TIDT277 April 2022
| A | B | |
|---|---|---|
| VIN | 13.5 | 13.506 |
| IIN | 0.1158 | 0.1909 |
| +VoutU | 17.27 | 17.29 |
| +IoutU | 0.0046 | 0.0045 |
| –VoutU | –4.133 | –4.134 |
| –IoutU | 0.0052 | 0.0052 |
| +VoutV | 17.11 | 16.8 |
| +IoutV | 0.005 | 0.049 |
| –VoutV | –4.292 | –3.974 |
| –IoutV | 0.0052 | 0.0489 |
| +VoutW | 16.74 | 16.78 |
| +IoutW | 0.049 | 0.0491 |
| –VoutW | –3.975 | –3.983 |
| –IoutW | 0.0487 | 0.0488 |
| PIN | 1.563 | 2.578 |
| Pout | 1.223 | 2.135 |
| Eff | 0.782 | 0.828 |
| Loss | 0.341 | 0.443 |
A ⇒ output W full load 50 mA; output V and U 10% load, so 5 mA (one channel fully loaded).
B ⇒ output W and V full load 50 mA; output U just 10% load, so 5 mA (two channels fully loaded).
Non-symmetrical loading 10% and 100% of channels results in voltage deviation 16.74 V to 17.29 V and –3.974 V to –4.292 V.
Non-symmetrical loading individually at a channel, so different loading at + voltage and – voltage could be a problem for a virtual ground using this Zener solution. Here the load at + and – output is almost similar.
If needed, this could be improved: by increasing the voltage drop across shunt resistors R7, R8, and R9 ⇒ needs rework on R6, try 73.2 kΩ.