Voltage Divider Help

[UncookedCorn]

I would like to set Node 1 to 2V with a Zenner diode.
Node 2 = 1.25V
Node 3 = 1.16V
Node 4 = 1.083V

What resistor values would I choose?

[radiolistener]

if you cannot solve so easy task by self, what is the reason to learn it?

[UncookedCorn]

That is why I am asking for help.

[ataradov]

For the resistor ladder calculations, assume V(1) = 2 V and simply write out the voltage divider equations for V(2), V(3), V(4). Some of the resistors in the upper or lower half of the divider would be a sum (series) of individual resistors. You will end up with 3 equations and 4 unknowns (Rn values). You must assume one of the values and then you would be able to solve that system to get the rest of them.

In practice, when I need to do something like this, I just make a Python script that loops over all possible values of resistors from the E12 or E24 value series. The script produces the most optimal value provided practical resistor values.

[Grandchuck]

Homework?
Here is an easy solution that avoids simultaneous equations.  Assume R1 is 100 ohms and find the total current at 22 mA.  Assume the zener current is half that and then just use Ohm's Law.

[MrAl]

I would like to set Node 1 to 2V with a Zenner diode.
Node 2 = 1.25V
Node 3 = 1.16V
Node 4 = 1.083V

What resistor values would I choose?

Hi,

There is a very easy way to do this but I am not sure if you have any other limitations.

First, for an example, assume your top resistor R1 is 100 Ohms as you have shown.
Next, assume that the total resistance of the other 4 resistors is 200 Ohms.  That's R2+R3+R4+R5.
Since the zener voltage is 2 volts, that means that the total current through those 4 lower resistors will be 2/200=0.010 amps.
Since you need 1.083 volts at the bottom, the bottom resistor must be 1.083/0.01=108.3 Ohms.
Next, your next voltage is 1.16 volts, and 1.16 minus 1.083 is 0.077 volts, and 0.077/0.01=7.7 Ohms, so R4 must be 7.7 Ohms.
The next voltage is 1.25 volts, and 1.25-1.16=0.09 volts, and 0.09/0.01=9 Ohms, so R3 must be 9 Ohms.
Now we have three of the lower resistors, and the sum is:
108.3+7.7+9=125 Ohms,
and since we assumed the total resistance was to be 200 Ohms, that means the upper resistor R2 has to make up the difference, which is 200-125=75 Ohms.
So you have all the resistors now:
100 Ohms
75 Ohms
9 Ohms
7.7 Ohms
108.3 Ohms

You can then go on to choose values that are near to these from the standard resistors, but if you do not like these values you can scale them up or down.

You also have to pay attention to the minimum and maximum current for the 'zener'  though.

A side note, zeners are not very good at regulating voltages in tight tolerances, and this seems like it might have to be more accurate than what a regular zener can provide.  Because of this, it would be wise to consider using a voltage reference diode instead of a zener, which will provide much better regulation.

Another consideration is the load values for each of the four voltages.  Resistor dividers are notorious for bad regulation with load.  If the load is fixed, you may be able to include that in the calculation for the resistors, but if the load varies you have to make sure the source impedances are much less than the load impedances.  Thus if you have a 100 Ohm resistor as the source impedance then the load should be at least 2k maybe higher depending on how good you need the regulation to be.  For precise regulation you should use a voltage reference diode or voltage regulator for each and every output voltage.

[CaptDon]

Are you going to be drawing any current from those nodes or is this just a homework problem we are solving for you?

[EPAIII]

Yes, there is insufficient information for a definitive solution. Any current that is drawn from any of the nodes will change the divider ratios. If there truly is zero current being drawn from any of the nodes, then you start with the 2 Volt assumption at node 1 and create a resistor string that is proportional to the Voltages you want, starting at the bottom. But there will be an infinite number of solutions.

The string of R2, R3, R4 and R5 in series will drop that full 2 V. So assign a total value of 2 to the four of them. Now, node 4 is 1.083V above ground. So R5 BY PROPORTION will have a value of 1.083. Notice I am not placing any units on this yet. I am just getting the RATIOS of the resistance values.

Next is R4. Now we need the Voltage difference between nodes 3 and 4 to find the proportional value of R4. 1.16V - 1.083V = 0.077V so the proportional value of R4 is 0.077.

Likewise for R3 which is between nodes 2 and 3. 1.25V - 1.16V = 0.09V. So, likewise we assign a proportional value of 0.09 to R3.

