What present-day substitutions are there for the previous PNP 2N107 and its NPN mate 2N170? These two formerly very popular transistors for experimenters now seem expensive and hard to find.

NTE 102A for the 2N107 and NTE 101 for the 2N170

Both should be available from Mouser Electronics but may still be a bit pricey due to the fact they are germaniums.
NTE's website should have all the data sheets.

I'm not looking to buy original germanium transistors.

From a little Internet research, I've learned that the best, cheapest substitutes for 2N170 and 2N107, are silicone BC547 and BC557 (also BC848 and BC858). They cost only pennies each from Mouser, and you can buy one or as many as you want.

(Mouser does list a 2N170 but cost is an impossible $23.00. No 2N107 is listed.)

But you'll have to fiddle with the bias resistor values to make silicon transistors work in place of germanium.

I think I'm up to it, after more than 40 years in hobby electronics.

If you do succeed in replacements of silicon for germanium, please tell us what futzing you did do to get it to work. How did you figure out the biasing and added or removed components. I understand that silicon has a wider bandwidth and gain so you could run into difficulties that I would be interested in learning about. I am thinking of this to me at least to be a barrier because of the difficulty of getting rare germanium's. I've had a though at the back of my mind that OP-amps are so much more easy to set the gain. As its just a ratio of resistors and voltage issues I haven't looked into.

And can a circuit that operated on 3 volts with Ge transistors be successfully converted to Si while still on 3 volts ?

Vince

It´s not that difficult to substitute Ge Transistor with Si ones, the main difference is the voltage from Base to Emitter (Vbe) is .2 -.3 for Ge and .5 -.7 for Si this causes that the Base Current Ib changes, and to restore the same value of Ib we need to compensate lowering the value of the resistor in the emitter Re to have the same base current Ib than before.

Remember that Ib=Vb-Vbe /(Hfe*Re + Rb) where Vb is the voltaje applied to the base and Hfe or Beta is the amplify factor of the Transistor. In that equation we can see that Hfe and Vbe change now Hfe and Vbe would be greater than before because Si Tr have greater Hfe and Vbe, thus if Vb-Vbe is smaller and (Hfe*Re + Rb) is bigger than before the net result would be Ib smaller too, so in order to have the same original value of Ib we need to lower the value of Re.

We can calculate the new value of Re but we also could measure the previous value of the voltage across Re and experimentally changing its value until we get the previous value.

Have fun doing it !