Two perfectly matched silicon transistors are connected

Two perfectly matched silicon transistors are connected as shown in the figure. Assuming the β of the transistors to be very high and the forward voltage drop in diodes to be 0.7 V, the value of current I is

Since Both thyristor are perfectly matched, so V_BE1 = V_BE2

∴	I_C1	= exp		V_BE₁ - V_BE₂		= e⁰ = 1
	I_C2			V_T

Since β for both are same, therefore I_B1 = I_B2 = I_B
Now by KVL in loop as shown,

I_R =	0 - 0.7 - (-5)	= 4.3 mA
	1 kΩ

By KCL at point M
I_R = I_C1 + 2 I_B

I_R = I_C1 + 2	I_C1
	β

∴ I_C1 =	β	I_R
	β + 2

For large β ,	β	≈ 1
	β + 2

∴ I_C1 = I_R = 4.3 mA
∴ I = I_C2 = I_C1 = 4.3 mA