Transistor noise test circuit from 'The Art of Electronics' doubt

[frilance]

Hello!

I'm planning to use a circuit attached (from 'The Art of Electronics 3rd Edition') to measure the noise of some fast low noise npn I got.



But it is not really well described in the book. I don't get the use of the "pre-insertion" and R4. Also I don't understand the use of the small resistors R2 and R3 on the emitter.

Any thoughts?

Thanks in advance

 

[erikl]

I suppose the pre-insertion switch S1 is closed before the DUT Q1 will be inserted, after this will be opened. This serves that the DUT's base potential initially is at GND, so the DUT cannot get destroyed by an initially possibly too high voltage from the charged cap(s) C1 and C2, if R8's tap wasn't at GND before. After opening the pre-insertion switch, the cap(s) get slowly charged via R8 .

R4 serves for a limited discharge current - good for the cap(s) and the switch.

R2 serves for a well defined ac base emitter voltage of Q1 , resulting from the current defined by the 200mV 50Ω ac source, used as reference during the gain test. Note that the ac source's screen node (its GND reference) is connected between R2 and R3.

R3 - with open or closed parallel jumper - probably is responsible for 2 different gain adjustments, due to its negative feedback effect.

 

[frilance]

That was extremely helpful. Thanks a lot erikl.

I'm planning to replace the upper part with just batteries, and forget about the noise from the supply lines. But I will use 9+3 V getting 12 V instead of 12.5 V as specified in the circuit. I don't fully understand how the feedback circuit sets the 2.5 V on the DUT's collector so I'm not sure what the effect of the 0.5 V less in the supply will affect that.

Any help is welcome.

Best regards

 

[erikl]

... how the feedback circuit sets the 2.5 V on the DUT's collector ...

It doesn't: it has to be set manually via the R8 pot: (set V_CE).

I'm not sure what the effect of the 0.5 V less in the supply will affect that.

It doesn't really make a difference if you use 12.5 or 12V at the top of the circuit. Fresh 9+3V batteries will have more than 12.5V at the beginning, and perhaps around 8V at the end of their life (so you could actually work with a single 9V battery and use it down to 7V). And you set the DUT's V_CE anyway by hand.