# Voltage Regulator Schematic Help...

#### mattd860

Joined Jan 9, 2016
58
Hi Everyone - I'm working on a voltage regulator schematic and I found one that I'd like to use but have a questions. Can you please identify why there is a 12 volt power source circled in red? I can understand why there is the 12v-raw but I'm not understanding this second 12v wire and why it's even there to begin with.

I'm trying to create a voltage regulator so I can safely use an Arduino in an automotive application.

Thanks

#### Kermit2

Joined Feb 5, 2010
4,162
It is flagged after it has passed through a diode. It will be a slightly lower voltage than the 12 volts labeled "raw".

#### dl324

Joined Mar 30, 2015
16,720
Power source can be AC or DC. Putting DC input after the rectifier saves a diode drop and allows it to be as low as 5V plus dropout voltage.

#### mattd860

Joined Jan 9, 2016
58
Sorry I should have been clearer - The power source is 12 volts DC.

So the 12v circled in red is just a Flag telling me that the voltage will be slightly lower at that point due to the diode drop? Or is this a second 12v input? And if it is a 2nd 12v dc input, then where do I get this input from if not from the same power source as the 12v-raw?? Does the diagram want this 2nd input to be raw or regulated?

I'm hoping it's just a simple Flag as Kermit2 states - and I'm sorry Kermit2 if I'm second guessing you but I'm just trying to better understand all this. Thanks!!!

#### dl324

Joined Mar 30, 2015
16,720
I'm trying to create a voltage regulator so I can safely use an Arduino in an automotive application.
You could use a cigarette lighter to USB adapter...

#### mattd860

Joined Jan 9, 2016
58
You could use a cigarette lighter to USB adapter...

Thanks - this is the most ideal method, however, the location of the device does not allow me to do this.

#### dl324

Joined Mar 30, 2015
16,720
The power source is 12 volts DC.
Then apply power at the input you circled input. Drop D15 and U15.

#### Kermit2

Joined Feb 5, 2010
4,162
You can input your 12 volts at that point as has been said. The symbol indicates a schottky diode so the voltage drop would be very small if it remains in series with the input.
Less than 1/2 a volt. Either way you choose to connect to 12 volt power would be okay.

#### mattd860

Joined Jan 9, 2016
58
Okay so either input is fine and I now understand this a bit more. Since it's an automotive application would it be safer to use the 12v-raw input or is it redundant because the voltage is probably already cleaned and regulated somewhere else?

#### dl324

Joined Mar 30, 2015
16,720
D15 would protect from reverse polarity and U2 would protect from high voltage transients.

If your supply can be disconnected, adding a bridge rectifier for DC input would give you auto polarity.

#### SLK001

Joined Nov 29, 2011
1,549
It's not an input. It is a 12V take-off to power other parts of the circuit that needs 12V to operate.

#### crutschow

Joined Mar 14, 2008
34,058
Okay so either input is fine and I now understand this a bit more. Since it's an automotive application would it be safer to use the 12v-raw input or is it redundant because the voltage is probably already cleaned and regulated somewhere else?
Automobile 12V is definitely not clean and regulated.

#### mattd860

Joined Jan 9, 2016
58
Alight I got it now. Perfect!!

Another question - If I wanted a 9.0v output instead of 5.0v dc, do I just change the voltage regulator in the diagram above to a LM2940T-9.0 and leave the rest of the circuit the same? Or will this require a different set of capacitors and components?

#### dl324

Joined Mar 30, 2015
16,720
It's not an input. It is a 12V take-off to power other parts of the circuit that needs 12V to operate.
That would only be true if 12V raw was around 12.4V.

#### dl324

Joined Mar 30, 2015
16,720
If I wanted a 9.0v output instead of 5.0v dc, do I just change the voltage regulator in the diagram above to a LM2940T-9.0 and leave the rest of the circuit the same? Or will this require a different set of capacitors and components?
Everything the same except for the zener on the regulator output; which I would remove anyway. If you want output over voltage protection, you need to use something that can take more current; like a TVS diode.

If you don't use an LDO regulator, you need to be concerned with drop out voltage.

#### #12

Joined Nov 30, 2010
18,224
it's an automotive application
That would only be true if 12V raw was around 12.4V.
Surprise! Automotive batteries are normally at 12.5 volts when properly charged.

#### dl324

Joined Mar 30, 2015
16,720
Surprise! Automotive batteries are normally at 12.5 volts when properly charged.
When the car is running, it will be closer to 13.8V. And it isn't clear that the circuit in question was only for auto powered apps...

#### #12

Joined Nov 30, 2010
18,224
I measured a car 2 days ago. The running voltage was 14.5V
Though I must admit, the battery was in perfect condition and Florida is much warmer than where most of our members live.
And I felt confident this is an automotive application because TS said that in post #9.

Joined Jul 18, 2013
28,521
All the auto's I have owned in the past 20yrs the alternator output was ~14.6.
Max.