hey guys

im trying to create a positive and negative 4.5V DC power supply but using only one 9V battery and i was wondering if that is possible as i dont know how to make a ground in reall life as to make one lead give 4.5V and the other give -4.5V

even if i try connecting it to my variable 12V DC power supply its the same thing thats why im not trying to use two batteries

im trying to power a circuit that is similar to push-pull transistors and i dont know how !(without the use of two batteries/with using my power supply)

can you please assist me with a schematic

thank you very much for the help

 

Post the circuit that you want to power . You might not need a dual supply
But just in case you need one in the future , try this :
http://sound.westhost.com/project43.htm
I haven't built and tested this circuit myself yet . It gives little current but you can probably get more by using power op amps .

 

The bottom supply on that page is much better than the top.

 

hey guys

im trying to create a positive and negative 4.5V DC power supply but using only one 9V battery and i was wondering if that is possible as i dont know how to make a ground in reall life as to make one lead give 4.5V and the other give -4.5V

even if i try connecting it to my variable 12V DC power supply its the same thing thats why im not trying to use two batteries

im trying to power a circuit that is similar to push-pull transistors and i dont know how !(without the use of two batteries/with using my power supply)

can you please assist me with a schematic

thank you very much for the help

popular electronics once ran a contest to come up with the best way to make a dual 4.5V supply from a 9V battery:

the winner was a guy who opened the battery and soldered a center tap wire at the middle point of the six cells inside.

 

Shagas : thanks for the schematic i've read the page and done my homework, with the attachment you will see (as you didnt test it so im showing you my results) :

' red wire represent positive voltage,green zero voltage and blue negative voltage '


A : is the circuit as it is just added balanced load to it.

B : added ground point to it.

C : made the simulation with unbalanced load.

D : added wires to the positive and negative leads of the power supply of the ICs

E : same thing but with load on positive lead

F : removed the ground

from this you can see that im still not getting any results until i add a ground to the circuit so im back at point one which is how can i get the circuit to give me positive and negative without using the ground symbol. as im not sure how to connect the circuit in real life

 

That point that you have connected between your two 10k resistors is ground .
0 volts . Now if you put your multimeter (black lead) on that point and then put the red lead to the positive terminal of the battery you will get +4,5 volts . If you put the red lead to the negative terminal you will get - 4,5 volts .. BUT . You will not use that as your ground because when you willl pull current though it , it will shift .

Your new 'ground' - 0 Volts will be the point between those 2 led's .
Then +4,5 volts will be at the positive terminal of the battery and -4,5 will be at the negative. It's all relative.
Look at figure 2 on the page I linked . It shows it clearly . +ve , gnd , - Ve


By the way , I see that your op amp power supply rails aren't connected . I don't know if that matters in the software or not . Try building the circuit on a breadboard and see what you get

 

Hello Zanac-X.

When you breadboard the circuit, you must add current limiting resistors in series with each LED.

Hobby LED'S typically have a Vf (forward voltage) of ~ 2 to 3.5 volts and need sometimes only a few milli-amps to light up. Voltage above the Vf results in increased current to the point where you will burn out components.

EDIT: Simulators aren't very good at simulating smoke.

Calculate resistor using Ohms law:
= (Volts Applied - LED Forward Voltage)/Led Current in amps
= (4.5 - 2.1)/.005
= 480 ohms, (a 470 ohm is just fine)

 

hey guys

im trying to create a positive and negative 4.5V DC power supply but using only one 9V battery and i was wondering if that is possible as i dont know how to make a ground in reall life as to make one lead give 4.5V and the other give -4.5V

even if i try connecting it to my variable 12V DC power supply its the same thing thats why im not trying to use two batteries

im trying to power a circuit that is similar to push-pull transistors and i dont know how !(without the use of two batteries/with using my power supply)

can you please assist me with a schematic

thank you very much for the help

If all you need is a 0V rail for an Op-Amp project; you can use a spare op-Amp as a voltage follower with the + input tied to a resistor divider across the 9V

If you're using a PP3 battery they don't have much Ah capacity, so I'm guessing that the bare Op-Amp solution is enough - If you do need some current handling; connect the Op-Amp output to complementary pair emitter followers - take the feedback to the negative input from the junction of the 2 emitters to clear up the non-linearity in the 2 transistors.

