How to calculate the bias of cryptographic hash output from biased input?
Let's call the output of your TNRG $X$—this is a random variable taking the values 0 and 1, with $\Pr[X = 0] = 0.4$ and $\Pr[X = 1] = 0.6$, so that $H_\infty[X] = -\log_2 0.6 \approx 0.737$ bits.  This induces the random variable $Y = \operatorname{SHA-256}(X)$, which has

\begin{align*}
  \Pr[Y = \operatorname{SHA-256}(0)] &amp;= 0.4, \\
  \Pr[Y = \operatorname{SHA-256}(1)] &amp;= 0.6.
\end{align*}

Since $\operatorname{SHA-256}(0) \ne \operatorname{SHA-256}(1)$, we clearly have $H_\infty[Y] = H_\infty[X]$.

What if we have a set of observations $\{X_i\}_i$, which are independent and identically distributed?  Then, e.g.,

\begin{align*}
  \Pr[X_0 = 0, X_1 = 0] &amp;= 0.4^2 &amp;&amp;= 0.16, \\
  \Pr[X_0 = 0, X_1 = 1] &amp;= 0.4 \times 0.6 &amp;&amp;= 0.24, \\
  \Pr[X_0 = 1, X_1 = 0] &amp;= 0.6 \times 0.4 &amp;&amp;= 0.24, \\
  \Pr[X_0 = 1, X_1 = 1] &amp;= 0.6^2 &amp;&amp;= 0.36.
\end{align*}

I leave computing the entropy of the tuple $(X_0, X_1)$ as an exercise for the reader.  What about $\operatorname{SHA-256}(X_0 \mathbin\Vert X_1)$?  Since SHA-256 has no collisions on $00$, $01$, $10$, $11$, once again the entropy is preserved.

In fact, the entropy is not preserved only if there is a collision in the possible inputs.  Obviously the entropy is at most 256 bits for the output of SHA-256 on a single input.  How many distinct outputs are there?  Nobody knows for SHA-256, but for a uniform random function $F\colon \{0,1\}^{256} \to \{0,1\}^{256}$, the expected fraction of distinct outputs is about half the total output space—specifically, $1 - e^{-1}$.  So you might estimate a maximum min-entropy of about 255 bits, in the best case of a perfectly uniform input.

However, you're not limited to feeding in 256 bits; you could feed in 512.  If you feed in $256 + k$ bits to a uniform random function 256-bit function, the expected fraction of distinct outputs is about $1 - e^{-2^k}$ which very rapidly approaches 1 as $k$ grows.  The min-entropy of $(X_0, X_1, \dots, X_{511})$ is obviously $-\log_2 0.6^{512} \approx 377 \gg 256$ bits.  While passing 512 bits through SHA-256 may not necessarily cover the entire 256-bit output space and may not attain the full possible entropy of 256 bits, it's probably a pretty safe bet that the entropy is close enough to 256 bits that the difference doesn't matter to you.  This is plenty to use as a key for a PRNG for cryptographic work.