I am trying to design a simple RF signal buffer using a MPSH10 BJT transistor. This is to operate at ~90MHZ and simply buffer an input RF signal to lower the out impedance of my oscillator which feeds it.
I am using LT Spice to simulate this circuit.
I have the transistor in a common collector setup biased midrail to a an 10mA quiescent current. The emitter resistor is 500 ohms. I AC couple a 3V pk-pk 90MHZ signal in from a source impedance of ~ 20k ohms. What I would expect to see is that same voltage signal on the emitter. Instead I see the expected 5 volts DC but only a very small ~100mv signal imposed on the output. When I look at the base current it is huge compared to the emitter current. It looks like the signal is being shunted through the BE and BC parasitic capacitance instead of through the BE junction.
How do RF amplifiers deal with these parasitic capacitance as even at the low 1pf BE capacitance of this transistor a 100MHZ signal sees a 1600 ohm shunt to the emitter. Would this not always form a huge voltage divider when trying to amplifer signals coming from a high impedence>
I have been reading "Radio Frequency Design" by Wes Hayward but have not seen any references to these parasitic causing such huge dividers. MOSFET's always have this capacitance so I don't see how using them makes it better.
Any guidance would be appreciated as I must be missing something fundamental. I have not taken a solid state devices course yet so my understanding of active devices is only from what I have learned from books and the internet.
Thank you
Best Answer
The gain.BW product for this component is min. 650 MHz. Let's use 900 MHz. So at 90 MHz, you can only expect a gain of 10x. Some of the reasons for limited Gain.BW are B-C and B-E capacitance, as well as base resistance. Your 500 ohm emitter R will then look like 10x = 5k when reflected to the base. When driven with a 20 kohm source, you will therefore only have a gain of ~ 5k/(20k+5k) = 0.2.
Now, your 3 Vp-p @ 90 MHz requires a slew rate of 3V*90M*2 = 540 V/us. With 10 mA bias current, you can calculate the max. capacitance allowed at the emitter to support this slew rate -- from I=C.dV/dt, you get C = 10m/(540V/us) = 19 pF. This is larger than the B-E capacitance, so you should be OK there.
Generally RF circuits don't use such (relatively) high impedances as you have (20k) -- then the relative effect of the parasitic capacitances is smaller.
You could use a somewhat higher emitter R -- to get slightly more gain; then add a 2nd stage (which would now be driven from a lower impedance (=re of the BJT, ~ 2.6 ohm @ 20 mA).