Let me examine this simple circuit

We know that for an ideal BJT the collector current will follow this equation:

$$I_C = I_S \times \left(e^{\frac{V_{BE}}{V_T}}-1 \right)$$

As you can see in this case we have \$I_S = 1E-14 = 10\text{fA}\$ and by default the ambient temperature is equal to \$ 27°C\$ thus \$V_T = 25.8649\text{mV}\$ and \$\beta = 100\$. And the emission coefficient/ideality factor \$NF = 1\$ by default.

And we can try to solve this circuit using the old method known as an iterative method.

First, we as usually assume some \$V_{BE}\$ value and solve for \$I_B\$ current.

$$I_B(1) = \frac{10V - 0.6V}{100k\Omega} = 94\mu A$$

Now I will use this equation to solve for the new \$V_{BE}\$ value.

$$ V_{BE} = V_T \ln \left(\frac{I_C}{I_S}+1\right)$$

But because you want to find the base current we must modify the equation to:

$$I_{SB} = \frac{I_S}{\beta} = 1E-16 = 0.1\text{fA}$$

$$ V_{BE} = V_T \ln \left(\frac{I_B}{I_{SB}}+1\right)$$

So, we finally can find the new \$V_{BE}\$

$$ V_{BE}(2) = V_T \ln \left(\frac{I_B}{I_{SB}}+1\right) = 0.7131V$$

Now I will use this new \$V_{BE}\$ value to find the new base current value.

$$ I_B(2) = \frac{10V - 0.7131V}{100k\Omega} = 92.869 \mu A$$

And we continue and find new \$V_{BE}\$ value and base current value.

$$ V_{BE}(3) = 25.8649\text{mV} \ln \left(\frac{ 92.869 \mu A}{0.1\text{fA}}+1\right) = 0.71276V$$

$$ I_B(3) = \frac{10V - 0.7128V}{100k\Omega} = 92.872 \mu A$$

$$ V_{BE}(4) = 25.8649\text{mV} \ln \left(\frac{ 92.872 \mu A}{0.1\text{fA}}+1\right) = 0.71276V$$

AS you can see because we are getting almost the same numbers we can conclude that.

\$V_{BE} = 0.7128V\$ and \$I_B = 92.872 \mu A\$ and \$I_C = \beta I_B = 9.2872mA\$

Of course, sometimes the equation does not converge, then we need to use for example the average value he average of the previous estimate and the calculated value.

Let me examine this simple circuit

We know that for an ideal BJT the collector current will follow this equation:

$$I_C = I_S \times \left(e^{\frac{V_{BE}}{V_T}}-1 \right)$$

As you can see in this case we have \$I_S = 1E-14 = 10\text{fA}\$ and by default the ambient temperature is equal to \$ 27°C\$ thus \$V_T = 25.8649\text{mV}\$ and \$\beta = 100\$. And the emission coefficient/ideality factor \$NF = 1\$ by default.

And we can try to solve this circuit using the old method known as an iterative method.

First, we as usually assume some \$V_{BE}\$ value and solve for \$I_B\$ current.

$$I_B(1) = \frac{10V - 0.6V}{100k\Omega} = 94\mu A$$

Now I will use this equation to solve for the new \$V_{BE}\$ value.

$$ V_{BE} = V_T \ln \left(\frac{I_C}{I_S}+1\right)$$

But because you want to find the base current we must modify the equation to:

$$I_{SB} = \frac{I_S}{\beta} = 1E-16 = 0.1\text{fA}$$

$$ V_{BE} = V_T \ln \left(\frac{I_B}{I_{SB}}+1\right)$$

So, we finally can find the new \$V_{BE}\$

$$ V_{BE}(2) = V_T \ln \left(\frac{I_B}{I_{SB}}+1\right) = 0.7131V$$

Now I will use this new \$V_{BE}\$ value to find the new base current value.

$$ I_B(2) = \frac{10V - 0.7131V}{100k\Omega} = 92.869 \mu A$$

And we continue and find new \$V_{BE}\$ value and base current value.

$$ V_{BE}(3) = 25.8649\text{mV} \ln \left(\frac{ 92.869 \mu A}{0.1\text{fA}}+1\right) = 0.71276V$$

$$ I_B(3) = \frac{10V - 0.7128V}{100k\Omega} = 92.872 \mu A$$

$$ V_{BE}(4) = 25.8649\text{mV} \ln \left(\frac{ 92.872 \mu A}{0.1\text{fA}}+1\right) = 0.71276V$$

AS you can see because we are getting almost the same numbers we can conclude that.

\$V_{BE} = 0.7128V\$ and \$I_B = 92.872 \mu A\$ and \$I_C = \beta I_B = 9.2872mA\$

Of course, sometimes the equation does not converge, then we need to use for example the average value of the previous estimate and the calculated value.