The circuit described in this article shows how you can use a capacitor and a transistor current mirror source to make a timer circuit.

Step 1: Design the Circuit

I have drawn this circuit in PSpice simulation software version 9.1:

The Db diode is used to protect transistors from buzzer discharging currents.

The buzzer and the LED turn ON when the Vbe2 (Q2 transistor base-emitter voltage) reaches the 0.7 V mark and the transistor turns ON, thus passing on the current to the next transistor cascaded stages:

Von = (Ve2 - Vbe2) = (2 V - 0.7 V) = 1.3 V

Where: Ve2 is Q2 transistor emitter voltage, Rv1 - Rv2 resistors connection node (2 V), and Vbe2 is Q2 transistor base-emitter voltage (0.7 V).

The switch is used for discharging the Ct1 and Ct2 capacitors to reset the timer.

Now we can do timing calculations:

Discharge time: 

We all know that the total value of parallel capacitors is the sum of all connected parallel capacitors. Let's assumed that Ct1 and Ct2 are both equal to 500 uF to simplify our calculations because you can also use a 1000 uF capacitor instead of those two capacitors.

The capacitors are fully discharged after 5 time constants:

5*Tc = 5*Rdc*(Ct1 + Ct2) = 5 * 100 * 1000 uF = 0.5 seconds

Charging time:

i = C dv/dt

Thus:

dv = (1/C) * (i*dt)

and

dt = C*dv / i

Imax = (Vs - Vbe1) / Rt2 = 2.3 V / 10 kohms = 230 uA

Imin = (Vs - Vbe1) / (Rt1 + Rt2) = 2.3 V / 1010000 ohms = 2.27722 uA

Maximum time to turn ON:

TimeMax = (Ct1 + Ct2) * (Vs - (Ve2 - Vbe2)) / Imin = 1000 uF * 1.7 V / 2.27722 uA

 = 1000 * 1.7 V / 2.27722 = 746.5243 seconds

You can double the charging time by increasing capacitor values or Rt1 variable resistor values. There are 2 Megohm variable resistors sold on the internet.

Minimum time to turn ON:

TimeMin = (Ct1 + Ct2) * (Vs - (Ve2 - Vbe2))/  Imax = 1000 uF * 1.7 V / 230 uA

= 1000 * 1.7 / 230 = 7.3913 seconds

Step 2: Simulations

Simulations show that our calculations match the results:

The green curve shows how the Q4 transistor pair saturates at about the 640-second mark. This is below the predicted 746.5243 seconds time to turn on because I used three cascaded transistor stages that amplify the Q2 transistor input signal. Thus the circuit is already ON a few seconds before the Vbe2 (Q2 transistor base-emitter voltage) reaches the 0.7 V mark.