Figure 1

Figure 2

For a clamped inductive load the turn-off waveforms are showed in Fig. 1. If we denote the switch voltage rise time withand the current () fall time withthe energy dissipated by the switch at turn-off is given by the formula:$E_{loss} = \frac {V_o*I_{in}} {2}*(t_r+t_{s off})$﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿Notice that the voltage rises to Vd before the current begins to fall, in fact the current path would only switch through the diode, but the diode has its cathode on Vo~=and so the current cannot flow through the diode until it is positively polarized.If make use of an RCD snubber as shown in Circuit 1 , to control the rate of voltage rise the result is the waveforms on Fig. 2, where the voltage now begins to rise slowly and it overlaps the current waveform transition. The key element in this type of RCD snubber circuit is the capacitor, the diodeand resistorare only to limit the capacitor discharge current during switch turn-on. The capacitorprovides another path to the switch current to flow through. You are invited to open the circuit project and follow the instructions to see the snubber effect on the circuit waveforms.﻿﻿﻿﻿﻿﻿The energy dissipated by the switch during turn-off with the snubber in place and approximatingwith the voltage across the snubber capacitoris derived as:$E_{loss}= \int_0^{t_f} i_D v _{DS}\,dt=\frac {1} {12}*\frac {I^2_D*t^2_{s off} } {2*Cs}$$P_{loss}= \int_0^{t_f} i_D v_{DS}\,dt=\frac {1} {12}*\frac {I^2_D*t^2_{s off }} {2*Cs}*f_{sw}$As can be seen from the formulas above the switching loss could be ideally brought to zero given a sufficiently large capacitor. Unfortunately there is a downside, during switch turn-on the energy accumulated in the snubber capacitoris dissipated through resistor. In other words increasing the snubber capacitance decreases the switch losses but increases the snubber losses. The value ofshould be chosen such that thetime constant is about one fifth of the switch minimum turn-on time, thus assuring thatgets discharged during this time interval. The energy stored in the snubber capacitor is:$E_{C_s}=\frac {Cs*V^2_{DS}} {2}$The case whereVmax whenis called a normal snubber,$C_n=\frac {I_D*t_{s off}} {2*V_{max}}$According to [1] it can be shown that there exists an optimal value offor which the total loss (switch + snubber) is reduced to 53% of what it would have been without the snubber present in the circuit. The current fall timeis mainly dependent on the mosfet driver voltage, driver resistanceand mosfet gate capacitance Cg=and can be calculated as described in [2] $t_{s off}= Rg*C_{iss}*log(\frac{\frac{I_D}{g_{fs}}+V_{th}}{V_{th}})$gfs =mosfet transconductance, Vth = gate threshold voltageAlthough it is possible to calculatefrom mosfet datasheet parameters and application parameters it turns out to be more complicated as it may seem. In fact the mosfet capacitances are not constant but are voltage dependent and Vth is also not precisely defined in datasheets. I will address this in more detail in one of the next posts. For this purpose we will rather determineusing a simulation or via oscilloscope measurement (using a shunt resistor, for example).The calculator at the bottom of the page calculates the optimum RCD snubber values and generates a spice netlist that can be used for verification. In fact due to device parasitic elements some tuning may be needed to find the optimal value. Additionally a test netlist may be generated for measuring(current fall time) for the actual circuit that can be used for calculating the snubber.﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿﻿Circuit 1