Two voltages for my breadboard

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ThomsCircuit

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From what i have read this is a linear voltage regulator. Id like to use this and a buck converter to power my BBoard.
The reason i would like to have 2 supply voltages is some of my projects have ICs that use less voltage than the rest of the circuit. The power source would be an unregulated DC 12v 2A wall wart. The buck converter is variable and was discussed here a few weeks ago. This circuit here below is a 12 to 5v @ 500ma. It believe this is more than enough to drive the low voltage ICs. Could you offer opinions as if this approach for my purpose is alright? I will be using the circuit below to make the regulator. I will however be changing out the 5v zener for a 5.6 and eliminating the diode. (it was suggest by the original author)
The finished circuit would be fitted into a project box with all the necessary input /output connectors and a mini VA meter.

Thank you.

 
Yes, that circuit will work. But the accuracy of your 5 Volts will vary with temperature. That is because both the Vbe of transistors, and the the voltage of zeners, change with temperature.

You far better to just use a linear voltage regulator such as one of the many LM7805 varieties that are temperature compensated.
 
Ok i can do that. sorry its so small. We got...
c1 220uf
c4 22uf
c2 & c3 are 0.1uf
and the 7805 in the center.
I'm wanting to use SMD equivalents if i can get them. So im substituting the 7805 for the LM78M05 and i have C2&C3 in SMD but i don't have SMD versions for c1&c4. Ill keep looking.
If there are any modifications that could be added pleases let me know.
Thank you.

 
ThomsCircuit said:
This circuit here below is a 12 to 5v @ 500ma.
Click to expand...
Not a good idea!
The transistor / 7805 will be dissipating 3.5W at full load and need a big heatsink.

If you use a switching regulator the losses will be a fraction of that. I'd just get an off-the-shelf module - there are thousands available and the modules cost less than the separate components!

eg. These LM2596 ones are widely available, under many names:

Pardon our interruption...

www.ebay.co.uk

I keep some of those and some small boost regulators in my parts boxes as they are so useful; also some tiny lower current ones.

(Always base the practical current rating on being half the advertised current though - the advertisers tend to make up figures or use the 100mS peak or whatever).

Or; we use these in products - a drop-in replacement for a 7805 circuit-wise, but again a switched mode buck module that dissipates very little power:

RS PRO Switching Regulator, PCB Mount, 5V dc Output Voltage, 6.5 → 36V dc Input Voltage, 500mA Output Current, 1 | RS

uk.rs-online.com

They are equivalents by several manufacturers.
 
Last edited:
rjenkinsgb said:
Not a good idea!
The transistor / 7805 will be dissipating 3.5W at full load and need a big heatsink.
Click to expand...
I understand. Thank you. So linear using 7805 or BD139 type is not as efficient as Buck (switching converters) I'm sure there's more to it then that but for where im at I will just use two buck converters setting one at 5v and allowing the other to be variable.
And you are correct about buying than building. But i did learn so much though going through the process in my buck converter thread.
 
ThomsCircuit said:
So linear using 7805 or BD139 type is not as efficient as Buck (switching converters)
Click to expand...
Yes, at least for a significant difference between the regulator input and output voltages.
The power dissipated by a linear regulator, regardless of the type or design, is simply the input voltage times the input current minus the output voltage times the output current.
The input current equals the output current plus any bias current the regulator requires.
 
LM2596 has a minimum load requirement to stay in regulation, before
it goes into discontinuous mode operation, also a f() of the L value used
and other parameters. So use wisely, you could damage circuits if load was
dropped with some circuits still connected to it.

Also efficiency various with output V -



Data sheet discusses these issues.

Three terminal regulators have thermal and short circuit protection, easy
to use and apply, and low noise if thats of concern. If using LDOs pay
attention to ESR of output caps, some have a min esr requirement to stay
stable. Also discussed in datasheets.


Regards, Dana.
 
Last edited:
Thank you Dana. I ended up designing a converter using MP2315 but i will be reading about how to design with this regulator as well.
 
Been reading about an adjustable Buck using this LM2596. (IMG1)
This circuit is rated at 3 amps but the inductors are L1 33uh 1.4a and the filtering inductor L2 is less than 1 amp. How does this circuit handle 3 amps? Could it be that the input is upto 40v? where the one in img2 is limited to 24v?

IMG1

I was working on another using MP2315 (IMG2) and its also adjustable and rated at 3 amps but the inductors are L1 4.7uh 3a and the filtering one L2 is the same so it makes logical sense.
FYI: The original design called for (L2) 220uh but the author said i could use any as it was just for filtering.
The mirror circuit on the right is to illuminate an LED at any voltage 4-24v

IMG2
 
ThomsCircuit said:
Hmmm. What to do?
I hope this clarifies why I question the questions. Thank you Nigel.
Click to expand...
Higher rated component's, or build a lower rated power supply?.

Or try it, and see if it goes BANG!!

Unless you're intending it for a specific high current purpose, then it will probably never get used much anywhere near 3A
 
Nigel Goodwin said:
Unless you're intending it for a specific high current purpose, then it will probably never get used much anywhere near 3A
Click to expand...
Correct. 1amp max. and that would be at 12 volts. the regulated 5v would be about 3-500ma. Driving low power ic's
It is just for my bread board.
 
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