Two regulated and fixed voltages +5 and +12

Thread Starter

AMIRAAM

Joined Dec 31, 2019
37
Hello everyone

I want to get out two regulated and fixed voltages +5 and +12

Can I do simply this?
ESCH02.png
Or I must separate each one alone like this

ESCH01.png

Or please if you know a better way then those tell me

And thanks for your helps
 

ErnieM

Joined Apr 24, 2011
8,181
Either way will work, you need to play against the tradeoffs.

In the first scheme the 12V regulator is supplying all the current, everything the 123V load AND the 5V load requires comes from the 12V regulator. Thus the total current the 12V chip can give to your loads is the max current minus the 5V load current. What you gain from this is the 5V chip runs a bit cooler as it only needs to dissipate 7 volts (12 - 5V) and in the 2nd scheme it has to drop even more voltage.

Do add the output caps to the 5V chip in the 1st way.

In the 2nd scheme you only need C1 or C2, not both as they are in parallel. Do choose a value for the combined load current.
 

crutschow

Joined Mar 14, 2008
27,761
Where is the AC coming from and what is its voltage?

If you use a center-tapped 12Vac output transformer you can get both about 15Vdc and 8Vdc from one bridge rectifier, to minimize dissipation in the regulators.
The 15Vdc comes from the bridge output, and the 8Vdc comes from the center-tap.
The two diodes to ground in the bridge act as a full-wave rectifier to the center-tap output.
 
Last edited:

sagor

Joined Mar 10, 2019
479
You can delete R13 and R14. But first, please explain why you think they are needed.

ak
I sometimes use things like R13 and R14 when the supply voltage is a bit too high. I calculate that at maximum current of 1A (for example), a reaonable voltage drop across the input resistor to keep the input voltages to regulators about 3V-4V above the regulator output. Simply put, I make the resistor dissipate the heat instead of the regulator. It does have limits, as the resistor has to handle a watt or two (or 3) of heat, which the regulator no longer has to handle. It makes heatsinking the regulator a bit easier. Higher wattage resistors are relatively cheap. Mind you, your enclosure has to handle the heat eventually, from both devices.
This approach makes a bit more sense with lower currents and smaller regulators like the 78L05 series, which can't have good heatsinks on them.
Just an opinion...
 

Thread Starter

AMIRAAM

Joined Dec 31, 2019
37
Thanks ErnieM , dl324, AnalogKid, crutschow, and sagor

ErnieM said:
Do add the output caps to the 5V chip in the 1st way.
I don't like to add the output caps to the 5V chip in the 1st way, that is why i place the 5v chip directly after 12 chip
Please can i do that and in the same time the output will keep the good regulation given by 12v chip and his caps !?

dl324 said:
What are the maximum loads required for each regulator?
The maximum loads required for each regulator:
1- +5v for three voltage divider there inputs are in lm324n both (inverting and no-inverting), 2 leds
2- +12v is for lm324n as comparator and amplifier, relay (400 ohm)
I don’t think that I have big loads!?

AnalogKid said:
you can delete R13 and R14. But first, please explain why you think they are needed.
Like he sagor Said:
I sometimes use things like R13 and R14 when the supply voltage is a bit too high………….Just an opinion...
I am using a center-tapped 12Vac, this supply voltage is a bit too high; I thought using the resistors for dissipate the heat instead of the regulator, I and because resistors are relatively cheap then regulator and easily replaced then ,
I don’t have big loads !?

crutschow said:
Where is the AC coming from and what is its voltage? .
Yes i use a center-tapped 12Vac I keep one for another separate high loads as small motors and other things like lamps!!

