You sure about that? The ‘Lil Professor says the junction temp is about 80C with 25C ambient. About 1/2 watt dissipated in the output typ. Not counting discharge pin dissipation but that’s pretty small. I couldn’t find ‘blows you away’ in the datasheet but the numbers are nicely under Absolute Max Ratings.What use is an output voltage of 3.5v ? You could only use it to power the motor on an electric bike at 20mph.
Test the chip at 12v and 15kHz and feel the heat. At 120mA you can just keep your finger on it and 200mA "blows you away."
The supply current isn't a continuous 100 mA. How can it be?You sure about that? The ‘Lil Professor says the junction temp is about 80C with 25C ambient. About 1/2 watt dissipated in the output typ. Not counting discharge pin dissipation but that’s pretty small. I couldn’t find ‘blows you away’ in the datasheet but the numbers are nicely under Absolute Max Ratings.
No. I am just making everything up. I just burnt my finger and all the skin has just peeled off.
How is it that: when you load the 555 with 100R on 12v, the output is 10v and the current is 100mA from the supply and the output is 50:50 The load is only getting an average of 50mA? Where is the other 50mA going?
This is the sort of questioning that no-one has ever thought of.
That's why the chip gets so incredibly hot.
It isn't "going" anywhere. the second "50" is the percentage of time when the output is low and there is almost zero current into a ground-connected load. No current = no power dissipation.How is it that: when you load the 555 with 100R on 12v, the output is 10v and the current is 100mA from the supply and the output is 50:50 The load is only getting an average of 50mA? Where is the other 50mA going?
I completely agree.This is the sort of questioning that no-one has ever thought of.
May be. So I’ll ask - where’s the other 50mA going? The chip will be hot in either case so we should dispense with that as an indicator.It isn't "going" anywhere. the second "50" is the percentage of time when the output is low and there is almost zero current into a ground-connected load. No current = no power dissipation.
You completely missed the point.
I haven't done the math, but I believe with the right timing components (namely, a big capacitor and a low ohms resistor), you could drive substantial current into them at higher frequencies. This would be current from the supply not appearing in the output.I’m not trying to provoke an argument but I am kind of a specs and numbers guy and I’m interested in how those are off by 100%.
I guess so but I was thinking more along the lines of a good design. Burning 40-50ma in a 15KHz timing RC says more about the designer than whether the chip is suitable for the task. And no, I am neither a fan nor an advocate of 555s. I just never experienced that level of internal power consumption when I did use them so I'm curious about that and want to ensure that questions get good answers.I haven't done the math, but I believe with the right timing components (namely, a big capacitor and a low ohms resistor), you could drive substantial current into them at higher frequencies. This would be current from the supply not appearing in the output.
The current in the timing RC circuit is not the same current at the pin-3 output.I guess so but I was thinking more along the lines of a good design. Burning 40-50ma in a 15KHz timing RC says more about the designer than whether the chip is suitable for the task. And no, I am neither a fan nor an advocate of 555s. I just never experienced that level of internal power consumption when I did use them so I'm curious about that and want to ensure that questions get good answers.
How about at "high" frequency, something over 100kHz?Why should the 555-timer chip get hot?
Test #1
LM555 @ 5V supply, 15kHz 50% duty cycle, 15Ω load at pin-3 to ground.
High voltage = 3V
Assuming average power is 50%
power through load resistor = 0.3W, temp = 40°C
power through IC = 0.2W, temp = 30°C
Test #2
LM555 @ 12V supply, 15kHz 50% duty cycle, 56Ω load at pin-3 to ground.
High voltage = 10V
Assuming average power is 50%
power through load resistor = 1W, temp = 70°C
power through IC = 0.2W, temp = 30°C
(Temperatures measured with IR thermometer.)
I don't think there will be any difference but I will check tomorrow.How about at "high" frequency, something over 100kHz?