Electronic – how to calculate the current in this circuit ( opamp)

The zener and 120 ohm resistor form one arm of a bridge. The 30K resistor and BJT form the other arm. The bridge will be balanced when the transistor \$V_{CE}\$ is the same as the \$V_{BE}\$. The op-amp forces said bridge into balance.

Under balance conditions, the current through the zener is about 5mA (held constant). The transistor \$I_C\$ is about 220uA.

The output voltage is the zener voltage plus \$V_{BE}\$. At the voltage the manufacturer has chosen for the buried zener (7.2V - \$V_{BE}\$), the positive temperature coefficient of the zener matches the negative temperature coefficient of \$V_{BE}\$ , so the overall temperature coefficient is close to zero and the output voltage is nominally 7.2V.

In the old days (before high-performance IC references became common), one of the best references you could buy was the 1N821A (and its cousins), which have a diode internally in series with a zener (you can tell this indirectly from the data sheet because the 'reverse' current is blocked by the diode to a maximum of a few uA).

Of course in the venerable LTZ1000, the tempco is usually further improved by stabilizing the die temperature at a temperature above ambient using the internal heater and temperature sensor.

Just another method. The key point is to judge if the diode is on.

Assume the LED is on, then the voltage drop on the LED and R2 should be 2V or so, then

$$ V_{D1} = V_{R2} = 2V $$

Then the current on \$R_{2}\$ should be

$$ I_{R2}= \frac{2V}{200\Omega}=10mA $$

Then the voltage drop on \$R_{1}\$ should at least

$$ V_{R1} = (I_{R2} + I_{D1}) \times R_{1} >10mA \times 1k\Omega=10V $$

Apparently, this voltage is greater than the source voltage, so the diode can NOT be on. So we think it's open circuit. Then it's a voltage source in series with two resistor now. And the math will be easy.