DALI schematic: help to analize reception circuit and limited current supply

Question

I am trying to understand how this schematic works. It is a physical layer implementation of the DALI specification. The schematic is in this document (page 11). I have mainly two questions:

1. In reception mode: when the bus is idle (voltage level between 9.5 and 22.5 V, according to the specification) if the lower limit (9.5V) is analyzed, the current through the optocoupler diode is: $$\frac{V_\text{bus (idle)}-V_\text{f (bridfe)}-V_\text{Z (D4)}-V_\text{f (U2)}}{R_6} = \frac{9.5-0.7- 5.1-1.25}{2.2\,\text K} = 1.1\text{ mA}$$ (approximately). With such current value, the U2 optocoupler transistor is not saturated and therefore, DALI_IN is not close to 0 V.
   What am I analyzing erroneously? For \$V_\text{bus(idle)}=22.5\,\text V\$, the current is approximately 7 mA, so DALI_IN has a voltage close to 0 V (the transistor is saturated).

2. Limited current source: The DALI specification limits the maximum current to 250 mA. In the schematic, the limited current source is created with Q1, D3, R2 and R3 but I can not understand how it works. How are the maximum current value and the output voltage calculated?

Answer 1

Let's start with the easy bit.

2. Limited current source:

The current limiter circuit is this one

^{simulate this circuit – Schematic created using CircuitLab}

D2 stabilizes a constant voltage on the base of the transistor, so the voltage on the base is V_B=V_D1 then the voltage on the emitter is the voltage on the base minus the base-emitter diode-voltage (V_BE) so V_E=V_B-V_BE. Now V_E is also the voltage across the resistor R1 so the current in the resistor must be I_R1=(V_D1-V_BE)/R1. it is obvious ofcause from this equation that since V_D1 AND V_BE are both constant that the current in R1 (I_R1) must be constant.

More coming up on the "1. ..reception mode" part thingie..

1. In reception mode:

..With such current value, the U2 optocoupler transistor is not saturated and therefore, DALI_IN is not close to 0 V.

Where do you get that idea?

If you look at the output side of U2 it is pulled up to 5v by a 10k resistor, if we calculate the current through the transistor in order to pull the output low we get that 5v/10k=0.5mA thats a very low current. Now we look at the datasheet of the TCLT1000 and we take a look at the graph of the collector-emitter saturation-voltage (y-axis) compared to the collector-current (x-axis). We see from this that if the current is below 1mA then the saturation voltage is below 0.2v what that means is that if the collector pulles 0.5 mA which is enough to pull the output completely low then the voltage across the output is going to be <0.2v so in other words it is going to be pulled to GND. Now we can estimate how much current that takes throught the led because we can look at the graphs of the CTR (current-transfer-ratio) compared to forward current in the led and from this we see that if the forward current in the diode is 1.1mA then the CTR is aprox. 50%, what that means is that 1.1mA throught the led = 0.55mA through the transistor. In other words the output is pulled low both in the case of 9.5v and 22.5v, but below 9.5v/1.1mA I would expect the output of the opto-coupler to start going high.