Voltage Divider Calcs - Page 1
Forum section: electronics / beginners

author: Mad Mart
author_rank: Contributor
timestamp: July 20, 2014, 01:13:00 pm
content: I'm trying to get a circuit to work which includes a 4026B (Texas Instruments) decade counter which hooks up to a seven segment display. The spec sheet gives a range of -0.5V to 20V for supply voltage, -0.5V to +0.5V for inputs and +/- 10mA current for the inputs. I have a 9V supply connected and I'm trying to use a voltage divider to get within the correct voltage range for the inputs. Can someone help me with the maths please?

author: Fank1
author_rank: Regular Contributor
timestamp: July 20, 2014, 01:18:55 pm
content: Read closerThe actual input range is -.5 to (VDD + .5)In this case with a 9V supply its -.5 to 9.5

author: mij59
author_rank: Frequent Contributor
timestamp: July 20, 2014, 01:34:08 pm
content: Hi,Please post the circuit diagram of your project.

author: edy
author_rank: Super Contributor
timestamp: July 20, 2014, 01:34:50 pm
content: I've attached the datasheet...I believe the 4026b can drive the 7 segment display directly, or would it be better to drive transistors which then turn on the LED's and have their own individual resistors as well? Then you would choose a voltage supply separately for the counter and your LED supply line (and choose appropriate resistors to get the right current).I am not sure if the counter can actually power the 7-segment display directly... 7 LEDs running at the same time, means it must output a good amount of current... and how the current output varies with voltage supply? Again I think you would still need resistors on your display unless the counter has correct output of V/mA on each pinout to the display... will need to read data sheet closer.

author: edy
author_rank: Super Contributor
timestamp: July 20, 2014, 01:42:50 pm
content: Look at page 9 of attached data sheet. You would supply your 7-segment display with a separate rail of your choosing as long as you figure out the voltage drop of the transistors and each LED and pick out a resistor to manage the current you need to limit for the LED. The counter gets its own supply which seems to recommend anywhere from 3v to 20v. You don't really need a voltage divider... just use 9v to supply both counter and display but choose the resistors to limit current through both.

author: mij59
author_rank: Frequent Contributor
timestamp: July 20, 2014, 01:43:52 pm
content: Quote from: edy on July 20, 2014, 01:34:50 pmNo you can't direct drive the led display.Take a look at page 9.

author: edy
author_rank: Super Contributor
timestamp: July 20, 2014, 01:45:24 pm
content: Thanks just noticed,look above...

author: mij59
author_rank: Frequent Contributor
timestamp: July 20, 2014, 01:46:31 pm
content: Quote from: edy on July 20, 2014, 01:42:50 pmRecommended power supply range is 3..18V.

author: Mad Mart
author_rank: Contributor
timestamp: July 20, 2014, 04:29:19 pm
content: I've attached an extract from my schematic and the actual data sheet for my IC. When I run this schematic in Circuit Wizard it works fine. On the breadboard, however, the display reads '0', which is what I would expect, but on pressing the switch the IC isn't incrementing to the next digit.

author: mij59
author_rank: Frequent Contributor
timestamp: July 20, 2014, 04:49:40 pm
content: Hi,You need to connect a pull down resistor to the clock pin and Vss.When pressing the switch the counter may increment more than one count.Check the state of the CE and R signals.The counter can not drive directly a led display check the data sheet.

author: Mad Mart
author_rank: Contributor
timestamp: July 21, 2014, 07:23:40 pm
content: Cracked it...or should I say I found the answer to why my counter wasn't incrementing. I didn't have pins 2 & 15 connected to 0V. I found the answer here...http://www.elektroniksforkids.com/projects/view/PRO116/Digital-Pulse-Counter-Single-Digit.phpI can confirm that the 4026 IC does power the seven segment display and there is no need for current limiting resistors to the segments (LEDs). I have connected the IC directly to the display and, with a 9V supply, I'm getting 1.92V with the current varying between 12.4 and 12.65mA on the outputs (varying from output to output, not varying on the same output).I stand to be corrected as I'm new to electronics.I would still like to know how to work out the voltage divider problem though. If I have Vin of 9V, how do I work out what resisters I need to achieve a Vout of 0.5V?

author: IanB
author_rank: Super Contributor
timestamp: July 21, 2014, 07:43:56 pm
content: Quote from: Mad Mart on July 21, 2014, 07:23:40 pmThe maths for voltage dividing is easy in principle. Voltages distribute according to resistances. So for your example,8.5 : 0.5 is as R1 : R2E.g. If you picked R1 = 8.5 k and R2 = 0.5 k you would have 0.5 V in the middle.In reality, you have to pick a total resistance of R1+R2 that does not draw too much current from the supply, but at the same time can provide enough current to the load sensing circuit. Also you may find that exact values are not available in the ratio you desire, so you have to experiment a bit to find close values.

author: Mad Mart
author_rank: Contributor
timestamp: July 21, 2014, 09:23:38 pm
content: Thanks Ian. Quite simple really. So you would just adjust the resistor values proportionally to get the current as near as required.