And R2: 2v - 1.25V = 0.75V so the proportional value of R2 is 0.75. And we have the proportional values of:

R2 - 0.75
R3 - 0.09
R4 - 0.077
R5 - 1.083

But, as I said, those are not resistances, yet. Next we need to establish the 2V at node 1 with a zener and that means choosing an actual zener and finding a value for R1. Every zener has a minimum current that it must carry in order to function. This value can be found on the spec sheet. But an actual zener has to be chosen. 2 Volts is a low rating so I picked the 1/2 Watt size because that is the most common. Then 5% and thru holes mounting - your choices can vary.

https://www.mouser.com/c/semiconductors/discrete-semiconductors/diodes-rectifiers/zener-diodes/?q=zener%20diode&mounting%20style=Through%20Hole&pd%20-%20power%20dissipation=500%20mW&voltage%20tolerance=5%20%25&vz%20-%20zener%20voltage=2%20V

And the data sheet for one of them:

https://www.taiwansemi.com/assets/uploads/datasheet/BZX55C2V0%20SERIES_E2301.pdf

It shows a test current of 5 mA so I will use that as the minimum current needed. To that we need to add whatever current the string of resistors R2 to R5 will draw. That is the current that will pass through R1. Your supply Voltage is 4.2 V so R1 must drop 4.2V - 2V = 2.2V. If the R2 - R5 divider draws zero current that 5 mA for the zener will be the minimum current through R1.

Then, by Ohms Law, the maximum value of R1 = 2.2V / 0.005A = 440 Ohms.

On the other hand, R2 - R5 will draw some current. So, some current must be added to that 5 mA value. But what? Here's where the infinite number of solutions comes in. From the data sheet the maximum zener current for this diode seems to be 20 mA: all the curves stop there. That is rather small, but I am not going to start over with another diode. That establishes the range of available current between 5 mA and 20 mA or a 15 mA range for the resistor divider R2 - R5. If most of that current is passing through the zener, that is 20 mA there. So R1 also must pass 20 mA. And R1 = 2.2V / 0.020A = 110 Ohms. That is the minimum value of R1 and, with no further design information, that is the value I would choose. BTW, 2.2V X 0.02A = 0.044W for the power rating of R1. And 2V X 0.02A = 0.04W power rating for the zener. This, by the way, is the general way that I calculate a zener regulator. Find the minimum series resistance that allows the maximum desired current to flow through the zener and then let the load bleed off what it wants down to the minimum level where the zener will continue to "zene" properly. The Voltage stays close to the rated zener value between those two extremes.

That allows a minimum resistance for the total series string R2 - R5 of R = 2V / 0.015A = 133.3 Ohms. Going back to the proportional values we calculated above, that would give us the following:

Rs   Proportion  Calculation                   Resistance Value
===================================
R2 - 0.75          133.3 Ohms X 0.75 =          100 Ohms
R3 - 0.09          133.3 Ohms X 0.09 =            12 Ohms
R4 - 0.077        133.3 Ohms X 0.077 =      10.27 Ohms
R5 - 1.083        133.3 Ohms X 1.083 =      144.4 Ohms

But those resistor values are only for the maximum current that the zener that I happened to choose allows. The current can be any lesser value in that Voltage divider string and it would be calculated in a similar manner by choosing the desired current (between 0 and 15 mA), calculating the total of the four resistor string and then multiplying by the proportional values to find the individual values. And, as I said, an infinite number of solutions are possible. Just as another example, if 10 mA current was chosen, then

R2 to R5 would be R = 2V / 0.01A = 200 Ohms.

And

Rs   Proportion  Calculation                   Resistance Value
===================================
R2 - 0.75          200 Ohms X 0.75 =          150 Ohms
R3 - 0.09          200 Ohms X 0.09 =            18 Ohms
R4 - 0.077        200 Ohms X 0.077 =       15.4 Ohms
R5 - 1.083        200 Ohms X 1.083 =      216.6 Ohms

If this is homework, I hope you get an A. But I also hope you learned something.

By the way, this is not the only way to calculate this. Instead of using a proportion you could use Ohms Law and Voltage and current totals all the way. I like the proportions approach because once they are established, the values for different currents are easy to calculate. Also, it would translate nicely to an Excel spreadsheet where cells could be used for any of the input values and you could play around with it easily.

And there could be other considerations in a real world problem.

Are you going to be drawing any current from those nodes or is this just a homework problem we are solving for you?

