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In the circuit diagram in the attached file,
HBTl amplifies RF signal, and the HBT2 base-collector diode is forward biased and linearizes the amplifier. HBT2 along with resistor Rb forms the base biasing circuit for HBT1. As the input power increases, the rectified dc current of the HBT2 base-collector diode increases and the DC voltage across the diode decreases. As a result, the DC voltage across the HBTl base-emitter diode increases slightly. Thus, the increased DC current drives HBTl more strongly, suppressing the gain compression and phase distortion of HBT1.
Since the dc voltage across the HBT2 base-collector diode decreases as the input power increases, the differential conductance of the diode decreases. Since the HBTl base-emitter conductance increases with the increase of the input power, the HBT2 base-collector diode can compensate the variation of the HBTl base-emitter conductance under large signal conditions, leading to suppression of the phase distortion.
Why Vbe of HBT2 decrease while Vbe of HBT1 increase? but they should equal Vbe voltage and same current becuase they are diode. Anyone can explain it? thx
Ref: An HBT MMIC Power Amplifier with an Integrated Diode Linearizer for Low voltage Portable Phone Applications from IEEE