Why the negative voltage with respect to E pole is needed for IGBT driving

During the IGBT turn-on process, the voltages at the G and E terminals of the IGBT are represented as:

$v(t)=A+Be^{-\frac{t}{\tau}}$

Using the three-factor approach,

$t(0)=A+B=V_L$

$t(\infty)=A=V_H$

Therefore, $v(t) = V_H + (V_L - V_H)e^{-\frac{t}{\tau}}$

The rise time for charging to the threshold voltage $V_T$ is:

$t_r=-\tau ln(\frac{V_T-V_H}{V_L-V_H})$

During the turn-off process of an IGBT, the voltages at the G and E terminals of the IGBT are represented as:

$v(t)=A+Be^{-\frac{t}{\tau}}$

Using the three-factor approach,

$v(0)=A+B=V_H$

$v(\infty)=A=V_L$

The result is: $v(t) = V_L + (V_H - V_L)e^{-\frac{t}{\tau}}$

The discharge time to $V_T$ is:

$t_f=-\tau \ln\left(\frac{V_T-V_L}{V_H-V_L}\right)$

The power consumption of IGBT includes the loss during conduction and the loss during turn-off;

Therefore, the total time $t=t_r+t_f$ must be minimized.

$t=t_r+t_f=-\tau\times ln(\frac{V_T-V_H}{V_L-V_H}\times\frac{V_T-V_L}{V_H-V_L})$

When $V_T$ and $V_H$ are fixed, choose an appropriate $V_L$ such that t is minimized, i.e.:

$f(V_L)=ln\left(\frac{V_T-V_H}{V_L-V_H}\times\frac{V_T-V_L}{V_H-V_L}\right)$ attains its maximum;

Differentiating $f(V_L)$ with respect to $V_L$, we obtain:

$\frac{f(V_L)}{dV_L}=\frac{2V_T-V_L-V_H}{(V_H-V_L)\times(V_H-V_T)\times(V_T-V_L)}$

When $V_L=-V_H+2V_T$, $\frac{f(V_L)}{dV_L}=0$

When $V_L<-V_H+2V_T$, $\frac{f(V_L)}{dV_L}>0$

When $V_L>-V_H+2V_T$, $\frac{f(V_L)}{dV_L}<0$

Therefore, when $V_L=2V_T-V_H$, $f(V_L)$ is minimized, meaning that the time required for the IGBT to turn on and off is minimized.

The IGBT also has the lowest power consumption and the highest efficiency of the power supply.

In this idealized scenario, the Miller effect and other factors are not taken into account. Adjustments should be made based on the actual waveform obtained through testing.