[Terry Bites]

V=IR!
Decide on the current flow you want in the divider. Lets say 10mA for arguments sake.
Then 10mA =Vnode1/(R2+R3+R4+R5) or (R2+R3+R4+R5)=1.25/.01 =125

R5=Vnode4/.01 =1.083/.01 or 108.3
R4= (Vnode4-Vnode3) =(1.16-1.083)/.01 =7.7
R3= (Vnode3-Vnode2) =(1.25-1.16)/.01 =9.1

If you were to slect from E24 series resistors you'd use:  R5=110, R4=7.5, R5=9.1 The sum of these comes pretty damn close to 125 with 1% reistors.

1.25V Zeners don't exist but shunt and series references do.
eg TS4061AILT-1.25. R1 must be low enough to provide the divider current and the reference operating current to flow R1<(V2-Vref)/(Idiv+Iref)

[MrAl]

V=IR!
Decide on the current flow you want in the divider. Lets say 10mA for arguments sake.
Then 10mA =Vnode1/(R2+R3+R4+R5) or (R2+R3+R4+R5)=1.25/.01 =125

R5=Vnode4/.01 =1.083/.01 or 108.3
R4= (Vnode4-Vnode3) =(1.16-1.083)/.01 =7.7
R3= (Vnode3-Vnode2) =(1.25-1.16)/.01 =9.1

If you were to slect from E24 series resistors you'd use:  R5=110, R4=7.5, R5=9.1 The sum of these comes pretty damn close to 125 with 1% reistors.

1.25V Zeners don't exist but shunt and series references do.
eg TS4061AILT-1.25. R1 must be low enough to provide the divider current and the reference operating current to flow R1<(V2-Vref)/(Idiv+Iref)

Check out reply #5.

[MrAl]

I would like to set Node 1 to 2V with a Zenner diode.
Node 2 = 1.25V
Node 3 = 1.16V
Node 4 = 1.083V

What resistor values would I choose?

Hello again,

Since the dividers will have some sort of load to ground for each node, it helps to fit that into the solutions for the resistors.  These solutions cover the case where there are four load resistors each connected to one node and the other terminal to ground.  That's the usual case, but we could consider other cases for load resistors also.

In the following, r2, r3, r4, and r5 are the load resistors for each node.  r2 connects to R2, r3 to R3, r4 to R5, and r5 to R5.
The voltage E1 is the source voltage (4.2v) and E2 is the chosen 'zener' voltage (original specified as 2v).

The solutions for the resistors are:
R3=((r3*v3-r3*v2)*R2)/(v2*R2-r3*E2+r3*v2)
R4=((r3*r4*v4-r3*r4*v3)*R2)/((r3*v3+r4*v2)*R2-r3*r4*E2+r3*r4*v2)
R5=(r3*r4*r5*v4*R2)/(-(r3*r4*v4+r3*r5*v3+r4*r5*v2)*R2+r3*r4*r5*E2-r3*r4*r5*v2)

and notice R2 is a parameter you choose for the total current draw.

The total resistance in parallel to the zener is:
RT=(r2*(R2*R3*R4*R5+r3*R3*R4*R5+r4*R2*R4*R5+r3*R2*R4*R5+r3*r4*R4*R5+r5*R2*R3*R5+r4*R2*R3*
R5+r3*r5*R3*R5+r3*r4*R3*R5+r4*r5*R2*R5+r3*r5*R2*R5+r3*r4*R2*R5+r3*r4*r5*R5+r5*R2*R3*R4+r3*r5*R3
*R4+r4*r5*R2*R4+r3*r5*R2*R4+r3*r4*r5*R4+r4*r5*R2*R3+r3*r4*r5*R3+r3*r4*r5*R2))/(R2*R3*R4*R5+r3*
R3*R4*R5+r2*R3*R4*R5+r4*R2*R4*R5+r3*R2*R4*R5+r3*r4*R4*R5+r2*r4*R4*R5+r2*r3*R4*R5+r5*R2*R3*R5+r4
*R2*R3*R5+r3*r5*R3*R5+r2*r5*R3*R5+r3*r4*R3*R5+r2*r4*R3*R5+r4*r5*R2*R5+r3*r5*R2*R5+r3*r4*R2*R5+
r3*r4*r5*R5+r2*r4*r5*R5+r2*r3*r5*R5+r2*r3*r4*R5+r5*R2*R3*R4+r3*r5*R3*R4+r2*r5*R3*R4+r4*r5*R2*R4
+r3*r5*R2*R4+r3*r4*r5*R4+r2*r4*r5*R4+r2*r3*r5*R4+r4*r5*R2*R3+r3*r4*r5*R3+r2*r4*r5*R3+r3*r4*r5*
R2+r2*r3*r4*r5)


The total current of the R2 through R5 string with loads r2 through r5 applied is then:
iT=E2/RT

and R1 is chosen as:
R1=(E1-E2)/iT

[Terry Bites]

So there is a harder way of doing it?

