Each pin can only sink or source up to 20mA. Assuming you're using LEDs with a (generous) 3V forward voltage drop with those 200 ohm resistors, the total current for 4 will be 36mA, so you cannot control them all from one pin. The AXE024 has no high current drivers built in. You can use a transistor to drive them.Iv decided I'm going to take out the 330r and replace it with separate 220 resistors for each led.
I don't think this is correct. According to Microchip, the output impedance of the average PIC chip (sink or source) is dependent on current, just as you would expect for a small MOSFET device. At the current you are talking about, the data sheet shows there would be about 10 ohms. 80 ohms into a short sounds about right, but that is a massive overload.Sometimes you won't need a resistor at all. The P channel (current sourcing) output impedance of a Picaaxe pin is about 80 ohms. Running a red led (e.g. 2.2 forward voltage) on a Picaxe running from a 3.3 volt supply gives 3.3-2.2/80=13.75 mA.
However, the N channel (current sinking) impedance is about 30 ohms, so at least an additional 25 ohms in series would be required for the above conditions.
It's certainly easy to measure. And I will admit that the values are somewhat variable from device to device. But in the case I quoted, they are tight enough to do without the resistor.I don't think this is correct. According to Microchip, the output impedance of the average PIC chip (sink or source) is dependent on current, just as you would expect for a small MOSFET device. At the current you are talking about, the data sheet shows there would be about 10 ohms. 80 ohms into a short sounds about right, but that is a massive overload.
Well,It's certainly easy to measure. And I will admit that the values are somewhat variable from device to device.
I have to respectfully disagree.But in the case I quoted, they are tight enough to do without the resistor.
I wouldn't say 620 ohm is all that odd a value, but your right. With this transistor you could easily get by with 4k7 or even more. I was just applying the over used 'Rule of Thumb' of 10:1. I am actually refreshed to see someone take the stance I usually take on this particular subject. In fact, with this transistor passing 100mA you'd likely be well into saturation with a 10K base resistor, especially if it was the 'C' version.There is really no need to get an odd value 620 ohm resistor. Just use a standard 1K or really just about any value between 1K and 4K7.
Billo, I think you're mis-reading the graph. While sourcing 14 mA, the high output pin can be anywhere between 4.2 volts (4.2/0.014=300 ohms) to 3.3 volts (3.3/.014=236 ohms).Well,
It clearly shows the output impedance increases (downward curve) as the current increases, and also shows that at about 14mA that impedance is around 10 ohms. I know you'd find variance between individual devices, but I wouldn't expect that value would ever be far outside the 9-11 ohm range.
I have to respectfully disagree.
No, I don't think I am. The slope of that graph at any given point is the internal resistance of the PIC and the variance shown in the 3 lines is not only supposed to take in different parts, but more importantly (by far) a huge temperature range.Billo, I think you're mis-reading the graph.
Again, its the slope of the line that is the measure of the internal resistance or dV/dI not V/I, which would give you the load resistance. Also, you might observe that the slopes of those lines are not hugely different at a given current.While sourcing 14 mA, the high output pin can be anywhere between 4.2 volts (4.2/0.014=300 ohms) to 3.3 volts (3.3/.014=236 ohms).
Let me apologize as I may not have made myself clear. I am talking about the output or internal resistance of the device, not the load resistance.To source 14 mA into a 10 ohm load, you'd only need a 0.14 volt output!
Billo, this is trivially easy to measure for a given condition. Don't do this for a 5 volt picaxe supply, as you will exceed the 25 mA Picaxe pin limit.No, I don't think I am. The slope of that graph at any given point is the internal resistance of the PIC and the variance shown in the 3 lines is not only supposed to take in different parts, but more importantly (by far) a huge temperature range.
Let's forget the fact that the slope of a point is zero, and try this your way. Use the 25C "typical" plot. Between 20 and 15 mA (dI=0.005) there is a dV of, roughly eyeballing, 3.8-3.2=0.6 volts. dV/dI=0.6/0.005=120 ohms. Granted, it's not a linear function, but you're still off by an order of magnitude!Again, its the slope of the line that is the measure of the internal resistance or dV/dI not V/I, which would give you the load resistance. Also, you might observe that the slopes of those lines are not hugely different at a given current.
Okay, I give up. Just one last question, have you ever heard of calculus?Let's forget the fact that the slope of a point is zero
Yes, indeed. Most EE's have, and even taken (and perhaps taught) the subject. Billo, this isn't a pissing contest, and by resorting to personal inuendo it appears that you have seen the error of your ways, as you seem to have run out of "eyeballing" resolution.Okay, I give up. Just one last question, have you ever heard of calculus?
The point itself has no slope. The line incorporating the point may or may not have a non-zero slope. I make no presumptions. I humbly ask you to demonstrate how, by eyeballing the graph provided, you arrived at a Picaxe MOSFET impedance of 10 ohms.@rq3.
No thanks. I have not seen the error in my ways and I am quite content with the knowledge that I have.
BTW, no innuendo at all. I was just asking because of your assertion. For someone to have be an EE and not know that a point on a continuous line has an associated slope is a little surprising though.
Also, please don't make presumptions on who I am or what my level of understanding is. Thanks. I leave you to your own.
In a continuous line, there is no point by "itself". Your statement is wrong.The point itself has no slope.
I estimated the slope of the load line. By looking at points on either side of, but close to, the point of interest Po = (Vo, Io) such that I had P1=(V1, I1) and P3=(V2, I2), where V2 < V0 < V1 and I2 > Io > I1. Now I estimate the slope (Ro) at Po as:I humbly ask you to demonstrate how, by eyeballing the graph provided, you arrived at a Picaxe MOSFET impedance of 10 ohms
And this is different from my exercise (following your methodology) in post #19 how? Can you provide points from that graph, arbitrarily close together (your choice), so as to yield a result of anything like the 10 ohms you claim? Didn't think so.I estimated the slope of the load line. By looking at points on either side of, but close to, the point of interest Po = (Vo, Io) such that I had P1=(V1, I1) and P3=(V2, I2), where V2 < V0 < V1 and I2 > Io > I1. Now I estimate the slope (Ro) at Po as:
Ro = (|V1-V2|)/(|I1-I2|)
Which, according to Ohms law, has the units of resistance.
You're getting there. Now you're up to 400 ohms! What happened to 10 ohms? The original point here was that the Picaxe (Microchip) I/O pins have an internal resistance sufficiently high to negate the need for current limit resistors when driving LED's under some conditions. A safe working value is the P channel MOSFETS are about 80 ohms (may actually be as low as 60 if pushed towards the 25 mA maximum), and the N channel about 30 ohms.Again, just eyeballing it, it looks a though the slope at 1.5V is much, much more than 10. There appears to be a .4V change going from 1.2mA to 1.3mA, so the slope is nearly 400, indicating about 400 ohms of internal resistance at that point. One more point, I would NOT expect this same internal resistance at the 50% mark with Vdd = 5V.
Billo, you are a gentleman AND a scholar. Many thanks for making me think and recheck my results.Actually, I can't. I printed out the graph (5V) and did a graphical analysis and I was off by a order of magnitude. The internal resistance at 14mA is more like 100 ohms. Sorry about that.
Are you happy now?
Yes, I do agree, and I apologize for taking this thread to where it went. I made a silly assumption based on another device and then misread the graph with my assumption firmly in place.The original point here was that the Picaxe (Microchip) I/O pins have an internal resistance sufficiently high to negate the need for current limit resistors when driving LED's under some conditions.