Emitter-follower as regulator with low voltage difference
Take the OnSemi D44H11 as an example. It's datasheet shows \$\beta=120\$ at \$1.5\:\text{A}\$ and at \$-40\:^\circ\text{C}\$. But that's "typical" at that cold temperature (worst case) and the actual value does vary from part to part. Let's say you decide to use this part and make a reasoned decision that you can expect to see at least \$\beta=100\$, over part and temperature variations. (You decide this is safe for your needs.) Then this means that you need a base current of \$\frac{1.5\:\text{A}}{\beta=100}=15\:\text{mA}\$. Of course, if your load needs less, then this figure would be smaller. And if your load requires more, then it would be higher.

The next step is to examine the zener diodes you might use. In this case, you make a reasoned choice for a \$5.6\:\text{V}\$ zener. Here's the Vishay datasheet on zeners. Here, you see that the 1N4734A is the right designation. It specifies two test currents: \$45\:\text{mA}\$ and \$1\:\text{mA}\$.



(In general, these ratings are specified for \$\frac{1}{4}\:\text{W}\$ operation. So if you multiply the larger test current times the rated zener voltage, you should find that this is about \$250\:\text{mW}\$.)

With \$45\:\text{mA}\$ through the zener itself, a variation of base current up to \$15\:\text{mA}\$ might not be too bad to try. You could consider supplying \$55\:\text{mA}\$ total; with let's say \$40\:\text{mA}\$ through the zener and another \$15\:\text{mA}\$ through the base of the D44H11 BJT. If the BJT has a very light load, then at worst you'd be sinking all \$55\:\text{mA}\$ through the zener. And if the BJT has a heavy load, requiring say up to \$20\:\text{mA}\$ base current, then the zener might be starved down to \$35\:\text{mA}\$. That's not too far from the specification, and considering the dynamic resistance this would only mean anything from \$5.55\:\text{V}\$ to \$5.65\:\text{V}\$ at the base of the D44H11 BJT. A variation of \$100\:\text{mV}\$ might be acceptable to you, over load variations.



So let's design this for the D44H11 BJT. At this point, I need to take note of the fact that you have a supply rail that may vary from \$6\:\text{V}\$ to about \$7\:\text{V}\$ (unloaded.) (My apologies for earlier not reading this detail.)

Here's the circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

On the output side I've added a bleeder resistor and an electrolytic capacitor. The bleeder resistor performs two functions: (1) it provides a minimum load on the design to keep it in an active mode; and, (2) it quickly bleeds charge from the output capacitor when the power is turned off. Both functions are useful.

Noting above that \$R_1=6.8\:\Omega\$, the worst minimum case current through \$R_1\$ is \$\frac{6\:\text{V}-5.65\:\text{V}}{6.8\:\Omega}\approx 51.5\:\text{mA}\$. The worst maximum case current through \$R_1\$ is \$\frac{7\:\text{V}-5.55\:\text{V}}{6.8\:\Omega}\approx 210\:\text{mA}\$.

This range of currents is a problem because it means the zener diode has too wide of a potential variation. Recalling from the zener datasheet, there is about \$5\:\Omega\$ impedance at \$45\:\text{mA}\$. Assuming this holds, this means the zener would range something over \$800\:\text{mV}\$ of variation, which I'm sure isn't acceptable.

Lowering \$R_1\$'s resistance would perhaps make the variation less, but at the expense of an increased output voltage and very excessive power dissipation in the zener. Raising \$R_1\$'s resistance would make the variation worse and might starve the BJT. Trapped between a rock and a hard place, so to speak. This is due to the fact that you do NOT have much "regulation headroom" here. You only have as little as about \$400\:\text{mV}\$, but this can vary to over \$1.4\:\text{V}\$, too. So a factor of 3.5X!!

Dealing with that kind of voltage variation with a simple resistor alone is what's causing the huge variation and making it difficult to manage the situation. You'll need to find another way. 



Let's say you want to keep this a linear (not a switcher) design. The hard part here is the voltage headroom, as just mentioned. But there is a way to do this. By allowing one of the BJTs of a current mirror to operate in saturated mode. It's harder still, because I want to deal with both \$V_\text{BE}\$ variation (most important of the two) and \$\beta\$ variation.

Here's the schematic:







simulate this circuit

That should work. The smaller saturated BJT used in the current mirror will not exceed about \$100\:\text{mW}\$. But it may vary from \$10\:\text{mW}\$ to \$100\:\text{mW}\$ over the solar powered voltage range. The other smaller active mode BJT used in the current mirror will be about a constant \$50\:\text{mW}\$ regardless of the solar powered voltage range. I've added some resistors to deal with the differential vagaries of the temperature variations (as much as \$\pm 10\:^\circ\text{C}\$ differential) between the parts and part variations of \$V_\text{BE}\$ and \$\beta\$ (equivalent to another \$\pm 7\:^\circ\text{C}\$ variation in temperature.) Ambient temperature variations are "common mode" so to speak, so the circuit should be fine on that point.

The details on the mirror resistor values are a little more complex and are also a matter of debatable judgment. Different designers may arrive at somewhat different values. (Not a lot different, though.) I'll avoid the long discussion and hope you can simply accept that there was some thinking behind their choice.

Note that it takes some extra work to deal with the 3.5X variation in headroom. But at least it's possible to manage the situation. The upshot is that you cannot really use your schematic (or the one I started with, at the outset.) You have to take an extra step as shown above to put the circuit into reasonable management.