What is the max power draw on the propeller
dr hydra
Posts: 212
Is the max power draw on the propeller under 500mA?
Comments
http://forums.parallax.com/showthread.php/153843-Propellor-Icc-Measurements
Edit: This doesn't include power supplied by pins to external loads, but that's maxed to 80 mA, which is probably why the data sheet says 300 to give a margin. In practice I've never even had a Propeller chip get noticeably warm.
Beware of reading datasheets, the absolute maximum does not have anything to do with maximum power draw for instance. The package is rated for 1W of which the operating power might only be a third and some may try to drive 7-segment LED displays directly from I/O for instance which while not "operating current" does have to go through the chip.
As mentioned the typical maximum is more like 100ma as it is very unlikely to have the chip itself draw 200ma although it would be possible but that would mean you are exercising every aspect of each cog including the counters and video.
Perhaps clarification of the original question is needed. I interpreted "max power draw on the propeller" to mean the maximum power draw of the propeller chip, all points included. In other words, operating power and IO current. If the question was intended as max power draw of the propeller as in operating current only, then yes, it would be much lower. However, depending on who you ask and in what setup, there are various answers as to what the operating max of the prop is. Not sure why you would want to know max current draw of operation only since you would have to include IO current to put it to practical use.
dr hydra: What is the nature or reasoning of your question? Are you looking to spec a max current draw for a project and need to include the props potential max? Or do you have a project that needs a lot of current and want to know if the prop can handle it?
Yeah, I was a bit surprised when you quoted the 1W figure because I was sure that you knew how to read datasheets. However the reply was mainly for other less experienced readers and I interpreted the "max power draw on the propeller" as pertaining to the Prop itself and not the loads. Also the theoretical maximum could be used but any kind of software that maxed it out would only be used to max it out, it wouldn't have any practical use and so typical maximum is perhaps a better figure to use.
Sometimes I like to limit the current draw from the 3.3V rail either because of the choice of regulator or because of other loads already on that regulator and so if my I/O loads are LEDs I tie them to 5V and sink them since the minimum voltage drop of a red LED is 1.6V but other LEDs are higher. btw, if I do use red LEDs in this configuration it will glow very faintly when it's off but that is easily fixed with a simple diode or parallel resistor.
You will have no trouble with this at all.
The Microchip EEPROMs dissipate negligible power when not being used and not much more for read-only use. During boot time, since only one core of the propeller is active and the it's running at a reduce clock rate, it's unlikely that the EEPROM access contributes to maximum power distribution.
If you read the EEPROM's data sheet, you can add that power demand to the Propellers. That number you gave is odd.
I think you will be happily surprised that EEPROMs use almost nothing in terms of power...
Operating current for the Microchip 24lc256 is 400microamps at 5.5VDC (less than 1 ma) for Reading and 3ma for Writing.
Pullup resistors also consume power. 3.3VDC/10,000ohms = ??
To measure power dissipation, you look at the worst case current consumption of every device in the worst case operational mode. Then you multiply that by the maximum voltage expected. That gives you worst case power.
For example, if a propeller pin is driving a 200 ohm resistor with a tolerance of 10% and an LED with a Vf of 1.0 volts with a Vsupply of 3.3 and 10% accuracy, then the power in the resistor is Vr squared / R. Worst case Voltage is at maximum Vsupply so we get Vr is (3.3 * 1.1) - 1.0 or 2.63V. Worst case R is at minimum resistance so we get R = 200 * 0.9 or 180 ohms. Power is 73mW.
Yes, some of that 73mW is dissipated by the propeller, but it's not possible to know how much and we don't really care. All we want to know is worst case power dissipation.
