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This derivation was adapted from the book "Tunnel Diodes" by Sylvester P. Gentile Appendix A:

The variable \$ R \$ is the load. The variable \$ r \$ is the differential negative resistance of the diode.

Take the change of voltage source \$ V \$ and place it across the load:

$$ V = i_1R $$

This is equation 1.

Then take the negative differential resistance and place it in series with the load. Then:

$$ V = i_2(R+r) $$

This is equation 2.

Now divide equation 1 by equation 2

$$ 1 = \frac{R}{R+r} * \frac{i_1}{i_2} $$

Multiply both sides by \$ \frac{i_2}{i_1} \$

$$ \frac{i_2}{i_1} = \frac{R}{R+r} $$

Let

$$ v_{in} = i_1R $$

This is equation 3.

$$ v_{out} = i_2R $$

This is equation 4.

Divide equation 4 by equation 3:

$$ \frac{v_{out}}{v_{in}} = \frac{i_2}{i_1} $$

then

$$ \frac{v_{out}}{v_{in}} = \frac{R}{R+r}$$

Multiply both sides by \$ v_{in} \$

Then:

$$ v_{out} = \frac{R}{R+r} * v_{in} $$

This derivation was adapted from the book "Tunnel Diodes" by Sylvester P. Gentile Appendix B:

The variable \$ R \$ is the load. The variable \$ r \$ is the differential negative resistance of the diode.

Take the change of voltage source \$ V \$ and place it across the load:

$$ V = i_1R $$

This is equation 1.

Then take the negative differential resistance and place it in series with the load. Then:

$$ V = i_2(R+r) $$

This is equation 2.

Now divide equation 1 by equation 2

$$ 1 = \frac{R}{R+r} * \frac{i_1}{i_2} $$

Multiply both sides by \$ \frac{i_2}{i_1} \$

$$ \frac{i_2}{i_1} = \frac{R}{R+r} $$

Let

$$ v_{in} = i_1R $$

This is equation 3.

$$ v_{out} = i_2R $$

This is equation 4.

Divide equation 4 by equation 3:

$$ \frac{v_{out}}{v_{in}} = \frac{i_2}{i_1} $$

then

$$ \frac{v_{out}}{v_{in}} = \frac{R}{R+r}$$

Multiply both sides by \$ v_{in} \$

Then:

$$ v_{out} = \frac{R}{R+r} * v_{in} $$

This derivation was adapted from the book "Tunnel Diodes" by Sylvester P. Gentile Appendix A:

The variable \$ R \$ is the load. The variable r is the differential negative resistance of the diode.

Take the change of voltage source \$ V \$ and place it across the load:

$$ V = i_1R $$

This is equation 1.

Then take the negative differential resistance and place it in series with the load. Then:

$$ V = i_2(r+R) $$

This is equation 2.

Now divide equation 1 by equation 2

$$ 1 = \frac{R}{r + R} * \frac{i_1}{i_2} $$

Multiply both sides by \$ \frac{i_2}{i_1} \$

$$ \frac{i_2}{i_1} = \frac{R}{r + R} $$

Let

$$ v_{in} = i_1R $$

This is equation 3.

$$ v_{out} = i_2R $$

This is equation 4.

Divide equation 4 by equation 3:

$$ \frac{v_{out}}{v_{in}} = \frac{i_2}{i_1} $$

then

$$ \frac{v_{out}}{v_{in}} = \frac{R}{R+r}$$

Multiply both sides by \$ v_{in} \$

Then:

$$ v_{out} = \frac{R}{R+r} * v_{in} $$

This derivation was adapted from the book "Tunnel Diodes" by Sylvester P. Gentile Appendix A:

The variable \$ R \$ is the load. The variable \$ r \$ is the differential negative resistance of the diode.

Take the change of voltage source \$ V \$ and place it across the load:

$$ V = i_1R $$

This is equation 1.

Then take the negative differential resistance and place it in series with the load. Then:

$$ V = i_2(R+r) $$

This is equation 2.

Now divide equation 1 by equation 2

$$ 1 = \frac{R}{R+r} * \frac{i_1}{i_2} $$

Multiply both sides by \$ \frac{i_2}{i_1} \$

$$ \frac{i_2}{i_1} = \frac{R}{R+r} $$

Let

$$ v_{in} = i_1R $$

This is equation 3.

$$ v_{out} = i_2R $$

This is equation 4.

Divide equation 4 by equation 3:

$$ \frac{v_{out}}{v_{in}} = \frac{i_2}{i_1} $$

then

$$ \frac{v_{out}}{v_{in}} = \frac{R}{R+r}$$

Multiply both sides by \$ v_{in} \$

Then:

$$ v_{out} = \frac{R}{R+r} * v_{in} $$

This derivation was adapted from the book "Tunnel Diodes" by Sylvester P. Gentile Appendix A:

The variable \$ R \$ is the load. The variable r is the differential negative resistance of the diode.

Take the change of voltage source \$ V \$ and place it across the load:

$$ V = i_1R $$

This is equation 1.

Then take the negative differential resistance and place it in series with the load. Then:

$$ V = i_2(r+R) $$

This is equation 2.

Now divide equation 1 by equation 2

$$ 1 = \frac{R}{r + R} * \frac{i_1}{i_2} $$

Solve for \$ \frac{i_1}{i_2} \$

$$ \frac{i_2}{i_1} = \frac{R}{r + R} $$

Let

$$ v_{in} = i_1R $$

This is equation 3.

$$ v_{out} = i_2R $$

This is equation 4.

Divide equation 4 by equation 3:

$$ \frac{v_{out}}{v_{in}} = \frac{i_2}{i_1} $$

then

$$ \frac{v_{out}}{v_{in}} = \frac{R}{R+r}$$

Multiply both sides by \$ v_{in} \$

Then:

$$ v_{out} = \frac{R}{R+r} * v_{in} $$

This derivation was adapted from the book "Tunnel Diodes" by Sylvester P. Gentile Appendix A:

The variable \$ R \$ is the load. The variable r is the differential negative resistance of the diode.

Take the change of voltage source \$ V \$ and place it across the load:

$$ V = i_1R $$

This is equation 1.

Then take the negative differential resistance and place it in series with the load. Then:

$$ V = i_2(r+R) $$

This is equation 2.

Now divide equation 1 by equation 2

$$ 1 = \frac{R}{r + R} * \frac{i_1}{i_2} $$

Multiply both sides by \$ \frac{i_2}{i_1} \$

$$ \frac{i_2}{i_1} = \frac{R}{r + R} $$

Let

$$ v_{in} = i_1R $$

This is equation 3.

$$ v_{out} = i_2R $$

This is equation 4.

Divide equation 4 by equation 3:

$$ \frac{v_{out}}{v_{in}} = \frac{i_2}{i_1} $$

then

$$ \frac{v_{out}}{v_{in}} = \frac{R}{R+r}$$

Multiply both sides by \$ v_{in} \$

Then:

$$ v_{out} = \frac{R}{R+r} * v_{in} $$