I want to add a fuse to a self-designed circuit, but for that I need the maximum current of the circuit. In this circuit I have some ICs, some resistors, some LEDs and some capacitors. When I want to calculate the maximum current for this circuit, is my approach correct so simply take the maximum current each IC wants (according to data sheets), and then add them up? Or do I miss something with this approach? Unfortunately the other answers (as far as I saw) are not valid for my problem.
How to calculate the maximum current a circuit needs
currentmaximum-ratings
Related Solutions
The current through the LED is limited by the "22R" resistor. Lowering the resistance value will increase the current and thus brightness. make sure the PNP transistor is also rated for the current. A "darlington pair" of transistors has a much higher gain and can be used for higher-current applications.
Current will be (6 - 2.0 - 0.2) / R.
- 6v supply
- 2.0v (approximate) drop across a red LED (blue may be up to 3.3v)
- 0.2v drop across the collector/emitter transistor junction
The cycle time will be a function of the 10u capacitor and 22R. Lowering the capacitance and/or resistance will make it cycle faster (I think :-).
Absolute maximum ratings are there to tell you what is (pretty much) guaranteed to damage your IC if exceeded, they are not recommended for normal use (notice the "at these or above..." in the first clip below)
All decent datasheets will have recommended operating conditions, which are the ones you want to go from rather than the absolute maximum values. For most small micros, 15-20mA is a typical maximum operating condition.
In the datasheet for your uC, it gives a table showing the OH and OL voltages over current, which both range up to 20mA. Note the voltage rise/drop at 20mA and you can see exceeding this will cause the OH and OL levels to go out of spec compatibility wise (example shown for 5V supply - things are different for 3.3V and 1.8V)
For example you can see at 25°C, the output low voltage is at around 0.5V @ 20mA (more with higher temp)
So, I would assume 20mA max for the outputs (at 5V - see other tables for 3.3V and 1.8V). Also take note of the little notes under the tables detailing the guaranteed hi/lo input values and combined current ratings.
Basically decide on your operating conditions (temp, voltage etc) read all the tables carefully and do the math to make sure you keep within specs.
If you are planning on driving some LED displays, either use an appropriately rated driver chip or add some discrete transistors on each pin to drive the LED. Unless you are only using a couple of low current indicator LEDs, this is generally the way to go.
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Best Answer
Adding all the worst case currents gets you your worst case current. Unless you need to trim the fuse as tight as possible to your actuall current, go with a little margin above this worst case.
Plain old melting-link fuses aren't for dealing with a little over current. They are for a catastrophic condition where there is so much current that something would break, catch fire, or the like.
These kind of fuses aren't for making sure a 1 A circuit doesn draw more than 1.1 A, for example. If that's what you're trying to do, you need a electronic fuse, possibly with a passive fuse as backup for the the really catastrophic case. Take a look at the datasheet of a melting-link fuse. You will see they have a wide tolerance and relatively slow response.