voltage dropping

[DACS46]

Hey everybody!

I am wanting to drop 12v d.c to 5v.d.c..
Someone told me to just drop a 2w resistor in series with output. That doesn't sound right, as resistors do not drop current.
Another said to use a 7805 regulator 12v to input and get 5v output.
What if I used a zener diode in series, as one will get 5.1 v output?
Then if needed, I could put a small watt resistor in series with the zener

Dave

 

[Minder]

In using a zener you will need a resistor and 5v zener in series wired across the supply source, then take the 5v from common and the cathode of the zener place an electrolytic cap across the 5v..
Zener regulators are practical if the If the current draw is minor.
M.

 

[DACS46]

Thank you for your quick response!
The current draw is coming from a 3mm led which draws .20 ma

Dave


Dave

 

[Minder]

Is that .2 ma or 200ma.
So Series R will need to be 35ohm for minimum Zener current so 27ohms should do it.
7/.2 = 35.
Or if .2ma then 35k or 27k for optimum.
Probably need to go to a lower value to reach forward current breakdown region for the zener.
M.

 

[DACS46]

Again, thank you for your quick response and the info. I was having a brain cramp. But, the air has cleared and my eyes opened to where I was heading.

Dave

 

[BobK]

Dave,

Can you give us a clue?

Bob

 

[davenn]

Again, thank you for your quick response and the info. I was having a brain cramp. But, the air has cleared and my eyes opened to where I was heading.

Dave

unfortunately it hasn't cleared for us....
what exactly are you wanting the 5V for ?
just a 20mA led ? .... or more than just a LED .... you haven't been very clear on that

for powering LEDs look at this resource here on our site

https://www.electronicspoint.com/resources/got-a-question-about-driving-leds.5/


it will tell you all about powering LEDs it's very straight forward


Dave

 

[DACS46]

My apologies. My lack of making its use more clear was rude of me.
I am a model railroader by addiction!
I have designed a fully operating signal system using negative logic. Not negative voltage.
Why negative logic?
The system had to meet certain criteria or I would toss it on the pile of other impracticalities.
1. Inexpensive
2. Minimum parts
3. Ease of assembly
4. Isolation from track power
5. Computer interface capability

The detectors run off 12vdc. The signal system runs off 5vdc. However, the input from detector to system, is 0 volts! I was wanting to run the entire system with one power source...12vdc to 5vdc.

But it actually adds parts, which in turn, begins to defeat the original system requirements.
I, at present, am using two power supplies, 12v 4a +/0 and 5v 4a +/0. I wanted to run the entire system from one supply. So, in response to the question if for led's only...that answer is of course, no.
I have petty much decided on building a power supply from which I can get both voltages with a common 0v. I think...

With this signal system, 0 = high/ true
. 1 = low/false
Rather than 1 = high/true
. 0 = low/false

Now with all this, I have, more than likely, opened a larger can of worms.

Again, my dull witted non response, was not intentional.

Dave

 

[duke37]

I do not see how the logic system affects the power supplies.

You have several alternatives.
1. Two power supplies.
2. A 12V supply with a buck convertor to give 5V also.
3. A 5V supply with a boost convertor to give 12V also.
4. A pre-built supply which gives both outputs (computer power supply).

I would think that 1 or 4 would be best.
2 would be better than 3 because of the large currents involved. 3 would need about 12A.

 

[DACS46]

The logic system doesnt affect the power supplies at all. But, when a train is detected, detector is 12v powered, it drives an opto-isolater which turns on 0v to the input of the signal board. This is what powers all the led's in the system. I have two separate power supplies. I was wanting to figure how to drop the 12v to 5v also.
Then I could drop it on the detector board without having the separate 5 v supply involved in driving the led's in all 40 three color signals. This way, I do not have to bring both supplies into the detectors. Just the 12v . Perhaps I am being too hard on myself and completely over thinking this.

Dave

 

[BobK]

You can make the LEDs run off 12V by changing the resistor. If only the LEDs are using 5V, this is the way I would go. You lose the same amount of power as you would with a linear regulator, but the power loss is distributed over all the LEDs, so it is more manageable.

Bob

 

[DACS46]

Thanks for the input everybody. But, as I stated in my post yesterday..."perhaps I am over thinking this!"
For some reason, I couldn't see the forest for the trees. a real "duh" gave brain blockage.
Not wanting to belabor the obvious here, but 0 volt is 0 volt (dc)! It doesn't matter.
I simply ran a jumper from common 5 v to common 12v, making the two separate supplies one with 0volt commons. 5v/12v/0v.
Now I connect 5v to the signal driver boards for power and the 12v to the detectors. As the 5v and 12 v commons (0 v), are now connected, I just take my 0 volt through the isolator and out to the signal board enable input. 0 v!
This removes cost and makes for economy of force, to coin a phrase.
I already made the change and it works perfectly.
Amazing how something so simple, can cause such a perceived difficulty

 

[DACS46]

Thank you for your input.
Both power supplies each, have the same transformers. Voltage and current identical. So all should be fine.
Doing this has gotten me to want to build a new power supply with one transformer though.

Dave