It's also important to contrive different operating scenarios. The EEPROM dissipation during reads has a max spec of 400 microamps. At 3.3V +10%, we get about 1.5mW. However, during boot, the propeller runs slowly and not anywhere near what the max power dissipation is during most application use. Once booted, the EEPROM draws only 1uA and the power dissipation drops to 3.7uW. I don't believe most applications ever will see appreciable power dissipated in the EEPROM - the exception are those applications that actively read or write the EEPROM during normal operation, but even those usually access the EEPROM so infrequently that the contribution to worst case can't be measured.
The standby current for the Microchip serial EEPROMs is only 1uA. The numbers you quote rarely occur because once booted, most applications never read or write the EEPROM again.
The standby current for the Microchip serial EEPROMs is only 1uA. The numbers you quote rarely occur because once booted, most applications never read or write the EEPROM again.
You still must factor that into the current draw properties of the chip. While the chip itself is not dissipating the power, it is still handling the flow of current. That's why individual IOs have current limits and why you can't utilize that current limit on all IOs at the same time; you would surpass the chips current draw capabilities.
Peter, agreed, we viewed the question from two different perspectives, but had the same overall view. I'll blame the opposite hemispheres we live on, LOL.
For example, current flows into some IO pins and out others. The only thing specified is the maximum power flowing into the Vdd pins. If the IO current has a net sum of zero, then that abs-max spec is inconsequential. But the power dissipation could be enormous.
Okay, I forgot the quiescent condition. Still, the EEPROM is next to nothing in this situation.
That'll give you an expectation of what to watch out for when experimenting.
This is nothing like your original question, why didn't you just say what you were trying to do in the first place instead of driving "us nuts"? However, you say that max output looks like 100ma, but what does that mean? Is the max for the STM32F board itself? You see you not being very informative.
There is no problem adding the Prop though, and if the regulator couldn't handle it (unlikely) you would just add your own, surely?
At first I was thinking about using one of my micro sticks but I think the stm32f4 discovery would work better for my project...but the 100mA power max might be a problem
It was about your saying "the power draw is driving me nuts" bit which I found amusing. All you had to do in the first place was think about what your problem was and what question to ask. However you asked "What is the max power draw on the propeller" which has nothing at all to do with what you need to know. Now however that you mentioned running from the STM32F4 Discovery board as a GPU and wondering if you can power it directly it becomes very easy to answer.
The answer? Yes, and no.
Yes because it has a "3V" output but that is from a 3.3V regulator through a Schottky diode which could be dropping that voltage by 0.4V but probably more around 0.3V which is why it is referred to as '3V". Because it is not a nicely regulated 3.3V and because the regulator has it's own limit of 150ma it is better not to use this and opt for using your own regulator off the "5V" supply which is also through a Schottky when powered from USB. There are plenty of 3.3V LDO regulators available and most supply at least 150ma which is more than enough for the Prop chip (the EEPROM is effectively "no load").
Presuming you're using a 5MHz crystal and running at 80MHz, the best approximation of maximum current consumption is 13mA per core + 0.8mA for the VCO and 0.3mA to drive the crystal. That adds up to 105.1mA. Presuming you're operating at 3.3V with a tolerance of 10%, then maximum power dissipation is 3.63V * 0.1051A = 0.382W.
If you use a linear regulator from 5V, keep in mind that the linear will dissipate a lot of power itself. In this case, the total power in the linear regulator and its load will be the 105mA * 5V or 0.525W. Since we know that 0.382 is in the propeller chip, the remaining 0.144W will be dissipated by the linear itself.
It's not really possible from the information you've provided to know what might happen if you exceed the available current. It may be that you violate the thermal solution's capabilities and something overheats. Alternatively, it may be that the regulator simply folds back, the voltage sags and the system fails. You could see both.
Finally, the power dissipation of any processor is a function of what its doing. The majority of the current consumption is the ~13mA per core at 80MHz. If you're using only 1, then your current consumption and power dissipation will be far, far less than these numbers. If you're running at a different frequency, then your current consumption will be affected.
It looks like my best bet is to use the 5v power supply on the discovery board and add a LM2937ET 3.3 LDO to power the propeller...thanks again for the help