Exercise with full wave bridge rectifier

Question

I study an example of youtube video https://www.youtube.com/watch?v=Kl8IOESVWlM an i want to make transfer function vo = f(vi) when i use the piecewise linear diode model.

I worked with the following way.

Replace the diode with the piecewise linear model. Then in every case apply KVL:

i > 0 D1 and D4 are reverse biased and D2 and D3 are forward biased. i < 0 D1 and D4 are forward biased and D2 and D3 are reverse biased. i = 0 All diodes are reverse biased!

and make a function with 3 parts.

My problem is that when i make a relation for example for the case when i > 0 in my relation i have the current of the circuit as unknown variable and i don't know how to remove this unknown parameter.

I want to use only cut voltage, forward resistance and load resistance and input voltage.

Answer 1

All diodes reverse biased is not possible with a bridge rectifier because of how they are connected.

With certain diode models, internal (to the model) ideal diodes may be in that state.

You don't specify directly your diode model, but I assume it looks like this:

^{simulate this circuit – Schematic created using CircuitLab}

The bridge looks like this:

^{simulate this circuit}

So if |Vin| \$\le\$ Vf \$\cdot\$ 2 the output current will be zero and therefore load voltage will be zero.

Iload = 0 |Vin| \$\le\$ Vf \$\cdot\$ 2

If |Vin| > Vf\$\cdot\$2 then the diodes conduct and we have load current of

Iload = (Vin - 2\$\cdot\$Vf)/(Rload + 2\$\cdot\$Rs) Vin > 0

Iload = (-Vin - 2\$\cdot\$Vf)/(Rload + 2\$\cdot\$Rs) Vin < 0 (load voltage is never negative)

In each case, load voltage is Iload \$\cdot\$ Rload