i want to convert 1.5 v 150 mamp to 5 v 2 amp. Is this possible ? if possible how to make circuit of this ?
thanks.
\$\begingroup\$
\$\endgroup\$
3
-
2\$\begingroup\$ not possible. Power can't be increased at the output. voltage can be stepped down or up (current will be inversely related). \$\endgroup\$– user19579Mar 10 '15 at 6:15
-
\$\begingroup\$ 150m Amp to 2 Amp. Its impossible if you don't have another secondary source apart from 1.5V, 150m Amp. Just calculate the power, your output power expectation is more more than input power fed. \$\endgroup\$– AKRMar 10 '15 at 6:15
-
\$\begingroup\$ If you used a 1.5V to 5V boost SMPS (switch-mode power supply) having 90% efficiency, then a load that draws 2A (10 watts) will require 7.3A from the 1.5V input (11 watts). Is your 1.5V 150mA energy source actually a standard AA, C, or D sized battery? Consider instead using 4 x AA batteries connected in series, to make 6V supply -- then use a low-dropout (LDO) linear regulator to regulate 5V. However standard AA battery will not be able to supply 2A of output current for an extended time -- 500mA is a lot for this type of battery. \$\endgroup\$– MarkUMar 10 '15 at 9:48
Add a comment
|
\$\begingroup\$
\$\endgroup\$
0
No, it is not physically possible.
1.5V at 150mA is equal to 0.225W (P=VI).
5V at 2A is equal to 10W (P=VI).
You can't magic the missing 9.775W from thin air.
It's a bit like having a glass of water that holds 0.225 litres of water, and expecting to pour it into a 10 litre bucket and have it fill the bucket. It's just not going to happen, is it?
To get 10W out you have to have at least 10W available (not taking into account any conversion losses of course). After all:
In physics, the law of conservation of energy states that the total energy of an isolated system remains constant—it is said to be conserved over time. Energy can be neither created nor be destroyed, but it can change form, for instance chemical energy can be converted to kinetic energy in the explosion of a stick of dynamite.
-- Wikipedia
So you will have to either provide at least 10W (plus conversion losses) of power at some voltage and convert it to 5V - be that from 1.5V say 7A at 1.5V, or maybe 12V at 1A.
