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### Topic: Even voltage output steps with parallel resistors (Read 8941 times)previous topic - next topic

#### jtpek

##### Feb 02, 2015, 10:39 pm
Hi,

I´ve been wondering for days how to build a circuit with resistors (parallel connection), which gives output steps with equal sizes. I would like to divide the maximum voltage of 5 volts into 10 steps..so the output voltages should be 0.5V, 1V, 1.5V, 2V, 2.5V, 3V, 3.5V, 4V, 4.5V, 5V.

---R1---?
---R2---?
---R3---?
---R4---?
---R5---?
---R6---?
---R7---?
---R8---?
---R9---?
---R10--?

Great illustration, but I think you know what I mean

Thanks in advance for any help !

#1
##### Feb 02, 2015, 10:58 pm
You need series connection, not parallel.
Put 10 1K resistors in series. The voltage any junction will be:
5V* (sum of resistors below the junction)/(sum of resistors above + sum of all resistors below)

Or simply, Vout = Vin * R2/(R1 + R2)

Vin -- R1 -- Vout -- R2-- Gnd

Vin - 0 ohm (5V ) - 1K - 4.5V - 1K -4V - 1K - 3.5V - 1K - 3V - 1K - 2.5V - 1K - 2V -  1K - 1.5V - 1K  - 1V- 1K - 0.5V - 1K - GND

1K works out a little low,  800 ohm works out better.  Try it.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

#### jtpek

#2
##### Feb 02, 2015, 11:37 pm
Thank you for your quick and correct response!

However it didn't answer my question, because I forgot to mention the most important part of it

I would need a circuit, in whitch I could add and remove resistors and the steps would be 0.5V. For example, if my circuit has only resistor R1 the output voltage would be 0,5V. With resistors R1, R2 and R3 in parallel the output would be 1,5V..with R1 to R5 it would be 2,5V and so on.

I can't get the maths right. Again help would be really appreciated!

#### jackrae

#3
##### Feb 02, 2015, 11:40 pm
Be aware that such a chain divider isn't suitable for the supply of a load current and is only suitable as a reference voltage.  Simply put, your external load should be at least 10 times the value of your total resistor chain to minimise any effect it will have on the chain step accuracy.  For the chain that CR has shown, your "load" should, ideally, be greater that 100kohms.

Also be aware that the accuracy of the steps is dependant on the accuracy of the resistors used, so if you buy +/- 5% resistors, the step inaccuracy may, under worst case conditions, be around 5%.  For example, say the first 9 resistors in the chain are all +5% and the last one is -5%, then your chain values for the 0.5v tap will be 9450ohms and 950 ohms so the output could be 5 x 950/10400 = 0.457 volts, not the 0.5 expected.

#### jremington

#4
##### Feb 02, 2015, 11:42 pm
What do you really want to do?
Your idea of adding or subtracting resistors is probably not the way to do it.

#### jackrae

#5
##### Feb 02, 2015, 11:43 pm
You simply connect 10 resistors in SERIES and the junction of each gives to the 0.5 volt taps.

#### Grumpy_Mike

#6
##### Feb 02, 2015, 11:50 pm
Quote
For example, if my circuit has only resistor R1 the output voltage would be 0,5V. With resistors R1, R2 and R3 in parallel the output would be 1,5V..with R1 to R5 it would be 2,5V and so on.

I can't get the maths right.
That is correct the requirements are not possible. That is why the maths won't work because it can't be done.

#### KeithRB

#7
##### Feb 02, 2015, 11:58 pm
I think what you need is a basic Op-amp mixer  or summing amplifier circuit:

The resistors are connected to a virtual ground so they don't influence each other, and you can choose to use any combination to get the voltage you want.

#### jtpek

#8
##### Feb 03, 2015, 12:01 am
I can't use serial connection in this case. Is it possible to somehow calculate the resistor values needed? I know that the accuracy of the resistors are not that good, but it would be good enough for me, though I just need steps big enough to properly measure which resistors in the chain are connected and which not. Thank you all for your responses.. But please help me with the calculations :/

I know I could use serial connection and maybe use someswitches or something, but only the parallel-connection way would suit my project.

#### jtpek

#9
##### Feb 03, 2015, 12:02 am
Is it really so that it can't be done ? :/ Has someone an explanation why not?

#### KeithRB

#10
##### Feb 03, 2015, 12:18 am
Maybe I don't understand what you want, but what is wrong with a summing amplifier:
https://en.wikipedia.org/wiki/Operational_amplifier_applications#Summing_amplifier

You can change the gain so that each output generates a different voltage if you want.

#### Grumpy_Mike

#11
##### Feb 03, 2015, 12:22 amLast Edit: Feb 03, 2015, 12:27 am by Grumpy_Mike
Look at the formula for resistors in parallel, you will see there is no way that it can do what you ask.
1/Rt = 1/R1 + 1/R2 + 1/R3 ..........

What is your real problem? Because this is sounding like an X-Y problem.

#### polymorph

#12
##### Feb 03, 2015, 12:44 am
You need something called an R2R Ladder. It is a simple DAC (digital to analog converter).

You must count up in binary to get 0.5V steps. To go all the way to 5V, you'll need a rail-to-rail output Op Amp.

http://www.tek.com/blog/tutorial-digital-analog-conversion-%E2%80%93-r-2r-dac
Steve Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - https://tinyurl.com/Technote8

#### larryd

#13
##### Feb 03, 2015, 12:52 am
@polymorph
I remember the days when you could buy resistors made from carbon.
I would take a Dremel tool and cut-off-wheel to trim the values to R an 2 R.