How to reduces nonlinearities due to the saturation of one of the BJTs in a differential amplifier
A diff-amp pair with collector resistors like this is usually considered good for maybe, under unusual motivations, a base voltage difference of perhaps \$\pm 100\:\text{mV}\$. Stretching the bases any more than that will often cause one of the BJTs to saturate, with more and more of the current normally in the collector shifting over to the base. (Though that does depend on the collector resistor values, too.)

If one of the BJTs does saturate, and while the base voltage on the now-saturating BJT continues even further upward, it merely hauls the collector voltage up along with the emitter. What else can it do? The emitter and collector are essentially smashed together now. So they both just track the base voltage upward, except by about a diode drop or so.



Let's look at a more realistic approach:







simulate this circuit &ndash; Schematic created using CircuitLab

Now you have a current sink that will act like one and this is designed for around the same value as you specified, \$200\:\mu\text{A}\$. Maybe a little less. But close.

Also, you now need headroom for it to operate. It will require at least two diode drops. At least \$1.5\:\text{V}\$. Add in the required diode drops for either of the diff-pair BJTs and you probably would be wise to keep \$V_1\ge 2.5\:\text{V}\$.

Also, since there is a 10-fold change in current with just a \$60\:\text{mV}\$ base difference, there really isn't any point in going much beyond that. So you should limit \$-100\:\text{mV}\le V_2\le +100\:\text{mV}\$.

To keep the BJTs from saturating, their collectors should always be higher than your maximum swept value for \$V_1\$, plus your maximum swept value for \$V_2\$. Let's say you will sweep \$2.5\:\text{V} \le V_1\le 3.5\:\text{V}\$ and \$-100\:\text{mV}\le V_2\le +100\:\text{mV}\$. Then this means the collectors should never go below \$3.6\:\text{V}\$. With about \$200\:\mu\text{A}\$ possible, this means \$\frac{5\:\text{V}-3.6\:\text{V}}{200\:\mu\text{A}}=7\:\text{k}\Omega\$. I used nearby values, above.

Now try your sweeps (N002 is \$Q_1\$'s collector, N003 is \$Q_2\$'s collector):





There really isn't a way to design a diff-amp pair so that the collector voltages can sweep completely through the entire range from ground to \$V_\text{CC}\$. It just cannot happen. Your original design assumed a balance of \$100\:\mu\text{A}\$ in each collector (half to the left, half to the right.) Each collector would be \$5\:\text{V}-25\:\text{k}\Omega\cdot 100\:\mu\text{A}=2.5\:\text{V}\$.

But that's not a sane setting. You have to reserve headroom for the \$V_\text{BE}\$ of your diff-pair and you also need to reserve headroom for whatever is acting as the current sink (current mirror BJT, or as in the above circuit.) There's no avoiding that reservation.

And if this is a discrete design, then there are even more elements of headroom required (degeneration for BJT variation and thermal changes, for example, at the expense of still more gain of course.)

So you must expect that the quiescent voltage at the collectors will in fact not be halfway between the rails. But somewhere else. That's life.