What I need please :

I don’t like to add the output caps to the 5V chip in the 1st way but I want to keep the good regulation given by 12v chip and his caps !?
Use R13 and R14 to drop voltage for both 5 and 12 v chip a condition that I have the suitable current on the outputs of both chips

(+5v for three voltage divider there inputs are in lm324n both (inverting and no-inverting), 2 leds

+12v is for lm324n as comparator and amplifier, relay (400 ohm))
Thanks again
 

dl324

Joined Mar 30, 2015
13,143
The maximum loads required for each regulator:
1- +5v for three voltage divider there inputs are in lm324n both (inverting and no-inverting), 2 leds
2- +12v is for lm324n as comparator and amplifier, relay (400 ohm)
I don’t think that I have big loads!?
Your information isn't easy to decipher, but having the 5V regulator input on the 12V regulator would probably be best.

What is driving the opamps?
 

Thread Starter

AMIRAAM

Joined Dec 31, 2019
37
Thanks Dennis
i have only one lm324n (quad ) A,B,C and D
A,B,C as amplifier there inputs are between +0.8 and +3,2 V (<10mA) to give an output (for 3 times) so +2.4 to +9.6 V
the D is used as comparator
the two LED (20mA) are supplied by the +5V
only the relay is actionned by the +12V ( the load of the relay is supplied by another sources)
ESCH03.png
 

AnalogKid

Joined Aug 1, 2013
9,362
I see that this is a preliminary schematic. Still there are some things to fix.

1. Unless the LEDs are an indicator type with built-in current limiting, you need to add a current limiting resistor in series with each one.

2. When Q2 is off, there will be 7 V applied to the LED, backwards. This might cause the LED to go into reverse conduction.

3. You should add a suppression diode across the relay coil, something like a 1N4004, to protect Q1 when the relay is turned off.

ak
 

Thread Starter

AMIRAAM

Joined Dec 31, 2019
37
Hello everyone

Maybe you only need one DC voltage supply.
I am sure that I need +5v regulated and fixed

+12v is for supplying the lm324n

1. Unless the LEDs are an indicator type with built-in current limiting, you need to add a current limiting resistor in series with each one.
Please can you help me to fix the problems of the reverse conduction of leds and about leds I want to add two leds one green and the second red

Green when the outputs D of lm324n get high

Red when the outputs D of lm324n get low

3. You should add a suppression diode across the relay coil, something like a 1N4004, to protect Q1 when the relay is turned off.
Thank you I can fix that

dl324 said:
Your information isn't easy to decipher, but having the 5V regulator input on the 12V regulator would probably be best.
I don’t like to add the output caps to the 5V chip in the same time I want to keep the good regulation given by 12v chip and his caps

please Is that possible!?

ESCH04.png
 

Thread Starter

AMIRAAM

Joined Dec 31, 2019
37
Why are you so sure that you need +5V?
Why does LM324N need +12V?
i am just a beginner in electronic all what i want to do is to go step by step
i have read some where that Absolute maximum rating of Lm324 is 32v
i just choose the +12v because i have 12 chip regulator, i have +12v relay and i have other loads which need +12v
i just choose the +5v because i have 5 chip regulator, and after tests i concluded that +5v is good for my applications
 
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MrChips

Joined Oct 2, 2009
24,238
i am just a beginner in electronic all what i want to do is to go step by step
i have read some where that Absolute maximum rating of Lm324 is 32v
i just choose the +12v because i have 12 chip regulator, i have +12v relay and i have other loads which need +12v
i just choose the +5v because i have 5 chip regulator, and after tests i concluded that +5v is good for my applications
I have >1001 components. I don't use them just because I have them.
You are just making life more complicated than it needs to be.
For your specific application, choose one voltage and stay with it.
You still have not told us what your are going to do with that circuit.
 

Thread Starter

AMIRAAM

Joined Dec 31, 2019
37
# MrChips
Don't you think
that this is a preliminary schematic !?
i have only one lm324n (quad ) A,B,C and D
A,B,C as amplifier there inputs are between +0.8 and +3,2 V (<10mA) to give an output (for 3 times) so +2.4 to +9.6 V
the D is used as comparator
the two LED (20mA) are supplied by the +5V
only the relay is actionned by the +12V ( the load of the relay is supplied by another sources)
 

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