Still looking for someone to tell me either "this might work"
This may work provided you will have 4 identical transistor in identical temperature.
Principles are based on dependency of Ic on Vbe.
Comment says "Ignoring base currents".
So for Q1 Vbe1 has a value as it is needed to make Ic1 (Ic of Q1) being equal to I1.
For Q2 Vbe2 has a value as it is needed to make Ic2 (Ic of Q2) being equal to Iscale.
I2 is Ic3 (Ic of Q3) and Ic4 (Ic of Q4) is Iout.
From simplified Ebers-Moll model we have:
Ic=Ies0 * e ^ (Vbe/(kT/q)).
I will use symbols I0 for Ies0 and Vt for kT/q.
So I get the equation for Ic current as:
Ic=I0*e^(Vbe/Vt).
Ic/I0=e^(Vbe/Vt).
Taking the logarithm of both sides we get:
ln(Ic/I0)=Vbe/Vt so
Vbe=Vt*ln(Ic/I0).
It was for single transistor.
Now our circuit is made that way that Vbe1+Vbe2=Vbe3+Vbe4.
So:
Vt1*ln(Ic1/I01)+Vt2*ln(Ic2/I02)=Vt3*ln(Ic3/I03)+Vt4*ln(Ic4/I04)
Vt=kT/q where T is junction temperature expressed in degrees Kelvin. When you have the same junction temperature you have Vt1=Vt2=Vt3=Vt4 = Vt so we get:
Vt*ln(Ic1/I01)+Vt*ln(Ic2/I02)=Vt*ln(Ic3/I03)+Vt*ln(Ic4/I04) and after dividing by Vt:
ln(Ic1/I01)+ln(Ic2/I02)=ln(Ic3/I03)+ln(Ic4/I04).
The sum of logarithms is the logarithm of the product, therefore we get:
ln( (Ic1/I01)*(Ic2/I02) ) = ln( (Ic3/I03)*(Ic4/I04) )
Logarithms are equal when the logarithm values are equal, so:
(Ic1/I01)*(Ic2/I02) = (Ic3/I03)*(Ic4/I04)
If transistors are identical Ies0 in their equations have the same value so I01=I02=I03=I04 = I0 and we have:
(Ic1/I0)*(Ic2/I0) = (Ic3/I0)*(Ic4/I0)
Ic1*Ic2/I0² = Ic3*Ic4/I0²
so:
Ic1*Ic2 = Ic3*Ic4
And using symbols from schematic:
I1*Iscale=I2*Iout
so:
Iout=Iscale*(I1/I2)