I've a question , regarding to the datasheet it cant source more than 60uA , is this means i should use a transistor which its base has a min for e.g 120uA to saturate ?
Roff, sorry we dont have this transistor at the egyptian market
I've a question , regarding to the datasheet it cant source more than 60uA , is this means i should use a transistor which its base has a min for e.g 120uA to saturate?
No, you need a transistor that saturates with a base current of only 60uA.Then the transistor's output current is only 1.2mA. It can drive a transistor that saturates with an output current of 24mA that can drive a darlington transistor.
Read my previous posts. I already suggested that. He says he doesn't have access to the one I suggested. It is likely that he will not have access to any you could suggest.
Since 89S52 is a CMOS part, its outputs should swing rail-to-rail with no load (the datasheet says 0.9*Vcc @ 10uA source current). The sink capability is 1.6mA at 0.45V, so you should be able to use a PNP as as a buffer. See below.
An IRF540 Mosfet needs 10V on its gate to turn on completely. A few out of 100 will turn on pretty well with only 5V on the gate. Try a few hundred of them and pick the most sensitive one.
Ahmedragia21, how many of these are you going to build? A good engineering solution is not required for a one-off. If you are going into manufacturing, then your concern is valid. However, if you are going into manufacturing, you shouldn't be concerned about your inability to find a local source.
Roff , why you made a voltage driver using the 10k and the 200 ohm , after calculating the output voltage ,it will be approx 4.9 V , what's the wise for that ?
Roff , why you made a voltage driver using the 10k and the 200 ohm , after calculating the output voltage ,it will be approx 4.9 V , what's the wise for that ?
The base voltage of the TIP120 darlington is a high of only 1.3V, not 4.9V. R1 isn't even needed since the darlington has it built-in.
The 200 ohm resistor R2 applies a base current of 18mA to the darlington to make sure it saturates.
The base voltage of the TIP120 darlington is a high of only 1.3V, not 4.9V. R1 isn't even needed since the darlington has it built-in.
The 200 ohm resistor R2 applies a base current of 18mA to the darlington to make sure it saturates.
I forgot about the internal resistor to GND.
I was using the presumably worst-case numbers mentioned previously: Vbe=2.5V@Ic=3A, and setting forced beta=250.
3A/250=12mA
(5V-2.5V)/12mA~200 ohms.
It is good engineering practice to provide a path for collector-base leakage current, and to discharge base capacitance, providing more rapid turn-off of the Darlington pair when the base current drive is removed.
I've two questions 1st but that scheme will make a voltage divider when the PNP is on , hence there will be a 4.9V on the base of the TIP121 correct ? but how the current will be calculated ?
2nd one is how a uC like 89S52 source and sink two different currents using only one output pin , is it not a simple transistor ?
I've two questions 1st but that scheme will make a voltage divider when the PNP is on , hence there will be a 4.9V on the base of the TIP121 correct ? but how the current will be calculated ?
2nd one is how a uC like 89S52 source and sink two different currents using only one output pin , is it not a simple transistor ?
Audioguru explained the first question. The typical Vbe(on) is around 1.3V (see fig. 2 in the Fairchild TIP120 datasheet). With the PNP driver in saturation (Vce(sat)~0.1V), the current through the 200 ohm resistor will be
Ib=(5-0.1-1.3)/200=18mA,
which will ensure saturation of the TIP120. This is more than the 12mA required, but as I explained previously, was chosen assuming the worst case Vbe(on) is 2.5V, as specified in the datasheet.
Regarding your second question, the 89S52 is a CMOS part, so it has a push-pull output stage. The N-channel pulldown transistor obviously (to me) is a much larger geometry device than the P-channel pullup (the "push" transistor). Larger geometry devices have more drive capability (lower on resistance). A second factor is that, for equal-sized devices, N-channel parts have about twice the drive capability of P-channel. This is why it can sink much more current than it can source.
You never answered this question, which I asked previously:
Ahmedragia21, how many of these are you going to build? A good engineering solution is not required for a one-off. If you are going into manufacturing, then your concern is valid. However, if you are going into manufacturing, you shouldn't be concerned about your inability to find a local source.
Roff, for your question i didnt understand it completley ,im not going to manfucture this product , its just a graduation project and yea we dont have local sources for most electronics parts in the egyptian market .