Low power chopper and power inverter
For the top circuit, I suppose you may want to demonstrate the simple relationship \$V_o = V_i\times dutycycle\$.

In that case, using a diode as one of the switch is a bad choice. When the diode is in the on mode, it has a significant voltage drop (comparing to your input of 1V), that means you end up with huge error. At low duty cycle, the current most likely would reverse, and then the diode would go off, causing even more error and additional issue.

If the current is limited to a low value like around 100mA, then one simple way could be to use something cheap and common, like the 2N7000 series N-channel MOSFET or similar for both upper and lower switches. With the input being 1V, you can simply drive both of the gates with 5V logic.

Addition notes in response to comment below:

For example, the relationship mentioned above is the basis for the output stage of a Class-D amplifier (i.e. power electronics), which is employed in some hifi equipment. I searched for Class-D amplifier articles, and here is one relates well to this discussion: Class D Amplifiers

It is probably even more instructive to try it with S2 being a "real" switch (constructed with a MOSFET for example) and then being a diode. At some operating conditions, the two versions may behave similarly. At other operating conditions, the two versions may behave quite differently, for example, at low duty cycle.

By the way, a P-channel MOSFET is most commonly used for the upper switch (as in the article linked above). But with the input (equivalent to Vdd) at 1V, you would need negative gate voltage to turn on a P-MOSFET. That is the reason I suggested a N-MOSFET because an easy to come by 5V CMOS logic output can rise 4V above the 1V Vdd to turn on the N-MOSFET.