Output Resistance BJT regarding Early Effect

Question

I have a doubt respecting how to calculate the output resistance on this question below.

I've searched at 4 different books and find the same way to compute this value but seems its not be right in that situation.

Translated the instruction:

For the circuit on current source as show the picture below, consider that the common emitter current gain (beta) is 160 and the Early voltage is 8V. The Vbe = 0,7V and the V_T = 25mV, the output resistance of circuit is:

Letter D is answer given.

Anyone have some tips?

My attempt. Like Sedra, Razavi and Boyleastad book said que the output resistance has a approximation expression given as: $$R_{out} = > \dfrac{V_A}{I_C}$$ as $$I_E = \dfrac{4,7-0,7}{20k\Omega} = 0,2\,mA$$

$$I_C \approx I_E$$

$$R_{out} = \dfrac{8V}{0,2 mA} = 40k\Omega$$

Answer 1

For regarding several tips and good insights coming from @jonk and @G36, I find the solution using small signal analysis.

^{simulate this circuit – Schematic created using CircuitLab}

DC bias:

$$I_E = \dfrac{4.7-0.7}{20k\Omega} = 0.2\, mA$$

As follow (small signal):

$$V_X = ro\cdot (i_x - v_{be}\cdot gm)+V\\ = ro\cdot (i_x - v_{be}\cdot gm)+ix\cdot (R_E||r_{\pi}) \tag{1}$$

but:

$$V = i_x\cdot R_E||r_{\pi} = -v_{\pi} = -v_{be}\tag{2}$$

Combine (1) and (2):

$$\require{cancel}\dfrac{V_X}{i_x} = ro\cdot (\cancelto{1}{i_x} +\cancel{i_x}\cdot R_E||r_{\pi}\cdot gm)+\cancel{i_x}\cdot (R_E||r_{\pi})\\ \boxed{R_{out} = ro + (1+ro\cdot gm)\cdot R_E||r_{\pi}} $$

So:

$$r_{\pi} = \dfrac{V_T}{I_B} = \dfrac{V_T\cdot \beta}{I_C};\quad\beta = 160;\quad ro = \dfrac{V_A}{I_C} = 40k\Omega\quad \text{(My initial mistake)}$$

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$$\require{cancel} R_{out} = 0,04M\Omega + 20k\Omega||\cancelto{20k\Omega}{\dfrac{25mV\cdot 160}{0,2mA}} \cdot \left (1+\dfrac{40k\Omega}{125\Omega}\right )\\ = 0.04M\Omega + 0,01M\Omega + \dfrac{400M\Omega}{125}$$

$$R_{out} = 0.04 + 0.01 + 3.2 = \boxed{3.25M\Omega}\tag{3}$$