# Make a Dual power supply from single supply need help

#### Sornpraram Xu

Joined Feb 24, 2019
21
Hi. I have a problem to make a Dual power supply from single supply.
I am trying to make a dual power supply that can give the minimum voltage at least 3.0 V +- 0.15 V and maximum voltage at 10 V +- 1 V
and when connect to a load it must give the Current NOT below 300 mA. but as you can see in the picture when I connect to the 40 ohm load
Current is drop to 260 mA. Did anyone have any Idea How I can rise the Current and the Voltage will not drop.
Thank you.

#### KeithWalker

Joined Jul 10, 2017
2,622
Hi. I have a problem to make a Dual power supply from single supply.
I am trying to make a dual power supply that can give the minimum voltage at least 3.0 V +- 0.15 V and maximum voltage at 10 V +- 1 V
and when connect to a load it must give the Current NOT below 300 mA. but as you can see in the picture when I connect to the 40 ohm load
Current is drop to 260 mA. Did anyone have any Idea How I can rise the Current and the Voltage will not drop.
Thank you.
Have you measured the voltage of your 24 volt supply when the circuit is fully loaded? Does it drop below 24 volts?
The The minimum voltage drop across each regulator is 1.25 volts so you are working very close to the limits of the regulators.
The unbalance in the outputs could be caused by the difference in values of R6 and R7.

#### LesJones

Joined Jan 8, 2017
3,905
You need to learn ohms law to understand why the current through a 40 ohm resistor with 10.3 volts across it is lower than you want. If you want more current through the resistor you either need to increase the voltage across it or reduce the value of the resistor. When I was at school I think ohms law was taught when I was about 13 years old. Ohms law is current is voltage divided by resistance. You can rearrange it to make voltage or resistance the subject the subject of the equation. So in your circuit the current is 10.3 volts divided by 40 ohms. Which is 0.2575 amps (Or 257.5 mA)

Les.

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#### bertus

Joined Apr 5, 2008
22,120
Hello,

I think there are two things that are playing you parts.
1) The voltage divider has to small transistors.
Take a look at the TIP120 / TIP125.
2) the LM317 / LM337 circuit is not correct:

There are fixed resistors between Vo and Vadj.

Bertus

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#### MisterBill2

Joined Jan 23, 2018
13,809
This looks a lot like a homework problem. But if it is a real situation then the very first thing I suggest is if the supply transformer has a center tap then use a second pair of diodes and create another source for the minus 12 volts. Then forget the balancing circuit because it will not be needed. Why chioose complex when simple will be much better?

#### crutschow

Joined Mar 14, 2008
31,135
!2V input is marginal for a 10V output with those regulators.
What is the output voltage with no load?

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#### sghioto

Joined Dec 31, 2017
3,713
Did anyone have any Idea How I can rise the Current and the Voltage will not drop
You need to increase the battery voltage. The LM317 needs 3 volts of headroom to regulate properly.
SG

#### Sornpraram Xu

Joined Feb 24, 2019
21
Have you measured the voltage of your 24 volt supply when the circuit is fully loaded? Does it drop below 24 volts?
The The minimum voltage drop across each regulator is 1.25 volts so you are working very close to the limits of the regulators.
The unbalance in the outputs could be caused by the difference in values of R6 and R7.
the voltage drop to 21 V.
You need to learn ohms law to understand why the current through a 40 ohm resistor with 10.3 volts across it is lower than you want. If you want more current through the resistor you either need to increase the voltage across it or reduce the value of the resistor. When I was at school I think ohms law was taught when I was about 13 years old. Ohms law is current is voltage divided by resistance. You can rearrange it to make voltage or resistance the subject the subject of the equation. So in your circuit the current is 10.3 volts divided by 40 ohms. Which is 0.2575 amps (Or 257.5 mA)

Les.
The problem is I can not reduce the resistance that is the requriment and if I increase the input voltage the output Maximum voltage will exceed 11 V

#### sghioto

Joined Dec 31, 2017
3,713
The problem is I can not reduce the resistance that is the requriment
You need 12 volts out to get 300 ma with a 40 ohm load. Your battery or supply needs to be at least 30 volts at the rated load.
SG

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#### Sornpraram Xu

Joined Feb 24, 2019
21
Hello,

I think there are two things that are playing you parts.
1) The voltage divider has to small transistors.
Take a look at the TIP120 / TIP125.
2) the LM317 / LM337 circuit is not correct:

There are fixed resistors between Vo and Vadj.

