Your circuit requires 2 diode drops of voltage to turn the transistor on, mine requires only one. Why have a second diode when you already have one in the form of the base-emitter junction of the transistor? (Just make sure it’s adequately rated for the current you’re putting through it.)
If you reduce the CT current sufficiently, mine will get to the same shape as yours, and if you carry on going, the bottom of the ‘square’ wave will round and you’ll eventually get half a sine wave hanging down from 5 V as there’s just enough to turn the transistor on at the peak of the wave.
My D1 clamps the reverse voltage to 1 diode drop. Note, it’s always safe to short-circuit a current transformer. It’s never guaranteed safe to open-circuit one (which is actually what you’re doing on the negative half-cycles, in the absence of R2). That follows from what I wrote above: the CT will generate whatever voltage is necessary to drive the current it needs to in the secondary circuit. A large CT without a burden can easily flash over and destroy itself. The SCT-013-000 has a transient voltage suppressor inside to clamp the voltage at a safe level. That is not necessary when an internal burden is fitted.
A current transformer is the dual of the voltage transformer. So everything you know about voltage transformers, you reverse for current transformers. So a VT generates a constant voltage at a current that varies according to the load, a CT generates a constant current at a voltage that varies according to the burden; a VT is lightly loaded when it is open-circuited, a CT is lightly loaded when it is short-circuited; an overloaded VT burns out because of too much current, an overloaded CT flashes over because of too much voltage; a step-down VT has most turns of thinner wire on the primary, a CT has most turns of thinner wire on the secondary; etc.