Power Examples

Real life examples of power calculations. For the theoretical background see the power page.

 
 

The device

The ULN2001 is a general purpose Darlington driver, there are seven drivers in a package. Each one is capable of switching 500mA and they can be paralleled up to share the load. So a beginner might think that if I parallel all seven drivers I can switch 7 * 0.5A = 3.5A or I can switch 7 loads of half an amp (500mA) each. Well the device would take it as far as current rating but what about power. A look through the data sheet shows the maximum junction temperature is 150 0C and that the thermal resistance junction to ambient is 70 0C / watt. Here they are assuming an infinite heat sink where the case is at the same temperature as the ambient. So how much power is being dissipated? We know that we are switching 3.5A so that is the current but what is the voltage? Well this is the voltage across the emitter and collector of the transistor when it is turned on fully, or as we say saturated. Here the data sheet only gives figures for currents of up to 350mA and we are switching 500mA. Fortunately there is a graph in the data sheets showing this voltage for all current values and we find that it can be almost 2V. Most parameters like this have a spread and to ensure every component works you have to take the most pessimistic figure. So each Darlington driver is dissipating 0.5 * 2 = 1W therefore for all the drivers going at once we have 7 Watts of power being dissipated.

So at 70 degrees per watt we have a junction temperature rise of 70 * 7 = 490 0C.

Now that’s a rise, if we are siting in a 20C ambient the rise gets added on to that, to give us a junction temperature of 490 + 20 = 510 0C. This is way in excess of the 150C maximum junction temperature, in order for this device to switch so much current and the junction temperature to stay within limits, we would need to be at an ambient of  150 - 490 = -340 0C which is colder than absolute zero. Clearly this is not on.


So how much then?

So we know that switching 3.5A is within the current capabilities but not the power dissipation for this device so how much current can I switch with it?

This is a much more difficult question as it involves deciding the maximum temperature you want the device to run at. That in turn depends on the ambient temperature, the heat sink you use and the amount below the maximum junction temperature you want to run. This reduction in maximum temperature is called a de-rating or a margin, and it affects the reliability and life of a component. So let’s make some design assumptions or simple guesses.

Suppose we make the maximum case temperature 50 0C that’s just about too hot to hold your finger on but not hot enough to burn. Next, suppose we don’t want to run the junction temperature hotter than 130 0C, let’s see where this gets us.


We can stand a 130 - 50 = 80 0C rise in temperature. With a thermal resistance of 70 0C / watt that gives us a maximum power of 80 / 70 = 1.14 Watts. We know that:-

W = E * I

So simply re arrange this to give us:-

I = W / E

Therefore with 1.14 Watts and 2V we get a current of 1.14 / 2 = 570 mA, that is total for all devices in the package. If we restrict the maximum switching current to 350 mA then the saturation voltage comes down to 1.6V and we get a current of:-

1.14 / 1.6 =  712 mA again over all drivers providing the maximum current in any one driver is 350mA. These values are no where near the 3.5A implied by simply adding up all the maximum current capabilities of each device. Remember we still have to ensure that the case gets no hotter than 50 0C so we need to measure this and use a heat sink if if gets hotter. Making different design assumptions will change these values, a spread sheet might be handy at this point.


All this assumes that the devices are switched on permanently, if they are only pulsed very occasionally then they can afford to dissipate more power in the short term, providing it averages out over a few seconds to be lower.

Darlington Driver Example

Powering an Arduino

The Arduino Diecimila has an on board regulator for use when powering from an external power supply. It uses the MC33269 series regulator chip. A question often asked is “can I run it off a 12V supply”. A look at the schematics show that C5 and C6 will be exposed to the full input voltage so the first thing is to see that those capacitors are rated at at least 12V, in this case they are with a 35V rating. So no trouble there. Next see if the regulator can stand that much, again looking at the data sheet shows this to be able to stand 20V. So next see if it can handle the power.

The regulator drops the difference between the input voltage and the output voltage. Also their is a series diode in front of the regulator so that drops 0.7V. This gives us a total voltage drop across the regulator of 12 - 5 - 0.7 = 6.3V.

The power burnt off by the regulator is a function of the current it is taking. Now the board takes about 120mA when it is running, add a bit extra for interface current LEDs and so on and let’s call that 200mA.

So now we can calculate the power, it is simply 6.3 * 0.2 = 1.26 Watts.

Is that too much? Well the data sheet for that device is a little more helpful than most. In includes this graph which shows what effect the copper on the PCB has on the thermal resistance.




If you look at the Arduino you will see that there is only the minimum pad, that is there is no extra copper surrounding the regulator. So in this case reading off the graph we have a maximum power of about 1.5W for a 50 oC ambient or a thermal resistance of 68 oC / W.  So that mans for our example we can only draw 1.5 / 6.3 = 0.238 A or 238mA. So it looks like we will be just about OK with 200mA. The data sheet also says there is a thermal cut out that operates at 170 oC so you can use that if you need in your calculations. However, it warns that the thermal cut out is not a substitute for an adequate heat sink.