Bertus
I don't have the TIP120 / TIP125
and I change the fixed resistors to adjustable resistors to change my minimum output voltage.

This looks a lot like a homework problem. But if it is a real situation then the very first thing I suggest is if the supply transformer has a center tap then use a second pair of diodes and create another source for the minus 12 volts. Then forget the balancing circuit because it will not be needed. Why chioose complex when simple will be much better?
I can only have 1 sorce of power from PSU.

!2V input is marginal for a 10V output with those regulators.
What is the output voltage with no load?

You need to increase the battery voltage. The LM317 needs 3 volts of headroom to regulate properly.
SG
I tried and my maximum voltage output is exceed 11 V any idea to fix that?

#### crutschow

Joined Mar 14, 2008
31,135
I tried and my maximum voltage output is exceed 11 V any idea to fix that?
You fix that by understanding how the regulators work.
The voltage set resistor values for the regulators would appear to be the wrong values otherwise it the outputs would not go above 10V.
If proper selected and connected, the output will be independent of the input voltage as long as the input is above the minimum required.

#### Wendy

Joined Mar 24, 2008
23,082

#### MisterBill2

Joined Jan 23, 2018
13,809
Getting 500mA fro
You could look up negative voltage generators. They use a capacitor diode arrangement to produce a negative voltage as seen in this AAC article
Getting 500mA from a charge pump device without a huge amount of ripple would be a very big deal and not likely to happen. So the next question is "what is the required current? (sorry, my internet connection vanished last night)
Getting much current from a switched capacitor arrangement is a big challenge because not only does every bit of power need to come from capacitor storage, but also the switch resistance will have a big effect as the current rises.

#### MisterBill2

Joined Jan 23, 2018
13,809
One more question: Is the circuit an actual circuit or just a simulation? And how important is that current to be 300mA? What is the purpose of having the split supply? Way too many unknown things here to give a really useful answer.

#### Wendy

Joined Mar 24, 2008
23,082
I prefer to synthesize a virtual ground as described in my article , as current requirements go away, the only catch is the base power supply must be floating which is generally the case anyway. you wind up with something that closely resembles a dual tracking power supply. I you have a specific question you can always tag me me by typing @Wendy in your post . this will work for any member here, as it happens I like mentoring people.

#### bertus

Joined Apr 5, 2008
22,120
Hello,

What is the source of the 24 Volts?
When it drops at 300 mA, it is likely to small.
When you want to have 2 X 12 Volts at the output of the LM's , you must have at least 28 Volts at the input (2 X 12 Volts + 2 X 2 Volts voltage drop on the LM's).

At a lower input voltage, the LM's will not regulate well.

Bertus

#### MisterBill2

Joined Jan 23, 2018
13,809
Still no answer as to how critical the regulation is or what the application might be. Thus no clue as to how much effort it may be worth to provide a pair of regulated voltages. Another totally different approach would be a 24 volt powered dual voltage switcher supply, available from Digikey, not cheap. And then there is my next very simple and totally non-orthodox cheap method of splitting the supply with a stack of forward biased 1 amp rated power diodes, enough diodes in the string so that they do not go into conduction. At 0.7 volts drop that would be 38 in series, or maybe 36, with the center being the defined common. As either load drew current the opposite set of diodes would conduct and drop half of the source voltage. Very unorthodox but it should work, and if both sides get loaded the same then the efficiency can approach 100%.But keep in mind that the stack has to have enough diodes in the string so that with no load they do not conduct. Strange but simple.

#### Wendy

Joined Mar 24, 2008
23,082
Problem with that scheme you want the Virtual ground to be regulated. The diodes values will wiggle all over the place.

#### MisterBill2

Joined Jan 23, 2018
13,809
Problem with that scheme you want the Virtual ground to be regulated. The diodes values will wiggle all over the place.
This is true, to an extent. Yes, the diodes allow some wiggle and wander, but in many cases it is OK to do that. It does keep the ground fairly close to the center, and if the loads are fairly stable that may be adequate. Since we were given no detailed requirements for this application I figure that it is an option to consider. It would be adequate for driving a motor in forward and reverse, or for a polarity sensitive solenoid to latch and unlatch. And with some big filter capacitors it could work with some audio amplifiers as well. So the "wandering neutral" may be all that is required. OR, it may not! The specific application being a mystery so far.