I am trying to implement the DALI TX/RX circuit below.

enter image description here

The description says:

The other transistor (T2) is controlled by the current flowing through the resistor (R1). If a DALI slave unit is connected, this current will be the same as that flowing through the power transistor. The value of the resistor is chosen in such a way that when the current exceeds 250 mA the voltage level across the resistor will open the transistor (T2) which in its turn closes the power transistor (T1). In this way the current is maximized to 250 mA.

The designed circuit is here below, the relative part is in the red rectangular box.

enter image description here

Could anyone explain me how does the resistor R1 (in the first picture) value 2.7 ohm opens the transistor when the current exceeds 250mA? On that design, the signal is transmitted as 12V (logic high) and 0V (logic low).

Here below is the circuit lab drawing in case it is needed.

schematic

simulate this circuit – Schematic created using CircuitLab

V2 is a pwm that gets 5V and 0V values.

datasheet of T2 and T3 transistor (BC337)

datasheet of T1 transistor (BD135)

  • Be careful to check that your design includes an edge rate limit so that you meet the 10us minimum requirement for DALI. Often example circuits from microcontroller manufacturers do not bother with this aspect but it is necessary in the final design. – Martin May 8 '13 at 16:22
up vote 4 down vote accepted

BC337 is a silicon transistor. Silicon PN junction starts conducting between 0.6V and 0.7 volts applied to it with P terminal being more positive than N. In your circuit, when current in 2.7ohm resistor reaches 250ma, it causes 0.675 volts across it turning on BE junction of transistor T2.

  • Mandar, thank you very much for the clear answer. May I also ask one more thing. Do you know what is the purpose of 390 ohm and 0.22 uF on the figuer B-1 between the output and input pins. – sven May 8 '13 at 15:13
  • I am not sure what X1-[1-4] input functions are. Parallel RC will provide lower impedance at higher frequencies. – Mandar May 8 '13 at 17:41

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.