mneary

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 Atmel AVR family does it right. Symmetrical drives like the PIC.

ahmedragia21

Thanks for all the replies .

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 ?

colin mac

The original 8051 could sink 16mA though, so it isn't true to the design.
The AT89Cx051 chips can sink 20mA.

audioguru

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.

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Uncle $crooge

Roff

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.

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Ron

ahmedragia21

what's the purpose of R1 in the darlington ? and what is its use in Roff's circuit ?

Roff

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.

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Ron

ahmedragia21

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 ?

Roff

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:

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Ron

ahmedragia21

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 .

ahmedragia21

Roff , how did you calculate the base resistor (3.9K) ?

i think it should be 2.25/1.6mA =1.4Kohm

Roff

The 89S52 can sink 1.6mA. You are not required to run it at that current. The 2N3906 collector current is about 18mA. It is a high beta transistor, and will easily saturate with the ≈1mA of base drive provided by the 3.9k. It will also work with 1.4k, but you will just be wasting power.

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Ron

ahmedragia21

thanks , i got this question in my , how the 89S52 sinks current like 1.6mA at 0.4VCC and it should output VOL is zero volt then at the base resistor there should be zero volt so that the PNP turns on , if this is true then how did you estimate your base resistor , you assumed that the base voltage is 0.4VCC should it be 0 volt ?

thanks

Roff

The exact amount of base drive current to the PNP isn't critical, so it is not important whether VOL is zero, or 0.4V, or somewhere in between. I just assumed the resistor has about 4V across it, and that about 1mA would be sufficient base drive. You could use any value between 2.4k and 4.7k and it should still work. Electronics is almost never an exact science.

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Ron

ahmedragia21

I just assumed the resistor has about 4V across it

how did you assume that and there's a zero volt in the base ?
im so confused