Thursday, April 10, 2014

Confirming Power Dissipation

As I go through to finalize the next rev of the PCB I though I would revisit my heat calculations.  There are two (three really) sources of heat in the regulator:
  • Darlington pre-voltage regulator (Q1)
  • Final Voltage Regulator (U1)
  • Field FETs (Q2 / Q4)

I calculated the Power Dissipated in Watts for each of these, and here are the results:







For the power supplies I assumed a current draw of 165mA, this is approximately  two times the expected maximum internal current draw. 






FET power dissipated is dominated by static losses (simple "P = I^2 * R" calculations).  The Ron of the FETs I adjusted for a 100c junction temperature.  And it should be noted, this is per-FET (remember, there are two of them).  Switching losses (dynamic Pd) is so small; primary because our PWM frequency is very low at 144hz - the max switching loss I calculated was 3.4mW, so we can in effect ignore switching losses.


At first glance the worst case appears to be 2.75W + 2.75W + 12W + 1.3W =  18.8 watts.  The heat-sink I selected for this design (Hongfa HF92B-120) has a Thermal Resistance of 1.1C/w.   At 18.8w we would expect the heat sink to gain 18.8*1.1 = 21c.  Put this into our 120f engine room we get to a nice warm 160f heat sink.


In reality  it is unlikely we will see both the system voltage AND the field current at max at the same time.  After all, if indeed we were running at 80V and drawing 30A of field current, this would be 2.4Kw of power JUST TO DRIVE THE FIELD!!   More likely is as system voltage increases we would see lower Amps drawn by the field. 



Lets take a look at a couple more realistic extreme examples, a single Alternator in a 12v:
  • 12v system
  • Single large frame alternator drawing 10A field  (This is about 2x what a common alternator will draw)
  • 100mA system draw (you will have to trust me on the new Power Supply figures I use below)
 Now total power becomes: 0.3W + 0.3W + 0.4W + 0.75W = 1.9W, or a 2c temperature rise. Now in our 120f engine room we are running at only 124f.  Will not even notice it.



Or, look at a this:
  • 48v system
  • Single large frame alternator drawing 3A field (a special 48v alternator - Higher volts means lower Amps)
  • 100mA system draw.
Comes to: 0.03W + 0.03W + 4.3W + 0.75W =  5.1W  comes out to perhaps 130f.




What about a realistic 'heavy' system?
  • 24v system 
  • 2x large frame alternators in parallel - 20A field current.
  • 100mA system draw.
     1.2W + 1.2W +  1.75W + 0.75W = 4.9W   --> gets us again  up to 130f in a 120f room.

What is the secret?  Looking at the power equation:  P = I^2  * R, to lower Power we can lower I or R.  The 1st was done by using two FETs.  Cut the field current in half between the two (FETs are largely self-balancing when ran in parallel).  We get a ton of bang from this due to the squaring of current in the power equation.   Next I selected a FET with low Ron.  With the low PWM frequency we can trade off higher gate capacantance to get a low Ron and not have to spend $7 per FET.

So, there it is.  Large Field currents, high system voltages, and looks like the Arduino Alternator Regulator will keep its cool.



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