Need Help In Combine 2 Voltage Source

Thread Starter

woon_h88

Joined Mar 25, 2009
49
Hi,
I need some help in combine 2 voltage source in parallel to increase the current flowing in the circuit.

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R1 = R2=5 Ohm
R3=4.7kohm ( To let most of the current to flow to the USB
I used source conversion method and i should get around 2A at Node A..But when i tried it out its only give me around 33mA...When i removed 1 of the source, its still give me 33mA ..Is it my circuit is wrong?

Thank you..
 

wayneh

Joined Sep 9, 2010
16,102
Why use R3 at all?

Current is determined by the load, which you haven't specified.
 

Vinetou

Joined Sep 5, 2014
12
You need to know what is the resistance of your load in order to solve this circuit properly. Remember that you have two resistors in parallel in this circuit. One is the R3 and the other is your load. You really have 2A on node A. One part of this current will flow through R3 and the remaining part will flow through the load. To calculate how much current will flow through each one, you must determine the resistance of your load.
 

MrChips

Joined Oct 2, 2009
19,270
Why do you need any resistors, R1, R2 and R3?

Where is the 5V coming from?

What is the USB device?
 

Thread Starter

woon_h88

Joined Mar 25, 2009
49
Hmm... My usb is something like mobile phone...it's resistance is less than 100 ohm...(ttested by using direct 5v to the phone and measured the current...my 5v is from other part of the circuit which providing a 5v dc from a 12v ac dynamo (with rectifier and voltage regulator)... Using 1 source is not enough to charge a phone.. So my group though of using 2 source..but now the problem is the 2 voltage source can't combine (parallel, cos we wan to increase the current but not the voltage.) the r3 is just to forced moat of the current to the load as the load resistance is lower...about why I use resistance... Cos I trying to change the source to current source so I can parallel it...
 

MrChips

Joined Oct 2, 2009
19,270
What is the required current and voltage to charge your mobile phone?
What is the voltage and current output of each of your charger sources?

You can put two voltage sources in parallel. One source is going to run flat out and its output voltage will drop until the second source starts to assist.
 

Thread Starter

woon_h88

Joined Mar 25, 2009
49
The requirements for voltage is 5v and the current need to be more than 1A.. But less than 2 A... And the voltage source is just 5v with 32 mA..
 

Thread Starter

woon_h88

Joined Mar 25, 2009
49
ya...i did tried but for some reason i still getting 32mA..which is very strange...
 

Thread Starter

woon_h88

Joined Mar 25, 2009
49
Hi..
Sorry to revive a post as i still seeking a way to increase my output current...


http://diyaudioprojects.com/Technical/Current-Regulator/

Hmm..If i addition added a Current-Regulator to my 5V output voltage regulator with the R1 ( on my current regulator side) to 0.5 ohm..Will i get at the load with 5V 2.5A? Say that if my load is just 1 ohm..
 
Last edited:

wayneh

Joined Sep 9, 2010
16,102
No, that would be 12.5W output from a 6W input. Not gonna happen. There can only be losses, not gains.
 

thodoris46

Joined Jan 29, 2013
10
What you are trying to make is very simple, it is difficult to you because you are doing electronics without ever opening a book to read.
Learn the basics , then read some more.After that read you're posts to yourself...
 

Thread Starter

woon_h88

Joined Mar 25, 2009
49
What you are trying to make is very simple, it is difficult to you because you are doing electronics without ever opening a book to read. Learn the basics , then read some more.After that read you're posts to yourself...
Hi,
I dont know what u r trying to say, but please dont try to put demoralized comment...Problem is im been capped to clean energy which is my dynamo motor..Its can give a constant 6V with regular cycling..

Thank you..

No, that would be 12.5W output from a 6W input. Not gonna happen. There can only be losses, not gains.
Thank for the answer...Will try find others way...
 
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