From voltage plot to diode circuit

Question

I'm trying to do this electronics exercise, but I can't tell if the question is wrong or I have to add something to the circuit.

I have to draw the diode circuit, starting from the input and output voltage graph, and calculate its function:

I've drawn this circuit:

The Vo voltage is the one of R3.

However, the corresponding plot of my circuit is this:

If Vi >= 0.7V, D1 is ON and D2 is OFF. R1 = 3/2 * R3.

If Vi <= -0.7V, D1 is OFF and D2 is ON. R2 = 2 * R3. EDIT: there is an error here and in the graph, R2 = R3/2

If -0.7V <= Vi <= 0.7V, they are both OFF.

I cannot understand how it is possible to have, as required by the exercise, a positive output voltage when the input voltage is negative.

Thank you

Answer 1

It is easy to satisfy the denominator of the voltage divider ratio R1/(R1+R2) which is the tangent or the opposite over adjacent in geometry.

This design spec requires a +ve only output for a bipolar input and this is called a fullwave (FW) rectifier.

But the voltage ratio is created by splitting open the +ve side of the FW bridge and inserting any R value such that the ratio of load to the sum of load + source = the slopes +2/3 and +2/5 for the negative input.

Teachers often not tell you why silicon diodes are 0.7V. This is typical for a small rated current and depends on power rating of diode. Thus there is not standard value of current for 0.7V. But since the current is low at 0.6 V, the internal electrode+semi junction "bulk" resistance, Rs does not influence the threshold voltage much so remember this. 1mA Silicon diode= 0.6V (same as Vbe) It might be 50mA to 1A get 0.7V depending on power rating so when using kiliohm resistors , assume the 0.6V @ 1mA instead.

In my diagram I put scope traces for the top diode V(t) above I(t) which shows a max of 623mV @ 29mA max which used even smaller resistors. I created the XY plot by creating Vin then Vout then "combine" then choose XY. You can rotate the plot by swapping part ends with a right mouse click.