dohzer
Member level 1
I'm a bit confused by an explanation I found in a Fairchild Application note: AN-6076 - Design and Application Guide of Bootstrap Circuit for Hign Voltage Gate-Drive IC
Specifically, page 3 at Fig. 9 and Fig. 10, and the explanation of why Dbs conducts.
I can understand the "example" part, but not the explanation above it.
The example suggests that the bootstrap capacitor exceeds the Dbs breakdown voltage, and it dies. I can understand this.
The explanation text before it talks about Dbs conducting which causes overcharging of the capacitor... but how can it charge if Dbs conducts. Wouldn't that limit Vbs (and therefore the capacitor voltage) to the diode's forward voltage drop? And regardless of this, if Vs is below Vb, how can Dbs conduct? Is it conduction via reverse breakdown?
Specifically, page 3 at Fig. 9 and Fig. 10, and the explanation of why Dbs conducts.
I can understand the "example" part, but not the explanation above it.
The example suggests that the bootstrap capacitor exceeds the Dbs breakdown voltage, and it dies. I can understand this.
The explanation text before it talks about Dbs conducting which causes overcharging of the capacitor... but how can it charge if Dbs conducts. Wouldn't that limit Vbs (and therefore the capacitor voltage) to the diode's forward voltage drop? And regardless of this, if Vs is below Vb, how can Dbs conduct? Is it conduction via reverse breakdown?