1st part
Example DC source of voltage U = 10 V supplies a circuit of current with series connected diode having parameters U 0 = 0.9 V, Rd = 30 mΩ, and resistor R = 0.5 Ω. Find current through the circuit, power losses of the diode, as also voltage across the resistor! Calculate a temperature of the diode if Θ0 = 35 °C and Rth = 0.7 °C/W! 1. Current through the circuit U −U 0 − IRd = I= R
10 − 0.9 − I ⋅ 0. 5
30 103 ;
hence I = 39.56 A. 2. Power losses in the diode ∆PV = U 0 I + I 2 Rd = 0.9 ⋅ 39.56 + 39.562 ⋅
30 = 82.56 W. 103
3. Voltage across the resistor U R = U −U 0 − IRd = 10 − 0.9 − 39.56 ⋅
30 = 7.91 V. 103
4. Temperature of the diode Θpn = Θ0 + ∆PV Rth = 35 + 82.56 ⋅ 0.7 = 92.8 °C .
Very essential are the dynamic operation processes of diode arising at the very beginning and end of its conductivity interval when the polarity of voltage across the structure changes. At the very beginning of conductivity interval a small increase of voltage across the diode can be observed but at the end (Fig.1.8) during a very short time interval trec0 diode in full extent conducts the current in reverse direction. The last situation can be explained with increased concentration of majority charges in opposite majority charge layers in duration of conductivity mode (in p layer n majority charges and otherwise). Therefore with the changing of polarity of the applied voltage the time interval — recovery time — arises during which the layers are «cleaned». In the interval trec0 the voltage across the diode is with polarity according to the direct conductivity and value of forward current flow but the value of voltage is gradually decreasing and at the end of recovery interval is at zero level. In the second part of recovery process the reverse current through the diode is very fast decreasing and reverse voltage across the diode is succeeding to supply network instantaneous meaning.
24
EnergoelektronikaEN BOOK.indb 24
07.05.2018 17:59:10