first circuit. LDR and charger.

iONic

Joined Nov 16, 2007
1,662
the goal is still the same and in both schematic i know i have used a light bulb but their is no light bulb that i want to lit. so i see where it can get confusing. ill try to be clearer with my schematic next time. so again i need to charge and use a 9v battery (400mAh) with a 12V (2.1/5.5mm) source. the x1 bulb in the "simple 9v charger picture" is the 9v battery.
O.K. Without know the application for use I could ask, "why not just use the 12V power supply run the light?" Can you explain what will be powered by the 9V battery you will have under charge?
 

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Dentsu

Joined Jul 7, 2011
47
that would be a fair question to ask but i rather focus on the end goal and hopefully i have clearly outline what i want to achieve. just want to give you enough technical info to get it working. 9v battery will power a LDR and a "load" that uses 9v <200mAh.
 

iONic

Joined Nov 16, 2007
1,662
that would be a fair question to ask but i rather focus on the end goal and hopefully i have clearly outline what i want to achieve. just want to give you enough technical info to get it working. 9v battery will power a LDR and a "load" that uses 9v <200mAh.
So the LDR will activate a load that requires 9V and 200mA, not mAH.
Unless you disconnect charger from the 9V battery, the charger will supply the power for the mystery load. So the LDR would have to provide two functions, turn off the charging voltage and apply power to the load.
 
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Dentsu

Joined Jul 7, 2011
47
the LDR only goes on when there is little to no light and does not regulate the charging or power to the main load. its only function is to turn an IR LED on/off. you would be correct to say that once there is no power from the main source (12v) the 9v battery kick in like how a battery backup work. once it is connected back to a 12v the battery can begin charging again and the main load can power off of it. please refer to this diagram below as i think the final schematic will look something similar to what i want. i guess i am doing a bad job so far as we are in the 3rd page and i am trying to explain what i want. :( don't know much else i can really do but i do hope i finish this soon. lurker please help out.

thanks for sticking around and helping a newbie like me and if anyone have any suggestion do share. i just uploaded a update HERE and hope the gate is what i am looking for to get it running from two source. what is needed in order to replace the 7805 with a 317 in my circuit? :confused:
 

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Dentsu

Joined Jul 7, 2011
47
i don't think i can. i tried measuring the output once with my multimeter but the 2.1mm plug make it difficult.

spec:

Connectors: 2.1*5.5mm DC plug
Output voltage: 12V
 

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Dentsu

Joined Jul 7, 2011
47
i don't think i have paper clip but will any metallic material like a nail be good enough? and isn't this dangerous?

i did it anyway and it measure in at 12.39V.
 
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iONic

Joined Nov 16, 2007
1,662
Can't promise you will get 10.5V out, but you are more than welcome to try.



This is just the charger with a charge current of 40mA for a 400mAH 9V rechargeable battery.

OR

The Switchmode Design.

 
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iONic

Joined Nov 16, 2007
1,662
i just notice that my unit for C2 was wrong and after changing the value from 1F to 1uF it made no difference. changing R1 + R6 produce more volatage on XMM2.

You misunderstood what I said. You placed (2) 1.8 ohm resistors in series, a total of 3.6ohms into your simulation.

I said to replace R2a and R2b with a 1.8K resistor and see what you get. In your sim circuit this equates to R1 and R6.
 
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iONic

Joined Nov 16, 2007
1,662
Why not use 7812 regulator instead? It must provide 12v output that is adequate to cost the battery.
Because with the drop out voltage and other loses you will not be able to regulate the output voltage and probably not even get 10.5V as needed.

P.S.

Please remove your adverts from the bottom of your messages.
 
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Dentsu

Joined Jul 7, 2011
47
probably doing it wrong but i just tried that and its output is under 3v. starting to think this isnt possible unless someone can show me other wise..might have to rethink my project now. just so you know the battery can be changed just as long as it equal 9V and compact.

In the end, my point is that you will probably not be able to get to 10.5V.

You could try this!

 

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iONic

Joined Nov 16, 2007
1,662
probably doing it wrong but i just tried that and its output is under 3v. starting to think this isnt possible unless someone can show me other wise..might have to rethink my project now. just so you know the battery can be changed just as long as it equal 9V and compact.
You are relying way too heavily on the simulation. + Remove the 12V 10W bulb from the sim circuit. Your not going to operate a 10W 12V bulb with a 9V battery. @ 9V to obtain 10W you will need more than 1A of current and the 9V batery can not manage this.

Zener Regulator Calculator
 
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