# Voltage divider circuit

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#### khan1994

##### Newbie
Hi Seniors i want to calculate R1, R2 and C .

How i can calculate if vin= 0-50v and Vout=0-3.3v and fc=5kHz.
I have attached circuit as below

Looking forward to kind response

Hi,

no circuit attached. Just press the "insert image" button.

A voltage divider basically is independent of frequency.
So what is fc used for?

Klaus Thank you for your reply. Fc is used to calculate the Capacitor value.

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First part
The circuit i provided i want to measure a signal that is expected to be
in the 0V-50V range. so what resistor values for R1 and R2 to allow an ADC with a +3.3V reference to
accurately measure this input.

second part

I want to Choose a filter cap for an analog input. if I used R1 and R2 from first Part
what will be the value for C1 if the sampling rate of the input will be 10kHz or 5Khz.

Thank you so much for letting me know about image size. In future I will take care of that.

Looking forward to your kind response.

Kind regards

#### Attachments

Hi,

Aah, you want to use C2 as low pass filter. That´s the information we need.
(You could use the capacitor in a voltag divider for a scope to adjust for stray capacitance...for example)

Also new and good information. You want to use it in front of an ADC.

1) For simple voltage dividers: you may do a internet search for "how to calculate a resistive voltage divider"
or "online voltage divider calculator".
I get 100 million hits. All basically tell the same: the voltages bahve like the resistors and vice versa.
So if 50V is input (voltage across R1 and R2) and you want 3.3V across R2 then there is 50V - 3.3V = 46.7V
across R1. Simple math, you should be able to do it on your own. (or use the online calculators)

Theretically you are free to choose the value for one resistor and calcuate the other resistor. There theoretically are no resitrictions for resistor values. Every resistor can be >0 Ohms (it can be microohms, milliohms, ohms, kiloohms, megaohms).

Like
* where does the (50V) signal come from? What is the source impedance, what´s it´s frequency range and what it´s voltage ripple, what ar the limits for current drawn... We can not give you any answer about this.
* where does the signal go to. Which ADC exactly? Is a buffer (opamp) used? what accuracy do you expect (in value, not as text), what resolution and precision do you expect ... over which bandwidth.

A voltage divider could be used for:
* to check battery voltage of a 48V driven stacker (huge battery). Mainly DC voltage, low ripple, no high precision needed, corrent not problematic
* get the average voltage of an PWM´d 48V signal. High voltage ripple that needs to be suppressed...
* check the output signal of an audio amplifier. Maybe you want to see signals in the range of 1kHz...(not to be suppressed)
* you may check the capacitor voltage of a tiny energy harvesting circuit. Current is critical, precision not that important, .. fast reaction...

There are countless applications with all different requirements.

*****
Btw: fc usually is used as cutoff frequency. With a RC circuit this means where the amplitude of the input sine is suppressed to 70%.
Fc of 5kHz and sampling frequency of 5kHz is (in my eyes) no good idea at all.
fc should be (much) lower than half of the sampling frequency. (unless you are explicitely using undersampling techniques)

Example:
Let´s say you want to get less than 1LSB ripple at a 10 bit ADC with a full voltage 1kHz PWM input....
then your (1 LSB is about 1/1000 on a 10 bit ADC) fc should be also 1/1000 of the input frequency
fc = 1/1000 of 1kHz = 1Hz (rule of thumb)

Klaus

• khan1994

### d123

Points: 2
ADC datasheets (e.g. for microcontrollers with built-in ADC) often have specification of maximal source impedance. Or input bias current can be specified. These info is necessary to decide about voltage divider impedance.

I agree that a low pass filter may reasonable. Besides sampling rate, the used input signal frequency range should be known.

Hi,

Quick maths, not knowing current needed out of divider:
3.3/50 = 0.066

Select Rtotal, e.g. 10k

10k * 0.066 = 660R, this is R2 (lower resistor in divider)

Therefore R1 = 9.34k (upper resistor)

Check, 50V * (660/10,000) = 3.3V.

--------------------

Klaus said aim for a low pass filter fc = 1/1000.

So 1/5 kHz = 5 Hz

fc = 1/(2 * pi * R * C)
So, C = 1/(2 * pi * R * fc)

C = 1/(2 * 3.14 * (9.34k//660R) * 5 Hz)

C = 1/(2 * 3.14 * 616R * 5 Hz) = 5,16736 E-5 Farads

In human, rather than scientific notation, C = 52 nF.
---------------------------------------

Bonus maths: I inrush = C * (dV/dt)

I inrush = 52 nF * (50V/? in seconds)

But, we know that 1 second/5 Hz = 0.2 seconds. As I have no idea if it's a 50% on/off square wave or a constant signal, I'll assume constant and not divide 0.2s to 0.1s. I'll throw in that dt depends greatly on signal rise time, so 13 uA is (much) lower than a fast rise time signal would generate.

So, anyway, 52 nF * 250 = 13 uA 'inrush'.

And, 50V/9.34k = 5mA.

Hope rough calculations help you get the calculating done and aid in considering aspects beyond merely Rdiv and fc. We haven't even factored in resistor and capacitor tolerances and their ppm/°C for the lowest, typical, and highest Vdivs and fcs we might encounter, unless the circuit only operates at that fairytale, elusive, perhaps even mythical place called 25°C.
--- Updated ---

Dohhh... dV the capacitor will see is 3.3V maximum, not 50V, my mistake. Excuse me, I'm very hungry...

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Hi
Klaus said aim for a low pass filter fc = 1/1000.
this is not generally true. it was for the PWM and for 0.1% ripple.

If it is used for audio you could use a much higher fc.

Klaus

• d123

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