wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
Just wanted to double check my understanding of bypass diode behavior:
Without any bypass diodes, if a single cell in a string were shaded, it would become reverse bypassed at its breakdown voltage (20-30 V ?) and would pass full string current (could be 10A). So it would be a little heater of on the order of 200W-300W.
If you have a bypass diode for every 20 cells (e.g. 3 diodes on a 60 cell module), then only those 20 cells are affected by the one shaded cell. Each cell has a Voc of around 0.6V, and the 19 * 0.6 V = 11V the lit cells can generate is less than the breakdown voltage of the shaded cell. That means the shaded cell will only pass at most its Isc.
If the shaded cell is fully shaded, its Isc = 0, and the full string current will go through the bypass diode, with a voltage drop of around 0.7V. With 10A string current, the diode is dissipating 7W. No power is dissipated in the shaded cell.
But if the shaded cell is say, 50% shaded, its Isc will still be positive, maybe 5A. The bypass diode fixes the voltage drop along the 20 cell substring at 0.7V, but the string current could still split as 5A through the bypass diode and 5A through the partially shaded cell. The 19 lit cells could each generate about 0.5V for a total of 9.5V, so the shaded cell could be reverse biased at 10.2V. Meaning the shaded cell would still be dissipating 50W. That's less than the 200W-300W figure with no bypass diodes, but more than I was expecting.
Is the above basically correct (within +/-20% on the numbers), or is a significant error in my analysis?
Thanks,
Wayne
Without any bypass diodes, if a single cell in a string were shaded, it would become reverse bypassed at its breakdown voltage (20-30 V ?) and would pass full string current (could be 10A). So it would be a little heater of on the order of 200W-300W.
If you have a bypass diode for every 20 cells (e.g. 3 diodes on a 60 cell module), then only those 20 cells are affected by the one shaded cell. Each cell has a Voc of around 0.6V, and the 19 * 0.6 V = 11V the lit cells can generate is less than the breakdown voltage of the shaded cell. That means the shaded cell will only pass at most its Isc.
If the shaded cell is fully shaded, its Isc = 0, and the full string current will go through the bypass diode, with a voltage drop of around 0.7V. With 10A string current, the diode is dissipating 7W. No power is dissipated in the shaded cell.
But if the shaded cell is say, 50% shaded, its Isc will still be positive, maybe 5A. The bypass diode fixes the voltage drop along the 20 cell substring at 0.7V, but the string current could still split as 5A through the bypass diode and 5A through the partially shaded cell. The 19 lit cells could each generate about 0.5V for a total of 9.5V, so the shaded cell could be reverse biased at 10.2V. Meaning the shaded cell would still be dissipating 50W. That's less than the 200W-300W figure with no bypass diodes, but more than I was expecting.
Is the above basically correct (within +/-20% on the numbers), or is a significant error in my analysis?
Thanks,
Wayne