Calculating VDS(off) for Electronic Converters

[D44]

Ah I see, so that's just the voltage drop across there (i.e. to get the diode 'working'), not that it limits the voltage in the circuit to 0.7V.

Ok, so starting with the Buck converter:
V_out = 12V
D = 0.25
Therefore V_in = V_out/D = 12/0.25 = 48V

V_ds(off) = 48-0 = 48V

Boost converter:
V_out = 12V
D = 0.25
Therefore V_in = V_out(1-D) = 12(1-0.25) = 9V

V_ds(off) = V_in - ground = 9 - 0 = 9V

Inverting converter:
V_out = 12V
D = 0.25
Therefore V_in = (V_out(1-D))/D = 36V

V_ds(off) = V_in + V_out = 36+12 = 48V

How's this looking?

Thanks

 

[berkeman]

Ah I see, so that's just the voltage drop across there (i.e. to get the diode 'working'), not that it limits the voltage in the circuit to 0.7V.

Ok, so starting with the Buck converter:
V_out = 12V
D = 0.25
Therefore V_in = V_out/D = 12/0.25 = 48V

V_ds(off) = 48-0 = 48V

Boost converter:
V_out = 12V
D = 0.25
Therefore V_in = V_out(1-D) = 12(1-0.25) = 9V

V_ds(off) = V_in - ground = 9 - 0 = 9V

Inverting converter:
V_out = 12V
D = 0.25
Therefore V_in = (V_out(1-D))/D = 36V

V_ds(off) = V_in + V_out = 36+12 = 48V

How's this looking?

Thanks

Good work. The only thing I'd change is say that for the Inverting converter, Vout is -12V, and the difference voltage Vds(off) is Vin - Vout, which gives the same answer. I can see why you did it your way, because of how the original figure is labeled.

Whew, that wasn't such a chore now, was it?

 

[D44]

Ahh right ok I see. That makes sense.

Ha, it does seem so much straight forward now I'm further down the line.

Thank you so much for your help. I'm sure you can see I would have never in a million years worked all this out myself.

You guys and this forum are great!