Need Help on this power supply.
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I am designing a supply with 6V battery and a 9 Volt DC Source, what
I want is that when I connect the circuit on the 9 V DC, the 6 V
Battery supply will diconnect and when I disconnect the 9 V DC the 6
V Battery will reconnect, this will happen without variation in the
output voltage of the circuit.
Please help.
I want is that when I connect the circuit on the 9 V DC, the 6 V
Battery supply will diconnect and when I disconnect the 9 V DC the 6
V Battery will reconnect, this will happen without variation in the
output voltage of the circuit.
Please help.
Comments
battery to the load thru a diode. This will leave about 5.3 volts available
to the load. When the 9 volt source is connected directly to the load, load
voltage will rise to 9 volts; the battery will be "off" because the diode is
reverse biased. When you disconnect the 9 volts, the battery will again
take over. The diode must be rated for your load current.
Hope this helps,
Ray McArthur
Original Message
From: mysemicon2000 <semiconductor@p...>
To: <basicstamps@yahoogroups.com>
Sent: Thursday, January 03, 2002 12:33 AM
Subject: [noparse][[/noparse]basicstamps] Need Help on this power supply.
> I am designing a supply with 6V battery and a 9 Volt DC Source, what
> I want is that when I connect the circuit on the 9 V DC, the 6 V
> Battery supply will diconnect and when I disconnect the 9 V DC the 6
> V Battery will reconnect, this will happen without variation in the
> output voltage of the circuit.
--- In basicstamps@y..., "Ray McArthur" <rjmca@u...> wrote:
> If you can afford approx. 0.7 volts diode drop, you can connect the
6 volt
> battery to the load thru a diode. This will leave about 5.3 volts
available
> to the load. When the 9 volt source is connected directly to the
load, load
> voltage will rise to 9 volts; the battery will be "off" because the
diode is
> reverse biased. When you disconnect the 9 volts, the battery will
again
> take over. The diode must be rated for your load current.
>
> Hope this helps,
> Ray McArthur
>
>
Original Message
> From: mysemicon2000 <semiconductor@p...>
> To: <basicstamps@y...>
> Sent: Thursday, January 03, 2002 12:33 AM
> Subject: [noparse][[/noparse]basicstamps] Need Help on this power supply.
>
>
> > I am designing a supply with 6V battery and a 9 Volt DC Source,
what
> > I want is that when I connect the circuit on the 9 V DC, the 6 V
> > Battery supply will diconnect and when I disconnect the 9 V DC
the 6
> > V Battery will reconnect, this will happen without variation in
the
> > output voltage of the circuit.
Rectifier site, these diodes comes 2 in a TO-220 package, common
cathode) and a low dropout regulator.
the 9v and the 6V sources at the anodes of the or'ing diodes.
--- In basicstamps@y..., "mysemicon2000" <semiconductor@p...> wrote:
> I am designing a supply with 6V battery and a 9 Volt DC Source,
what
> I want is that when I connect the circuit on the 9 V DC, the 6 V
> Battery supply will diconnect and when I disconnect the 9 V DC the
6
> V Battery will reconnect, this will happen without variation in the
> output voltage of the circuit.
>
> Please help.
another post suggested, you can gain a few tenths of a volt with a schottky
diode.
Ray McArthur
Original Message
From: mysemicon2000 <semiconductor@p...>
To: <basicstamps@yahoogroups.com>
Sent: Thursday, January 03, 2002 2:30 AM
Subject: [noparse][[/noparse]basicstamps] Re: Need Help on this power supply.
> Have you tried this circuit already?
>
>
> --- In basicstamps@y..., "Ray McArthur" <rjmca@u...> wrote:
> > If you can afford approx. 0.7 volts diode drop, you can connect the
> 6 volt
> > battery to the load thru a diode. This will leave about 5.3 volts
> available
> > to the load. When the 9 volt source is connected directly to the
> load, load
> > voltage will rise to 9 volts; the battery will be "off" because the
> diode is
> > reverse biased. When you disconnect the 9 volts, the battery will
> again
> > take over. The diode must be rated for your load current.
**************** etc.