# Voltage Divider

#### shibin_varghese

Joined Jan 14, 2019
56
Dear Friends,
I have a doubt. It may seem to be very simple for you.
What is the problem in using a voltage divider in Ohmic range other than in Kilo Ohm range?.
Or can I choose a resistor combination in parallel as resistor R1 and R2 in the voltage divider?.

#### MrAl

Joined Jun 17, 2014
6,633

#### DickCappels

Joined Aug 21, 2008
5,944
1. The output impedance is a function of the resistors that make up the divider and the voltage source impedance.

2. Power dissipation is a function of the resistance of the divider.

3. At higher resistances an equalizing capacitor or two may be needed to flatten the frequency response if that is required.

4...

• narkeleptk

#### shibin_varghese

Joined Jan 14, 2019
56
In my case
Vin = 8V
R1 = 330 Ohm
and R2 = 220 Ohm
since my voltage at the controller side is 3.3 V.

#### Picbuster

Joined Dec 2, 2013
979
Dear Friends,
I have a doubt. It may seem to be very simple for you.
What is the problem in using a voltage divider in Ohmic range other than in Kilo Ohm range?.
Or can I choose a resistor combination in parallel as resistor R1 and R2 in the voltage divider?.
Ohm, KOhm , mOhm or M Ohm is no difference however; a load will disturb the ratio.

Uout = U*(r2//load) / (r1+r2//load) the // will tell you parallel connected.
When you work in Ohm a load of 1M will not disturb a lot.
But the trouble occurs when U is constant with a frequency shift.
When the load is an impedance the output voltage will change with the frequency.

Picbuster

#### MrAl

Joined Jun 17, 2014
6,633
In my case
Vin = 8V
R1 = 330 Ohm
and R2 = 220 Ohm
since my voltage at the controller side is 3.3 V.

#### shibin_varghese

Joined Jan 14, 2019
56
I am not intended in making a source using a voltage divider but I am using it as a reference voltage.
I am using a Current Transmitter XTR111AIDGQR and its output current is then passed through a voltage divider as mentioned earlier.
When I use a voltage divider in Kilo Ohm range my output voltage to the controller is different from what it is when it is connected to a voltage divider in Ohmic range.

#### LesJones

Joined Jan 8, 2017
2,359
As the output of the XTR111AIDGQR is a current you do not need a potential divider. You just choose the load resistor to give you the desired voltage at the full scale output current

Les.

• BobTPH

#### shibin_varghese

Joined Jan 14, 2019
56
As the output of the XTR111AIDGQR is a current you do not need a potential divider. You just choose the load resistor to give you the desired voltage at the full-scale output current

Les.
You are correct. But my controller works in 3.3V logic so I need to scale down the voltage.
Can someone please explain what is the problem in using a voltage divider in Ohmic values

#### MrChips

Joined Oct 2, 2009
19,406
A voltage divider has limitations which you must be aware of. The voltage divider formula is:

Vout = Vin x R2 / (R1 + R2)

1) A voltage divider consumes power:
P = Vin x Vin / (R1 + R2)

2) A voltage divider to be used as a voltage reference cannot supply current.
In other words, any load connected to Vout must have resistance much larger than R2.
If the application can allow some degree of voltage loss, then a rule of thumb is R2 should be 1/10 of the load resistance or even less.

3) An alternative to a voltage divider to be used as a reverence voltage is to replace R2 with a zener diode. The value of R1 is calculated using:

R1 = (Vin - Vz) / Imax

where Imax is the maximum current demanded by the load and Vz is the zener voltage.

The power dissipated by R1 is:
PR1 = Imax x (Vin - Vz)

The power dissipated by the zener diode is:
Pz = Imax x Vz

In both cases, select devises that are rated for twice the calculated wattage.

A zener voltage regulator circuit itself has its own set of limitations (which is another topic).

• narkeleptk, MrAl and absf

#### iimagine

Joined Dec 20, 2010
388
If you know your load resistance, you can use it in place of R2.
Vout = Vin * R2 / (R1 + R2)
3.3V = 8V * 100 / (R1 + 100)
Using algebra to solve for R1:
3.3V = 800V / (R1 + 100)
3.3V * (R1 + 100) = 800V
3.3VR1 + 330 = 800V
3.3VR1 = 800V - 330
3.3VR1 = 470V
R1 = 470V/3.3V
R1 = 142.42
All you need is R1 connected in series from Vin to Vout

#### BobaMosfet

Joined Jul 1, 2009
836
Dear Friends,
I have a doubt. It may seem to be very simple for you.
What is the problem in using a voltage divider in Ohmic range other than in Kilo Ohm range?.
Or can I choose a resistor combination in parallel as resistor R1 and R2 in the voltage divider?.
Let's put this in English for you. The purpose of a resistor is to resist the flow of current. The lower ohmic value, resists LESS current. The higher ohmic value resists MORE current. Still with me? That means that if you use low-value resistors, the divider itself is a larger load, consuming more energy itself.

You need to understand that the ohmic value of the resistors isn't chosen arbitrarily (except for teaching examples). The value of the resistors is chosen based on the amount of current the 'load' beyond the divider will need, and what voltage it needs as well. You always gotta see the bigger picture in electronics.

• SamR

#### BobTPH

Joined Jun 5, 2013
2,051
You are correct. But my controller works in 3.3V logic so I need to scale down the voltage.
Can someone please explain what is the problem in using a voltage divider in Ohmic values
It does not output a voltage, it outputs a current of 0 to 20mA.

If you want a voltage output of 0 to 3.3V you use a single resistor calculated by Ohm's law:

3.3V = 20mA * R

R = 3.3 / 0.020 = 165Ω

Bob

• TeeKay6

#### MrChips

Joined Oct 2, 2009
19,406
It does not output a voltage, it outputs a current of 0 to 20mA.

If you want a voltage output of 0 to 3.3V you use a single resistor calculated by Ohm's law:

3.3V = 20mA * R

R = 3.3 / 0.020 = 165Ω

Bob
You meant scale the voltage down by (8 - 3.3) = 4.7V
R = 4.7V / 0.020A = 235Ω

#### BobTPH

Joined Jun 5, 2013
2,051
You meant scale the voltage down by (8 - 3.3) = 4.7V
R = 4.7V / 0.020A = 235Ω
No, I meant exactly what I said. The datasheet states that the output is a currennt in the range of 0 to 20mA. Using a load resistor of 165 Ohms will convert this to a voltage of to 3.3V.

Perhaps I misunderstand something. Where did you get 8 - 3.3?

Bob

• #### MrAl

Joined Jun 17, 2014
6,633
Here are a couple of basic regulators. They help to keep the output voltage more constant.

In the series regulator Vb is made about 0.7v or so higher than Vout.

R4 in the shunt regulator makes the output a little more 'soft' if needed.

#### KeepItSimpleStupid

Joined Mar 4, 2014
3,637
maybe useful info. Maybe not. 5 V/250 ohms = 20 mA

An industrial standard is 0-20 mA and 4-20 mA; Coincidence?

There is another problem that might occur. Voltage sources may only source a few mA because the PLC might have a 1 meg input Z.

With any voltage divider application, there are constraints.