• State TI Thinks Resolved
  • Locked Locked
  • Replies 1 reply
  • Answers 1 answer
  • Subscribers 20 subscribers
  • Views 220 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Original question:

NE555: the exact power dissipation of NE555P for a normal working condition.

NE555: Power consumption

Cedrick
Expert 8270 points
Part Number: NE555

Hi, 

Good day. I hope that you are doing good.

Our customer want to know how to calculate the power consumption of NE555 and I came across to this E2E post (https://e2e.ti.com/support/clock-and-timing/f/48/t/851806) . I just want to clarify the information provided on that post. 

Base on the post, we can estimate the power consumption by adding the positive value factor. Let say for example for Vcc = 5V.

Does it mean that for low level output the computation would be: (5V * 6mA) + (0.4V *8mA) = 33.2 mW

And for high-level output, the computation would be: (5V * 5mA) + ((5V - 3.3V) * 100mA) = 195 mW





I would like to know if my understanding is correct. IF not, can you please let me know the right way. 

Furthermore, I would also appreciate if you could let me know the formula for astable mode. It seems that the frequency is missing. 

Thank you and have a great day.


Regards,

Cedrick

  • 0
    TI__Guru*** 140106 points

    Cedrick,

    Power dissipation and power consumption are not the same thing. The dissipation is the power in the device that makes internal heat. Power consumption is power drawn from the power supply. 

    For monostable, the output will likely be low most of the time. Power consumed is VCC * IQ = 5V * 6mA[max] + VCC^2 / RA. Output current must also be added. For a RL also connected to VCC, add VCC * IRL or  VCC^2 / RL.

    For output high, power consumed is VCC * IQ = 5V * 5mA[max] + VCC * IRA. IRA (resistor current) drops from VCC/RA to 1/3*VCC/RA as the capacitor charges. Output current must also be added, VCC * IOUT.