: Once you go over the hill, you pick up speed

Hi,

Can you advice schematic or a sketch?

John

If the battery (12V) connects to a charger (12 to 15V) on the identical connector that a 3.3V circuit uses, you ought to redesign the system so those are NOT possible to interconnect.

Or, maybe your '3.3V circuit' includes all the necessary stepdown regulators to safely connect to higher voltages?

Won't this leave the poor 7805 output stage facing the charger?

Surely a resistor from 7805 output to ground as a dummy load and a diode from 7805 output to the charger socket pin would be safer.

I guess it probably won't kill a 7805 to find it's "regulated" output hard clamped to 2 diode voltage drops above its input and at nearly three times the notional target output voltage but it can't be ideal.

Regards, Martin Brown

After a while. Right now I must do some fabric marking for SWMBO'd ;-) ...Jim Thompson

-- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at

| 1962 |

Remember: Once you go over the hill, you pick up speed

So, that's exactly what Dave has provided you a circuit to do. What don't you like about it?

I might have put D2 between the output of U1 and the common between D1 and the connector -- that would be the most conservative way to make sure you don't damage the regulator, and if you're concerned about the diode drop you can fix it with a diode in the ground lead of the regulator. But I think Dave's circuit should work -- check your device data sheet to see if it says anything about the reverse voltage between Vout and Vin.

This is, by the way, an incredibly inefficient way to power electronics from a battery. You'd be much better off with a switching regulator instead of the 7805; with just a bit of care you could configure it to turn off altogether when Vout raises above 5V after which power would go to your charging circuit.

--

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

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

Well, he's got something like a female connector on the battery and male connectors on both the circuit and the charger, so it's not that bad.

--

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

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

dump the 7805 and use a zener diode

just replace d2,d3,x1 in my schematic with short circuits.