AT89C2051 current and voltage

[colin mac]

Guys can I get some help understanding current and voltages. Pins 7 to 9
are connected to a 1k resistor and base of NPN transistor so a high output turns on transistor.
What is the current through those resistors and what is the voltage at the base of the transistor?

 

[Papabravo]

Start with the assumption tha pin 19 is at ground. It is trying to turn on one of the segments on the display. The display segment has a voltage drop of maybe 1.8 volts (check the datasheet). The base emitter junction has a voltage drop of 0.7 volts, which puts the base at 2.5 volts. One side of the resistor is at 5 volts and the other side is at 2.5 volts so the base current is:

(5 - 2.5)/1K = 2.5 mA

Check the processor datasheet; the output at pin 7 or 9 won't be 5 Volts if the pin is sourcing 2.5 mA, it will be something less.
If the transistor has a beta of 100 then the collector current can be up to 250 mA. The actual current will depend on other factors. The 330 Ohm load resistors will cause a collector current of

(5.- 1.8 - 0.2)/330 = 9.09 mA

which makes sens since LEDs typically like about 10 mA to produce reasonable brightness. This circuit has what we call a "forced beta" of 3.6 which is just the 9.09 mA/2.5 mA. The transistor wants to have a beta of 100 but the external circuit elements dictate a beta of 3.6. This means even if your transistor beta was out past the 3 sigma point on the normal distribution it would still work in this circuit, because it only needs to be capable of doing 3.6

 

[colin mac]

Thanks a lot, major help. So as long as the transistor is saturated you can take Vce to be 0.2V?

 

[Papabravo]

Thanks a lot, major help. So as long as the transistor is saturated you can take Vce to be 0.2V?

It was just a guess as was the forwad voltage of the 7 segment and the Vbe of the transistor. The guesses will get you close enough so you understand what is going on when you make actual measurements. Your mileage may vary.

 

[colin mac]

I see, thanks again

 

[colin mac]

Actually won't this circuit only light one segment. There won't be enough current
for all segments to light at once?

 

[Papabravo]

There is enough gain in the transistor to allow eigth segments to be illuminated at 9 mA each.

8 * 9 mA = 72 mA
"forced beta" = 72 / 3.6 mA = 20

As long as each of the processor pins can sink 9 mA AND the processor package can disipate the 72 mA total you should be just fine. If the processor has a lower limit for total current the drop the current by raising the resistor value.

 

[colin mac]

I asked because I simulated this in ISIS and found your original calculations to be right for on one segment but all segments wouldn't light together. I built it with 7 LEDS instead and found that the reason it wouldn't work in ISIS is that the emitter current is being split between each branch while the collector current stays fixed. Each branch is getting just under 2mA for 7 LEDs.

I then found, that if I get rid of the base and collector resistors, and put a 150 Ohm resistor in series with the LED at the emitter, with 5V to the base, I get
(5 - 0.7 - 2) /150 = 15.3mA, thru each branch that I add, with the collector current increasing by around the same amount as I do.