Programmable Unijunction Transistor Circuit

The 10K and 1K resistors in the circuit form a voltage divider when the switch is pressed closed. With the +12V supply this divider nominally sets a transistor base bias voltage at about 1 volt. Very little base current flows due to the fact that the emitter of the NPN transistor is held above ground and as such the NPN base emitter voltage never gets high enough to let the transistor turn on. In a simulation of such circuit with a 2N3904 transistor model it shows that the presence of the 1K resistor keeps some bias across the LED of about 0.7V due to very low level currents in the transistor . If the 1K resistor is removed and when the switch is closed to GND then the bias across the LED drops to essentially zero because the transistor becomes fully turned off.

From a functional standpoint to get an LED to turn on and off from the switch there is no need to have the 1K resistor as relates to just this simple circuit. On the other hand if this circuit was used in a more complex system that had a monitor circuit across the LED looking for the above mentioned bias it could be an indicator that all the wiring from the switch to the LED was intact and in place. In an real burglar alarm system where the switch and LED may be located far apart this residual bias detection can play a role to ensure that the wiring has not been tampered with.

Assuming that you are talking about the gain at low frequencies ("DC gain") a derivation of the gain will result in an expression like $$ Av = g_m (r_{o2} || r_{o3}) $$ where \$g_m\$ is the transconductance of Q2 or Q3 and the \$r_{o2}\$ and \$r_{o3}\$ is the output resistance of these transistors.

The gm of the transistor is easily found to be $$ g_m = \frac{I_C}{V_T} \qquad V_T = \frac{kT}{q} $$ The output resistance is a property of the transistor and usually described by the Early voltage \$V_A\$. Using the Early voltage the output resistance is $$ r_o = \frac{V_A + V_{CE}}{I_C} \approx \frac{V_A}{I_C} $$ since usually \$V_A \gg V_{CE}\$. Assuming that both transistors have the same ro the gain is $$ Av = g_m (r_{o2} || r_{o3}) = \frac{I_C}{V_T} \frac{V_A}{2 I_C} = \frac{1}{2}\frac{V_A}{V_T} $$ With an Early voltage of 100V and a kT/q of 26mV we get a gain of about 2000.