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Substituting a 2n for a 2n unijunction transistor. Thread starter johnnyinwa Start date Jun 25, Search Forums New Posts.
Thread Starter johnnyinwa Joined Jun 24, Heh guys -- new user here, Well I am working on an electronic organ project from Forrest M. The circuit calls for a 2n unijunction transistor which is now obsolete and very expensive. I know I can substitute a 2n PUT that will do the job but I need to know the values of the programming resistors to make the substitution work. Please explain in depth as I am new to the field of electronics.
An attached schematic would be nice. Thank you so much for your help with this issue! Scroll to continue with content. Below is the basic circuit using the 2N PUT which should have characteristics reasonably close to the 2N Rpb is the timing resistor selected by the switch. If you have a problem with it not oscillating for certain resistor values you may have to tweak the values of the resistors and consequently the value of the timing capacitor Cc in proportion.
Good luck. They were given to me by my friend. Thanks for the replies, Please look at the schematic on the original post and tell me if there is a better substitute for the 2n I was thinking a might do the trick but I don't know how to hook it up? What is the output waveform for the 2n? By the way I am using a 9 volt supply for the circuit. Another question -- why is the 10k resistor "R10" needed in the circuit?
As far as the coupling capacitor c2 goes is there a standard value to use or can it be calculated? Thank you so much for your help here guys. A connected as an astable multivibrator may work for you. You would vary R1 to change the frequency. But note that the output is more of a square-wave then a pulse. You can make that into a short pulse, similar to the unijunction's output, by running the signal through C2 with the value of C2 reduced to give the pulse width desired.
You also need to add a diode to ground at C2's output anode to ground to suppress the reverse voltage pulse from the square-wave's negative slope. The output of a unijunction is a short pulse. R10 is needed to provide DC current for the base of Q2.
The C2 capacitor can not pass this DC. The value of C2 is somewhat arbitrary. It just has to be large enough to pass the pulse from the unijunction with sufficient current to turn on Q2.
Likely it could be larger or smaller without significant change in the sound. I still don't understand why r10 in the attached schematic is needed. I understand that its dumping a small amount of dc current to the base of q2 but why does q2 need the current. Isn't the dc current just noise that gets in the way of the signal from q1? You must log in or register to reply here. Similar threads substituting capacitors? You May Also Like. Continue to site.
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