 

ian field :you know.......a picture is worth a thousand ward......because i didnt really understand anything of what you said.........sorry about that

 

Hello Zanac-X
For the first part, ian field is describing what you are showing in post #5...
You should post a diagram of the circuit "similar to push-pull transistors" that you are trying to power mentioned in post #1

 

"That point that you have connected between your two 10k resistors is ground .
0 volts . Now if you put your multimeter (black lead) on that point and then put the red lead to the positive terminal of the battery you will get +4,5 volts . If you put the red lead to the negative terminal you will get - 4,5 volts .. BUT . You will not use that as your ground because when you willl pull current though it , it will shift .

Your new 'ground' - 0 Volts will be the point between those 2 led's .
Then +4,5 volts will be at the positive terminal of the battery and -4,5 will be at the negative. It's all relative.
Look at figure 2 on the page I linked . It shows it clearly . +ve , gnd , - Ve"






I still haven't figured all that about + and - voltages. Thanks for this post, I'll read through it several times.
I have a brain meltdown every time I read about the negative leads/supplies.

 

tubeguy : im sorry i should have showed the schematic earlier, im sorry for that

let me explain : it first started of by the idea of "transistor test circuit" as i put a transistor in the circuit and use the switch to identify if its NPN or PNP and though of improving it with the ability to work as a switch (within the circuit(not shown in the pic)) and test if it works well and for the PNP i just couldn't figure out how to supply negative voltage and thats why i started the post, but now i need the schematic that im requiring to use it in a future project(next one actually) and a friend of mine do really need it too so its not just the transistors

know that i do know how to do it with two batteries but i dont know how to get negetive voltage using the two leads of my power supply and thats the problem

thank you all for your time it really means a lot

P.S :the picture dose not represent the "fully functional" circuit or its "accurate numbers" that i have in mind. its just one circuit of my project "component tester board" and this is my sketch for this circuit thats why i didnt show it to you guys...sorry about the mess

 

Metalmann : ammmmmm...... yea about that i know what you just said, i just put the ground there because i was in a hurry when i was building the circuit and i just put a ground to explain what is going on in the circuit and for the sake of avoiding any mistakes i changed the point right after my post and still the same results thats in the attachment, and it really isnt a big deal as my problem is the circuit dose not work without a ground symbol


side note :when a changed the schematic to PCB board simulation (and of course no ground symbol exists ),when the battery was attached the circuit's(current flow) simulation did not indicate any negative voltage at all unlike doing it with two batteries which works without any single problem

 

Zanac, to power a PNP you don't need 'negative voltage'
The way a transistor works is that for an NPN to start conducting , the BASE has to be at least ~ 0.6 volts HIGHER than the emmiter.
For a PNP the BASE has to be at least ~0.6 LOWER than the emitter . Now if your emmiter is at ground or somewhere there (which you assume ) then you will need negative voltage , yes .
BUT .. you don't need to do that . You can just connect your emmitter (arrow on top)on your PNP to to + voltage of your battery (let's say 9volts) and now if your base is Less than about 8,4 then the transistor will start to conduct .
In your last circuit that you showed , all you have to do is flip around the PNP.(don't forget to flip you right led around aswell and power it with +9v)

 

 

Shagas : thank you for the idea....i some what know this but here is my restrictions :

1- since its a test circuit i only have three holes to put the collector emitter and base so the switch is used to ether put negative voltage or positive voltage to base pin

2- im supposed to have no idea if the piece im holding is PNP or NPN and with that said you can see in the pic that I've connected two LEDs in opposite direction so that one of them represent PNP and the other NPN.

3- the entire them is to make one simple circuit within the board that enables me to identify this component and test it to see if it works properly so any other ideas like identifying them by using voltmeters or two small circuits for each type are not excepted by me as what is the point of making the project then ?

4- i cant use two batteries and i do need the knowledge of making negative power rail in my future circuit using only two leads so its not just about this PNP

 

ScottWang : well thank you its very much suitable for me but im not using it until all hope is lost

 

 

ScottWang : thanks agian fig.4 is what i need for my circuit thank you very much im going to use it.

there is one thing though about what Metalmann said , i hope this attachment helps you to know what im trying to say, im sure im doing something wrong here

the attachment is just a simple circuit i made just to point out the negative voltage and current flow. when i simulated fig.4 i just had the same thing,positive and ground(no negative voltage in either case)

 

i think im going to use this circuit .