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I’m learning to build clocks using sequential logic. One thing I don’t quite understand from the data sheets is the max input and output current that comes out of ICs like the CD4060 and CD4026.

I’ve chosen to run my project on 5V DC. I want to operate the ICs within their recommended currents but I don’t really understand all of the values listed on the data sheets.

For instance if I power my CD4026 on X amount of current, what will the current be on those output pins which go to my LED displays? Equal to the inputs?

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    \$\begingroup\$ Have you attempted to read the datasheets? The recommended drive circuit for those 7-segment outputs to drive an LED includes a transistor per segment either as an emitter-follower (CC) or a common-emitter amplifier (CA display). With a transistor you can easily get the few mA a modern LED display requires. The 4000 series has very little drive capability with a 5V supply, less than 500uA with 0.5V drop over temperature. \$\endgroup\$ – Spehro Pefhany Jan 1 at 7:21
  • \$\begingroup\$ The display i’m using recommends 10mA. So if I use 15v instead could I get away with 6.8mA without transistors? I read the data sheets many times. \$\endgroup\$ – StrugglingStudent117 Jan 1 at 7:28
  • \$\begingroup\$ HC4511 on 5V would probably be a better choice. The visual difference between 5mA and 10mA is not that great. \$\endgroup\$ – Spehro Pefhany Jan 1 at 7:30
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I am referring to this datasheet. It is CD4060B part from TI

You can read the current capability of the IC from the specification table in the datasheet.

enter image description here

All examples are for \$ VDD \$ of 5V supply.

  1. Outputs sinking current - marked in yellow color (Example: for common Anode LEDs)

The typical current the IC can sink is only 1mA while the output voltage can raise up to 0.4V. If you need more current to be sinked the output voltage will rise. The remaining voltage will be available for the load.

  1. Outputs sourcing currents - marked in green color (Example: for common cathode LEDs)

On the same lines as the source current demand increases, the output voltage will start falling from 5V up to 2.5V when the current load is about 3.2mA.

Hence, you have to consider these currents and the voltage drop while choosing the LEDs or the driver ICs.

  1. Display type - as the typical current recommended for the display is about 10mA, a FET or a BJT can be driven by the IC. The BJT will support easily in sourcing the current needed by the display.

Please refer to this graphs, which will you give an idea of capability of the IC with respect to the source and sink currents

  • with 5V supply, the current cannot be expected to be more than couple of milliamperes.
  • with 15V supply, there is a possibility to support 10mA for sure. also, note the drop in the output voltage (Drain to Source voltage drop).

enter image description here

enter image description here

If the output current is 6.8mA at 15v and my display requires 10mA, is there even a need for a resistor between the IC and the output pin?

The display needs 10 mA --> The 10 mA current you should control. The display is nothing more than a diode. It is like connecting a diode to a output pin of the driver. It is similar to shorting the output pin of the driver IC to ground.

The current should be controlled using a resistor in your case.

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  • \$\begingroup\$ " If you need more current to be sinked (6.8 mA), the output voltage will raise up to 1.5 V. " - the chart does not show this (6.8mA is with Vdd=15V). fig. 4 shows minimum sink current of about 3mA at 2.5V. Where did the graphs come from? \$\endgroup\$ – Bruce Abbott Jan 1 at 8:28
  • \$\begingroup\$ @BruceAbbott fixed. it was my mistake \$\endgroup\$ – User323693 Jan 1 at 8:29
  • \$\begingroup\$ Excellent information. While it seems I could power my 10mA display with a 15v supply, I would like to look into providing more current for my future projects which may not use the same display. The issue I now face is that I would need a total of 7x4 = 28 transistors. Another post on here was recommended using the UDN2981 in place of 8 transistors, but can this be confirmed? My setup is basically: Quartz crystal with CD4060 + D flip flop to generate the signal, then around 4 CD4026 to drive my displays. So the driver would have to be able to accept input from the CD4026 \$\endgroup\$ – StrugglingStudent117 Jan 1 at 17:01
  • \$\begingroup\$ @StrugglingStudent117 the driver ICs are definitely recommended. I am not aware about your constraints. As per your question it was needed to check whether the CD4060 can be made to work or not. Actually, I too feel that using s driver IC is good idea. If you can post another question with details of your schematics and the LEDs definitely me or someone will help you verify your design \$\endgroup\$ – User323693 Jan 1 at 17:05
  • \$\begingroup\$ How does everyone post those schematics on here? Is there software to draw them? I will gladly post the final design once I get there. \$\endgroup\$ – StrugglingStudent117 Jan 1 at 17:11

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