Roff
Well-Known Member
Oops! Too much wine, I guess!Should be cathode to +12V.
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Oops! Too much wine, I guess!Should be cathode to +12V.
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.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?
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.A logic N-mosfet turns on fully at 5V?
So why not use one of those?
A typical IRF540 will pass 20A with the gate at 5V so I don't think he'll have any problem with it passing 3A, even with a bleow average MOSFET.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.audioguru ,Try a few hundred of them and pick the most sensitive one.
i think this is not a good engineering soultion
The original 8051 could sink 16mA though, so it isn't true to the design.The poor high-side drive isn't unique to Atmel, it's a 'feature' of the 8051 family. So the Atmel 8051 clones are true to the family's features.
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.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 ?
I forgot about the internal resistor to GND.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.
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.what's the purpose of R1 in the darlington ? and what is its use in Roff's circuit ?
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 beI'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 ?
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.