Post History

<h2>First Circuit</h2>

Let's start by looking at the conventional current mirror circuit:

<img src="https://electrical.codidact.com/uploads/iZBwQStavQWc9YH28gpetkgN">

The base of Q1 (Not sure which one that is? Use component designators next time!) is driven to whatever it takes to pass the current dumped onto the collector.  Q2 is assumed to have identical properties, so its base is also driven to support the same collector current as Q1.

The current onto Q1 is switched to be either 0 or 2 mA.  Q2 will therefore sink either 0 or 2 mA.  Since a bias of 1 mA is always supplied to C1, this will therefore switch between +1 mA and -1 mA onto C1.  You say the switching frequency is 1 MHz, so 500 ns per phase.  The voltage change on Q1 per phase is:

 &nbsp; &nbsp; (1 mA)(500 ns)/(470 pF) = 1.1 V

Since the voltage will change linearly with time each phase, the waveform on C1 is a 1.1 Vpp sawtooth.

As for speed, 1 MHz feels fast for discrete parts.  One advantage is that Q1 never saturates, but you are still stuck with nothing removing the charges in the bases other than by causing collector current when the current is switched off.  Also, care should be taken to make sure the collector voltage of Q2 stays high enough to keep Q2 out of saturation.

A current mirror like this is much more useful in integrated circuits where the two transistors will be physically close to each other on the same die, and likely well matched.  Two separate discrete parts out of a bin won't be as well matched.  One trick to deal with that in discrete designs is to add small emitter resistors.  That cuts into the compliance range a little, but makes the circuit more tolerant of part variations.

<h2>Second circuit</h2>

<img src="https://electrical.codidact.com/uploads/VeQQqqximLzLqZzkCmdxprXc">

In this case, the original current mirror is the other way around.  Q2 is being driven, and Q1 follows.  Q3 is an emitter follower.  The input current drives Q3 to whatever it takes to ultimately cause Q1 to sink the input current.  That also causes Q3 to sink that same current, so the whole thing does work like a current mirror when looking only at the three external connections.  The inaccuracies due to the finite gains of the transistors will be a little less, but that doesn't seem to matter in this case since you asked about speed, not accuracy.

In general, more parts make things slower, unless those parts are specifically for driving the other parts harder and faster.  That doesn't seem to be the case here.  Draining the charges in the bases of Q1 and Q2 when the current turns off is still the same problem as before.  Now there is yet another part with a non-zero response time to add to the mix.  I don't see this as being faster.  In fact, I suspect it will be slower in real life, especially with discrete parts.

However, the best way to answer the question is to try out both circuits and see.