The Current Load: Why Do Two Diodes in Parallel Not Share It?

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SUMMARY

Two diodes in parallel do not share the current load evenly due to their highly non-linear I-V characteristics, specifically the equation Id = Is*exp((Vd/Vt) - 1). The saturation current (Is) is temperature-dependent and varies between diodes, leading to one diode drawing more current as it heats up, further increasing its Is. This results in a scenario where it is highly unlikely for two paralleled diodes to share current equally, often leading to one diode hogging the majority of the current. To improve current sharing, connecting each diode with a small valued series resistor can help balance the load.

PREREQUISITES
  • Understanding of diode I-V characteristics
  • Knowledge of saturation current (Is) and its temperature dependence
  • Familiarity with dynamic resistance in diodes
  • Basic circuit design principles involving resistors and diodes
NEXT STEPS
  • Research diode I-V curve analysis and its implications in circuit design
  • Learn about temperature effects on semiconductor devices
  • Explore methods for improving current sharing in parallel diode configurations
  • Investigate the behavior of transistors in relation to non-linear characteristics
USEFUL FOR

Electrical engineers, circuit designers, and students studying semiconductor physics who are interested in diode behavior and current sharing techniques.

iScience
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why do two diodes in parallel not share the current load?
 
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iScience said:
why do two diodes in parallel not share the current load?
They may, but the current sharing depends on:
  • The forward voltage of the diodes
  • The "dynamic resistance" of the diodes (dVf/dI)
 
They don't? The shares may not be equal but how could they not share? You mean the current goes all through only one of them?
 
nasu said:
They don't? The shares may not be equal but how could they not share? You mean the current goes all through only one of them?
Yes, 1 device tends to hog most of the current. Diodes have a highly non=linear I-V curve, i.e. Id = Is*exp((Vd/Vt) - 1). Important to know is that "Is" is a strong function of temperature, and varies among parts. Due to mismatches in parameters, as soon as 1 diode draws more current, it gets hotter. Then Is goes up with the higher temperature, which increases diode forward current Id further. Then temp goes higher, Is increases, etc. It is a slim chance that 2 paralleled diodes share current 50%/50%. Possible, but very unlikely.

If, however, each diode is connected to a small valued series resistor, then the diode-resistor pairs are connected in parallel, the resistors force current sharing. Does this help at all?

Claude
 
"Sharing" does not imply 50-50. As long as one portion is not zero, they share.
 
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cabraham said:
Yes, 1 device tends to hog most of the current. Diodes have a highly non=linear I-V curve, i.e. Id = Is*exp((Vd/Vt) - 1). Important to know is that "Is" is a strong function of temperature, and varies among parts. Due to mismatches in parameters, as soon as 1 diode draws more current, it gets hotter. Then Is goes up with the higher temperature, which increases diode forward current Id further. Then temp goes higher, Is increases, etc. It is a slim chance that 2 paralleled diodes share current 50%/50%. Possible, but very unlikely.

If, however, each diode is connected to a small valued series resistor, then the diode-resistor pairs are connected in parallel, the resistors force current sharing. Does this help at all?

Claude

Definitely, thanks

and since transistors are also nonlinear this accounts for their observed behavior as well?
 
iScience said:
Definitely, thanks

and since transistors are also nonlinear this accounts for their observed behavior as well?
Yes.

Claude
 

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