Potential Divider

what value of current you gonna handle?

4.4 + 3.2 + 1 = 8.6

am i missing something here???

Total should be 6 7 its multiples 60, 600, 6K, 60K .. &submultiples

0.6, 00.6

tap at 4.4 2nd tap @ 3,2 & the 3rd @ 1.0 [Ohms, K, or M]

Depends on the current required.

Basically this may be used only to get Ref V [Which should be followed by an 1x1 Amp to get more Fan-outs] @ every tap.

V1= RI/(R1+R2+R3).VT.....................................................1

V2= R2/(R1+R2+R3).VT.....................................................2

V3=R3/(R1+R2+R3).VT......................................................3

What ever current through the circuit, V1+V2+V3=VT=6.

Thus in voltage divider the sum of output voltages must be equal to the total voltage.

6 can not be divided into 4.4, 3.2, 1 but you can divide into 1.8, 3.2 and 1V OR any combination.

Use 3 seperate pairs of resistances, not suitale for any major current draw though.. Use resistor & zenner diode combination for a stable voltage tap.

Nice homework question. If you cannot figgure this out by researching passive sites, then you shouldn't be playing with electricity.

There appears to be a great deal of inaccuracy in the replies here...unless I misunderstand the question, the solutions and assumptions provided are incorrect. If I've misunderstood the question, I apologize for my comment. Lets start by clarifying the question:

You want to know the resistor values for a voltage divider that provides taps at 1V, 3.2V and 4.4V. You also have virtually no current draw from the taps or you would not be using this circuit. Therefore, this circuit is being fed to high impedance op-amps or some other high impedance IC so current draw from the taps is of no concern.

Based on these assumptions, you can approach the problem from 3 different ways: Simultaneous linear equations, standard resistor divider equations (one previous reply attempted to use them but they are not correct and contain the wrong number of resistors) or use a simplifying assumption. The third approach is easiest: With no current draw from any of the taps we can make the bias current anything we want. Assume a bias current of 1mA. Then do the following:

Let:

V1 = 1V, V2 = 3.2V, V3 = 4.4V NOTE!!!! These are tap potentials NOT resistor Vr drops.

Set I = 1mA

Therefore, using ohms law:

R1 = Vr1 / I = (V1 - 0V) / I = 1.0V / 1mA = 1k ohm

R2 = Vr2 / I = (V2 - V1) / I = 2.2V / 1mA = 2.2k ohm

Now note the pattern: With a 1mA bias the resistance is equal to the voltage drop across the resistor X 1000 ohms.

R3 = (4.4 - 3.2) / 1mA = 1.2V / 1mA = 1.2k ohm

R4 = (6.0 - 4.4) / 1mA = 1.6V / 1mA = 1.6k ohm

Note, there are 4 resistors in a divider network to produce 3 taps!

Also note that V1+V2+V3 does not equal VT!!! The sum of the voltage drops across the resistors = VT = Vr1 + Vr2 + Vr3 + Vr4! That is 1V + 2.2V + 1.2V + 1.6V = 6V!

If you want to halve the bias current then you simply multiply every resistor by a factor of 2. If you want to double the bias current (make is more resistant to small current draws from the taps) then you can simply divide each resistor value by 2. You can use any mult/div factor you like as long as you follow this pattern.

Finally, if you anticipate having significant current draw from the taps (> I / 100 or for a bias of 1 mA a tap current of > 10 micro amps) then all you need to do is use a high impedance unity gain follower with its input connected to the tap (one follower for each tap). This isolates current draw effects from your bias network.

Good luck and again, my apologies if I've misunderstood what you want and for calling other conclusions inaccurate. Based on my interpretation of the question, this IS the correct solution.

thats what i have in mind!!!! but this guy only post to make us fight and then he disappears from this discussion and here were are us fighting for someone wo doesnt care!!!! hehehe im just kidding and talking seriously i have a similar solution but with different values of resistances (2.2 KOhm, 2.2 KOhm, 1 KOhm, 590 Ohm) the current is the same (1mA aprox),

The circuit is simple but one more time its important that we know what kind of devices we gonna connect to the resistors, if that devices are OPAMPS or any high impedance IC as say the last comment theres no problem. in the circuit shown, the devices are represented by the voltimeters (which have a very high impedance too) so if you want to connect the devices, just connect it instead of the voltimeters.

Destroyer's method requires a differential connection, complicating the interface between the bias network and devices requiring the reference voltages discussed. Guests method uses a common ground which cuts complexity in half.

"hi, i have a divide the 6volt into three parts 4.4volt, 3.2volt and 1volt then what value of resistance should be added in the circuit"

Sorry i dont read the part that the guy says "all the pottentials must have the same ground" (maybe because he dont say that), i dont know what he needs.......

Here ya go:

100% marks.

Not really any current draw to measure the voltage at any of the taps will alter the voltage division no matter how small, unless it is a purely theoretical exercise and I am being a little bit too specific. But then a lot of questions on this forum are rather ambiguous

Voltage means zero current. If one can not measure it, then that is another problem all together. Put a capacitor, charge the capacitor for very long period and then sense using 0.0001% charge out of it. Even if voltage drop is taking place, the measurement can be corrected using what is lost. More than 1ppm measurement is not some one looking here all that seriously.

Hello All.

How right or wrong is some-one, the question was simple "Potential Divider" with resistors. & we should keep close to the Q. May add some simple comments & thats all.