f28335 Current Consumption

Blake,

The full current consumption information is in Section 6.4 of the datasheet (p 119), Table 6.1. 

The good news is that it is under 500mA.  But not by much.  However, the F28335 has a number of ways to reduce power consumption - the easiest being disabling most of the peripherals (the ones not in use).  And there are also low power modes as well.

Tim

yep i see that now, about 442mA (max) so that really doesn't allow me to do what i need to. 

In this case, i have purchased an isolated DC-DC converter to get the 5V i need, Do you see any reason why i wouldn't be able to keep my c2000 connected to a computer and then have my isolated 5V supply and connect that up to the ground and 5V of the c2000 to give the extra current that i require?

Sorry I don't quite understand what you are trying to get at with the circuit.  Maybe a schematic or even rough picture will help.

However, I wouldn't necessarily write off the 5V USB supply for a number of reasons:

1) The current given is for F28335 all guns blazing.  Most applications will use less peripherals and thus less current.  What peripherals are in use?

2) The F28335 needs 3.3V and 1.8V supplies.  The USB gives 500mA @ 5V = 2.5W of power.  If you use a small DC-DC converter chip (switching) with approximately 85% efficiency you can get maybe 1.2A on the 1.8V supply (IDD) and slightly less on the 3.3V (IDDIO, IDDVFL).  This is more hardware effort than a quick linear reg but will give way more than the required current.

In the end it may be a toss up between simplicity and efficiency.  

Tim

Sorry it was a bit vague, i attached a picture,

effectively what i am thinking of doing is hooking an isolated 12 to 5V dc-dc converter up to the circuit to increase the amount of current on the 5V rail.

That should work.  However I would suggest a couple of things:

1) If the DC/DC converter can take the load by itself I would use it stand alone and not draw current from the USB.

2) If it can't then you can't just connect the 5V lines together.  Either use diodes to connect the 5V lines together or just simply use the USB for any 5V you need and run the 3.3V and 1.8V regs directly off the ISO psu.

Tim

the PSU only provides 600mA so i will need both the USB and the ISO PSU due to a couple of the parts i am using. Can you explain the connection of the lines with diodes, I am not entirely sure what is happening there?

Here is the first option - with the diodes.  Note there should be some capacitors there as well for the regulators but I have omitted them for speed of drawing.

The diodes are there so that there isn't any current flow from the ISO PSU to the USB 5V line when it is not operating.  This could potentially harm the USB host.  It also protects the ISO converter from the USB as well.  Note that the 5V line will be slightly below 5V due to the diode drops.  This isnt a problem when using 3.3V and 1.8V regulators but could be if you need a clean 5V for other signals.  I would however recommend doing something like the second option.  This will eliminate any need for placing the diodes as the two 5V lines are never connected.

Hope that clears things up,

Tim

yea i see where you are going with it, the only thing i am thinking about is that i am using the tmds28335dock which won't really allow me to use the second option, so in that case i will be have to use the first option with the following set-up:

You shouldn't need to.  Mine was just an example of how to split the power effectively.  What I would do is draw the 5VUSB to power just the F28 and nothing else.  As you have found out it should be able to do this.  Now use the ISO 5V to power the 5V circuitry.   This way you shouldnt need to combine the 5V lines.  In your picture there is very little current budget left on the 5V to be sent down the diode anyway so I would just ignore the connection.  Heres a quick drawn schematic to illustrate it:

Tim

that was my initial idea anyway i was just looking for a second opinion and now that i have gotten yours i am more than confident about doing what you have just pointed out.

It will work fine as everything that comes out of the chip is 3.3V so it  is only the inverters i need to power due to my choice of optocoupler being an inverting style optcoupler.

thanks for the help