Is it possible to use two voltage regulators of different values with one power source?
Can you elaborate?
I presume you wish to make a power supply with 2 different output voltages?
Yes. If I have, say, a 12V DC wall wart, can it supply both a 9V and 3V voltage regulator? Or would there be a short or some other deleterious outcome?
Ok, I have put together a schematic for you, just be aware that the input has to be DC and that the output current will depend on your choice of ic (they come in 500mA, 1A, 1.5A and 2A versions) and what your 12V supply is rated to ie if you chose both the outputs to be capable of 1A then you will need at least 2A coming from your 12V supply
Hope this helps
That's very helpful but I have a few questions if you can spare the time:1). I assume R2 is a variable resistor or potentiometer- is that right?2). What's the value of R1? And what does 240R refer to?3). For the three volt supply you've indicated an adjustable power regulator. Why not a 7803?4) And just for curiosity's sake, if I use a 1A power source and 1A regulators, will the current divide evenly between them?Thanks for the diagram!
I can always spare the time to help out a friend (however obscure)
1) Yes you are correct R2 is a variable resistor (which is the same as a potentiometer) you use this to adjust the output voltage of the regulator
2) R1's value is 240 Ohms, thats what the 'R' indicates, its the same reference method as 150K or 2K2, where the k means thousand; the R means 0 and M means millions (or more commonly mega) its called si notation
3) the reason not to use a 7803 is that as far as I am aware it does not exist the lowest I know of is a 7805 5 volt reg
4)if you use a 1 amp supply the best you can get is a little less than 1 amp total output (each reg uses about 80mA to function), it will give the power to the reg that needs it. ie you could draw 800mA at 9V and 40mA at 3V, or 200mA at 9V and 640mA at 3V no worries, the sum of the output current has to be less than 1A and it makes no difference what amount you pull out of each reg so long as you are below 1A total (minus the reg's sink current of appx 80ma each)
Does this clear it up?
Out of curiosity what do you intend to use this supply for?
I want to do a few of these (https://www.instructables.com/id/Battery-Powered-Portable-VU-Meter/) except instead of the loudness being quantified in the number of LEDs lit, it will be quantified in the brightness of the LEDs. I think this can be done by, instead of the LM3914 driving each LED individually, the LED pins will open transistors that will shunt the power signal through a lower-resistance path. I drew a diagram. The 9v supply is because I may want more than one LED. I think I want three meters, each with a filter: bass, mid and treble. Does this seem feasible? Or is my approach all wrong?
I would use a Digital to Analouge converter to set the brightness of the LED, a simple R 2R resistor ladder would do a good job, and you can use a cap in parallel with the led to smooth out the transition between input levels changing,
Having now seen you project, it is not necessary to have two supply voltages, it would be cheaper and considerably simpler to get a power pack that gives anywhere between 5V and 12V DC out, the circuit for the VU meter will run happily anywhere in that range, just put your filter before the audio amp (lm386) so as to produce a better output, if you want more than one LED per channel then just put them in parallel and adjust your choice of resistors to compensate for the extra current, you will then also have to use a transistor to drive the DAC inputs as the display driver will only sink 13mA on its pins (active low), something like a BC548 will do the trick as it has a max 200mA emitter current.
Ok, a couple more questions:
1) I'm actually thinking about something like 30 LEDs per channel. Can I use the same plan and keep adding LEDs in parallel? Would I need a cap for each one?
2) One transistor for all the DAC inputs? Do each of the display driver pins run through it?
Thanks again for all your help so far.
Actually one more question: does the term "to sink" mean the same as "to consume"? As you can see my understanding of the nomenclature is limited.
And I believe I've answered 2) for myself: No. A transistor for each input.
I was just having a think about this last night, it would be simpler still if you don't use the LM3916 at all, just go for the filter, take that into your LM386 (set it up for adjustable gain), use 4 1N4004 diodes to rectify the output and then take that trough to the base of a 2N2222, use the 2N2222 to control the current flow through your diodes, select a suitable resistor and your off. The 2N2222 will allow you to run up to around 50 LEDs depending on which package you go for.
Use the gain adjustment to set the output of the amp into the 2N2222's knee region
Sink current is the amount you can put through that pin before you blow it up, source current is the amount you can draw before the same result
Oh aha. I guess that would be pretty simple- and cheap. I'll take a crack at drawing that.
How did you go?
I've been a little busy but I may have something later tonight if you want to take a look thanks!
Would this do it? I'm assigning the values now.
Almost there, you have a problem with your rectifier (the diodes are backwards and the grounds are not isolated) and 2N2222 placement, i would do it this way:
How neat! Now I see how it works! Rad- I'll have one with values up later.
Wait a second- is your bridge lined up correctly? The IC signal is AC, no?
Yep your right :(
Stupid mistake to make
Here is another
Oh and the 3V supply will need to be adjusted at the 5K resistor.
The LM317 is adjustable between 1.2V and ~10V if running off a 12V supply - and the output is short circuit protected on both devices!