If the clock in the diagram above is a wrist watch, even a 1.09mA current may overload its alarm circuitry. In that case, you may use two transistors as shown in the diagram below. This is called a darlington.

A darlington can be considered as a single transistor with the following characteristics:

V_{BE,darlington}=V_BE1+V_BE2

I_B,darlington=I_B1. I_C1= h_FE1∙I_B1. I_C2=h_FE2∙I_B2=h_FE2∙I_E1=h_FE2∙(h_FE1+1)∙I_B1. I_C,darlington=I_C1+I_C2=h_FE1∙I_B1+h_FE2∙(h_FE1+1)∙I_B1=(h_FE1+(h_FE1+1)∙h_FE2)∙I_B1. h_{FE,darlington}=I_C,darlington/I_B,darlington=h_FE1+(h_FE1+1)∙h_FE2. h_FE1>>1 => h_{FE,darlington}=h_FE1+h_FE1∙h_FE2. h_FE1∙h_FE2>>h_FE1 => h_{FE,darlington}=h_FE1∙h_FE2.

Let's now recalculate a proper value for RB. Assume T1's current gain is 300 and T2's current gain is 100.

I_C,darlington = 9/100=90mA. h_{FE,darlington}=300∙100=30000. V_{BE,darlington}=1.2V.

The voltage across RB equals 3V-1.2V=1.8V => RB < 1.8V/3µA = 600k. 560k is a safe value. I_B,darlington will be 1.8/560k = 3.21µA.