[MrAl]

So there is a harder way of doing it?

Hi,

Ha ha, that's funny thanks for the humor I do love comedy really.

Maybe an iterative method?

The methods I used are straightforward but sometimes the solutions come out very long.  One time I did a circuit and ended up with an equation that had so many characters in it it would fill up two pages in a large book.  Maybe 5000 characters maybe more.
The key is that the solutions allow for a one-time calculation that provides the exact answers without pussyfooting around the bush trying this and that and this and that over and over again until we get something close.
This full problem (as invoked later in the thread) involved a set of load resistors and that complicated the solution by a great extent.  That happens sometimes but sometimes there is just no way around it.

To make the calculation even harder but still possibly necessary, we would have to figure out the sensitivities of each voltage with respect to several variables such as the resistor tolerances.  We could also consider the variations if the resistor loads have some variation themselves due to the application requirements.  That could get ugly

[EPAIII]

Sounds a lot like how I approached it. I wasn't aware of how long my method was going to be until I had posted it.

I was trying to show the method and logic involved as the OP was not sure how to approach the problem. If you just provide the shortest way, then the logic behind it may not be obvious, particularly to a novice. So, when a similar problem comes up they may try to follow the simple way but if it fails they are still at a loss as to how to solve that new problem.

[MrAl]

Hi,

Yeah I know what you mean

I guess my shortest answer would be to use Nodal Analysis and then solve for the resistor values.  That goes for the full problem or the no-loads problem.

[Zero999]

I would go for the method given in reply #7.
https://www.eevblog.com/forum/beginners/voltage-divider-help/msg5174904/#msg5174904

If I wanted a practical circuit, I would put the value into a spreadsheet, with the potential divider formula, each time setting the value of one of the resistors to an E24 value and look at the values it generates. Another method would be to multiply all values by 100, like Mr Al did and use an online calculator to work out near-exact values by connecting any combination of E24 values in series, or parallel.

Thinking about it again, multiplying by 200 would be better.

R2 = 0.75*200 = 150
R3 = 0.09*200 = 18
R4 = 0.077*200 = 154
R5 = 1.083*200 = 216.6

Now R2 to R4 are all standard E24 or E96 values. An online calculator can be used to get the exact value of R5 from two resistors.
https://www.qsl.net/in3otd/parallr.html

Gives R5 = 130+86.6 = 216.6

[MrAl]

I would go for the method given in reply #7.
https://www.eevblog.com/forum/beginners/voltage-divider-help/msg5174904/#msg5174904

If I wanted a practical circuit, I would put the value into a spreadsheet, with the potential divider formula, each time setting the value of one of the resistors to an E24 value and look at the values it generates. Another method would be to multiply all values by 100, like Mr Al did and use an online calculator to work out near-exact values by connecting any combination of E24 values in series, or parallel.

Thinking about it again, multiplying by 200 would be better.

R2 = 0.75*200 = 150
R3 = 0.09*200 = 18
R4 = 0.077*200 = 154
R5 = 1.083*200 = 216.6

Now R2 to R4 are all standard E24 or E96 values. An online calculator can be used to get the exact value of R5 from two resistors.
https://www.qsl.net/in3otd/parallr.html

Gives R5 = 130+86.6 = 216.6

Hi there,

This is where the idea of what the load resistances would came up from.
What is to prevent us from using 1 Megohm for R1, or 10 Megohms, or 100 Megohms, or 1 Gigohm?  It's the load resistance values.
Conversely, what is to prevent us from using 10 Ohms for R1, or 1 Ohm, or 0.1 Ohm, or 0.01 Ohm, etc.?  It's the load limit on the source voltage (4.2 volts).

This of course means to do this and get any results that are actually useable in real life, we have to know more about the system.  Without that additional information, we are doomed to a lifetime of trial and error, and probably never get it right.  We would have to be able to change all the resistors for each new test to see if it worked out right.

Speaking about reality, a resistive voltage divider is almost never a good answer when it comes to voltage regulation.