Although this has been studied before, I find it interesting to apply basic techniques. Here are two.

First, let $p_i$ be the probability that the first $i$ batches don't cover everything, and let $I_i$ be the $0-1$ indicator for that event. The number of batches required to collect all $n$ is $\sum_{i=0}^\infty I_i$, so the average value is $\sum_{i=0}^\infty p_i$.

Each probability $p_i$ can be computed by inclusion-exclusion because it is easy to calculate the probability that $i$ batches will all miss a particular subset $S$.

$$p_i = \sum_{S \ne \emptyset} (-1)^{|S|+1} \left( \frac{n-|S| \choose b}{n \choose b} \right)^i = \sum_{s=1}^{n-b} (-1)^{s+1} {n \choose s} \left( \frac{n-s \choose b}{n \choose b} \right)^i$$

We can change the order of summation:

$$\begin{eqnarray}\sum_{i=0}^\infty p_i &=& \sum_{i=0}^\infty \sum_{s=1}^{n-b}(-1)^{s+1 }{n \choose s} \left( \frac{n-s \choose b}{n \choose b} \right)^i \newline &=&\sum_{s=1}^{n-b} \sum_{i=0}^\infty (-1)^{s+1} {n \choose s} \left( \frac{n-s \choose b}{n \choose b} \right)^i \newline &=& \sum_{s=1}^{n-b} \frac{(-1)^{s+1} {n \choose s}}{1- {n-s \choose b}/{n \choose b}} \end{eqnarray}$$

This is an exact answer, but because the sum alternates, it's hard to pick out the largest contribution. It takes some work to simplify the sum to $nH_n \approx n(\log n + \gamma)$ if $b=1$.

A second approach is to start with a standard coupon collector who picks coupons one at a time, and mark off batches of $b$ distinct coupons (ignoring coupons before the start of the batch). It takes an average of $c(n,b)= \frac{n}{n} + \frac{n}{n-1} + ... + \frac{n}{n-b+1}$ coupons to get a batch of $b$ distinct coupons, but after you find the last coupon you continue to the end of the batch. So, the average number of batches is between $nH_n/c(n,b)$ and $1+nH_n/c(n,b)$.