12 volt DC to 5 volt DC regulated

[wyrebiker]

I wish to connect a permanant lead to my motorcycle battery (12 Volt DC) reducing output to 5 volt DC. Cheapest and easiest method please.

 

[Faizan Jawaid]

I wish to connect a permanant lead to my motorcycle battery (12 Volt DC) reducing output to 5 volt DC. Cheapest and easiest method please.

U should also specify the current capacity too...if u require current less than 1A a simple 78L05 will also regulate the 12V to 5V...if the current requirments are high i suggest a and open-loop (without feedback) flyback converter can do that...But cheapest is 78L05...

 

[me_guitarist]

What u gonna do with the 5V output?

 

[Khabree]

Cheapest solution is 7805 IC, if your current is 1A or less.

 

[hbaocr]

1. You can use regular IC KA7805 ; or use Zenner diode!

 

[manu]

What's your output current?

 

[hkBattousai]

I have a question about 78xx voltage regulators?

Assume that the input DC voltage is Vi and the output of the IC is Vo volts.

ΔV = Vo - Vi
I : Current drained from IC
P = ΔV * I

Is this P power dissipated by the IC, or some does some other conversion mechanism work, like swith mode?

 

[horzcat]

Like most of the guys said, use a 7805 IC. But what they didn't say is, you must use a heat sink with the IC. Otherwise, the IC won't last more than a few months. Both the IC and the heatsink are cheap. You can find them in any radioshack.

 

[Faizan Jawaid]

Like most of the guys said, use a 7805 IC. But what they didn't say is, you must use a heat sink with the IC. Otherwise, the IC won't last more than a few months. Both the IC and the heatsink are cheap. You can find them in any radioshack.

We would have definitely advised a heat sink if we all know the current required which the post maker dint mention....if current is near to 1A we would require a heatsink if current is less than 500mA its is not necessary...

 

[manish12]

a series of 10 si diode give more current than 7805 , and it is chep too !

what do you think

12-5 = 7

number of diode = 7/0.7 = 10

and you have a option 5v, 5.7v , 6.4v , 7.1v , 7.8v etc .

 

[Faizan Jawaid]

a series of 10 si diode give more current than 7805 , and it is chep too !

what do you think

12-5 = 7

number of diode = 7/0.7 = 10

and you have a option 5v, 5.7v , 6.4v , 7.1v , 7.8v etc .

Motorcycle battery is also connect to a Dynamo...Which charges the battery...High rpms produces higher voltages near to 13 14 volts...Diodes will give high current but voltage will not be regulated...And the requirement is of DC regulated....Diodes will not work here...

 

[hkBattousai]

a series of 10 si diode give more current than 7805 , and it is chep too !

what do you think

12-5 = 7

number of diode = 7/0.7 = 10

and you have a option 5v, 5.7v , 6.4v , 7.1v , 7.8v etc .

Series of 10 diodes dissipate different voltage values for different current values you drain.

For instance, if you drain no current, the ouput voltage will be equal to input voltage, which is equal to 12V.
If you drain 1mA, the output voltage will be, approximately, 12V - (10 x 0.6V) = 8V.
If you drain 1A, the output voltage will be, approximately again, 12V - (10 x 0.7V) = 5V.

You can connect a bleeding resistor, but it would be a waste of energy for most applications.

So your method is not right. 7805 is a better solution.

 

[hbaocr]

hi; you can use this sche below to increase the current ouput of IC7805.

 

[bayramonur]

Can you post the values of all equiments in the reglated circuit? And all transistor alternatives? What maximum current value is?

 

[hbaocr]

hi;
R1,R2 is register to protect transistor and IC7805 (limit the current cross them)

R1 is power resister to protect IC7805 about 2 Ohm 2W
R2 depend on the max output current from (E pole of transistor).But Q1 is chosen to get the expected current output from that circuit.Because the output current of that schematic is current output from E of Q1.So you can choose Q1 to get the expected current !

 

[Audioguru]

Your booster circuit has a diode in the ground connection of the 78L05, so its voltage regulation will be poor.
It has an emitter-follower on its outpput so its voltage regulation will be poor again.
Two poors equals a very lousy regulator.

 

[Faizan Jawaid]

Your booster circuit has a diode in the ground connection of the 78L05, so its voltage regulation will be poor.
It has an emitter-follower on its outpput so its voltage regulation will be poor again.
Two poors equals a very lousy regulator.

Can u please explain how the regulation will be effected by these 2 mentioned points?

 

[Audioguru]

The 78L05 has a very high gain and has negative feedback for excellent voltage regulation.

Your circuit has a diode in the common leg of the regulator that changes its voltage when the temperature changes and changes its voltage when the current through it changes when the input voltage changes.

The emitter-follower on the output has more voltage drop when the load current increases and has less voltage drop when the load current decreases.

The voltage regulation will be much worse than the regulator by itself but the regulation is better than no regulator.

 

[fsmyth]

For a 5A example, see:
**broken link removed**

There are other examples on the same page that go
up to 20A.
<als>

 

[grittinjames]

use LM 2576