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