This constant current circuit consists of two PNP transistors, four resistors (R55, R56, R57, R58), and three capacitors (C96, C97, C98) for RF decoupling. Resistor R59 acts as a current limiter, and U3 is an enhanced RF MOS tube.

 First, let's look at the circuit formed by Q8, R57, and R56. Assuming the transistor's B-E voltage drop is 0.8V and Q7's base current is negligible, we can obtain the formula 10000i + 0.8 = 3 - 1000i. This gives Q8's emitter current i as 0.0002A and Q8's base voltage as 2V.

Let's look at the circuit composed of Q7, R55, U3, R59, and R58. It is known that the base voltage of Q7 is 2V and the BE voltage drop is 0.8V, so the emitter voltage of Q7 is 2.8V and the current flowing through R55 is 0.0111A.

Since R58 is a large resistor, most of the current flowing through R55 flows to the drain of U3. Even if the collector voltage of Q7 is equal to the emitter voltage, the maximum current flowing through Q7 is only 0.00028A.

The circuit can produce constant current, mainly because the loop composed of Q8, R57 and R56 clamps the base voltage of Q7 and stabilizes it.

The gate of U3 is connected to the collector of Q7 through R59, and the voltage across R58 is taken; its value is determined by the current distribution balance of the R55, Q7, R58, and U3 loops, and varies between 0 and 2.8V depending on the MOS tube.