I simply set my power supply to 4.20V and 1A limit and done...
What actually happened is that the battery charged about 0.9C and it
was slowing down the closer it got to 4.2V.
If the power supply was truly capable of 4.20 V at up to 1 A then it should have delivered that voltage when the current was less than 1 A. If it could not put out 4.20 V at less than 1 A then it must have poor voltage regulation. If the lower voltage measurement was at the battery (with 4.20 V at the power supply) then there must have been excessive resistance in the wires and/or connectors.
This means that to charge at a constant 2C current the whole way I
have to use voltage much higher than 4.2V, but a fully charged lipo
cell must not be over 4.2V or it does the FLAME ON thing.
You must not do that. The voltage at the battery must never exceed 4.20 V. When it reaches 4.20 V in constant current mode, the charger must hold the voltage down and let the current taper off, then it should shut off when current drops to ~1/10th of the set current. Current tapers off due to the battery's internal resistance and any resistance in the wiring between the battery and charger.
The equivalent circuit looks like this:-
simulate this circuit – Schematic created using CircuitLab
The charger's current regulation is represented by I1 and voltage regulation by D1. The wiring between charger and battery has a total resistance of 0.1 Ω, and the battery has an internal resistance of 0.1 Ω. In this scenario the battery has already charged up to 4.10 V internally.
So what happens now? The difference between the charger voltage and battery internal voltage is 4.2 - 4.1 = 0.1 V. The total resistance between them is 0.05 Ω + 0.05 Ω + 0.1 Ω = 0.2 Ω. Ohm's law says that current = voltage / resistance, so the charging current is 0.1 V / 0.2 Ω = 0.5 A. Voltage at the charger terminals is 4.20 V. At the battery terminals it is 4.2 V less voltage drop in the wiring, ie. 4.2 - (0.1 Ω * 0.5 A) = 4.15 V.
As the battery continues to charge the voltage difference between charger and battery reduces, which causes the current to taper off. If left on charge for long enough the battery will eventually reach 4.20 V when the current drops to zero.