Engineering 6806: Design Project 9/10/02 Engineering 6806 Power and Heat Michael Bruce-Lockhart

Engineering 6806: Design Project 9/10/02 Power & Heat 2 Modeling Heat Flow Use resistive equations Power = Heat = Current Temperature = Voltage Heater is a current source Fixed temperature is voltage source Insulation is a resistance

Engineering 6806: Design Project 9/10/02 Power & Heat 3 Heating a Building Heat flow through walls Outside temperature T in = T out + P x R Inside temperature Power dissipated in heaters = Heat generated by heaters P = (T in – T out ) / R

Engineering 6806: Design Project 9/10/02 Power & Heat 4 Piercing the Wall Piercing creates parallel heat paths Lowers resistance More power to maintain T in Calculable: R = RSI/A

Engineering 6806: Design Project 9/10/02 Power & Heat 5 Heat in Semiconductors Thermal resistance junction-to-case θ conventionally used instead of R Ambient temperature Thermal resistance case-to-air Case temperature Junction temperature Average power Dissipated In semiconductor

Engineering 6806: Design Project 9/10/02 Power & Heat 6 Temperatures Thus the case temperature is T c = T a + θ ca x P av Notice it fluctuates up with ambient temp Junction temperature T j = T c + θ jc x P av = T a + (θ jc +θ ca ) x P av If T j > 155 °C, junction melts!

Engineering 6806: Design Project 9/10/02 Power & Heat 7 Adding a Heatsink Thermal resistance rating of heatsink Thermal gunk used to attach heatsink to case If θ ch + θ hs << θ jc Then T c ≈ T amb & T j ≈ θ jc x P av Best you can do!

Engineering 6806: Design Project 9/10/02 Power & Heat 8 Supply Bypassing The supply pins on the H-Bridge must be bypassed Large inductive loads V = L di/dt Cutoff the current => high voltages Need bypass caps to handle recirculating currents 100 µF for every amp of current

Engineering 6806: Design Project 9/10/02 Power & Heat 9 Power Calculations Overheating is caused by power dissipation. A switching element dissipates power when it’s off, when it’s on and when it’s switching The off power is the peak voltage times the leakage current (which should be very small) times the proportion of the time the switch is off– where D is the duty cycle. The on power is the switched current (which we’ll call the peak current although it may vary depending upon what the motor is doing) times the switch drop times the proportion of the time it is on–

Engineering 6806: Design Project 9/10/02 Power & Heat 10 Switching Power The switching power takes a little more work. Suppose the voltage goes linearly from 0 to V pk in t s seconds while the current starts from I pk and drops linearly to 0 in the same time. Then the instantaneous power dissipated in the switching device is By integrating this over the switching time we can calculate the energy dissipated tsts V pk I pk

Engineering 6806: Design Project 9/10/02 Power & Heat 11 Shape Factor Assuming the positive and negative going switching is symmetric, this occurs twice per switching cycle, so that the average power over one cycle is The 1/6 represents a shape factor–that is it depends upon the exact shape of the currents and voltages during the rise and fall times. The details may differ (the waveforms might be more exponential, for example, and the rise and fall shapes might differ) but the only difference to the analysis is in the exact value of the shape factor. Thus the total power dissipated in the switching element is I off small General shape factor