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Half bridge rectifier

Analysis

Oftentimes in analysis for a nonlinear systems, we choose to linearize the system about a specific operating point. This leverages the fact that for a small perturbation $V_{new} = V_{old} + \delta V$ , the series expansion of a nonlinear function will be primarily linear for small $\delta V$ . This comes from the calculation of the powers of $V_{new}$ ; for instance,

\left( V_{new} \right)^{2}\ = \left( V_{old} + \delta V \right)^{2} = \left( V_{old} \right)^{2} + 2V_{old}\delta V + (\delta V)^{2}

2\delta V \ll \ V_{old}

then

\left( V_{old} \right)^{2} + 2V_{old}\delta V + (\delta V)^{2} \approx \left( V_{old} \right)^{2} + \epsilon

where epsilon is some error term. Applying the same logic to the ideal diode equation gives us the response.

I + \delta I = I_{0}\left( \exp\left( \frac{V + \delta V}{K_{B}T/q} \right) - 1 \right)

Rearranging to subtract out the original current I

\delta I = I_{0}\left\lbrack \exp\left( \frac{V + \delta V}{K_{B}T/q\ } \right) - \exp\left( \frac{V}{K_{B}T/q} \right) \right\rbrack

Calling

\frac{V}{K_{B}T/q} = V_{0},\frac{\delta V}{K_{B}T/q} = V_{\delta}

\delta I = I_{0}\left\lbrack \exp\left( V_{0} + V_{\delta} \right) - \exp\left( V_{0} \right) \right\rbrack

Applying a Taylor expansion on all terms

I=I_0\left(\sum_{n=0}^{\infty} \frac{\left(\frac{V}{K_B T / q}\right)^n}{n !}-1\right)=\underbrace{I_0\left(\frac{V}{K_B T / q}\right)}_{\text {linear }}+\underbrace{I_0\left(\frac{1}{2 !}\left(\frac{V}{K_B T \backslash q}\right)^2+\frac{1}{3 !}\left(\frac{V}{K_B T \backslash q}\right)^3+\cdots\right)}_{\text {non-linear }}

Cancelling like terms being subtracted in the brackets gives

\delta I = I_{0}\left\lbrack V_{\delta} + \frac{\left( V_{0} + V_{\delta} \right)^{2}}{2!} + \frac{\left( V_{0} + V_{\delta} \right)^{3}}{3!} + \ldots - \left( \frac{\left( V_{0} \right)^{2}}{2!} + \frac{\left( V_{0} \right)^{3}}{3!} + \ldots \right) \right\rbrack

Finally applying the approximation

\left( V_{0} + V\_\delta \right)^{2} \approx \left( V_{0} \right)^{2}

and cancelling the resulting terms

\delta I \approx I_{0}V_{\delta} = \frac{I_{0}}{K_{B}T/q}\delta V

At this point, the perturbation can be make to look like ohm’s law, and thus the perturbation is linear in behavior. This is equivalent to approximating the I-V curve of the diode as a tangent line approximation, and is a theme that is used extensively in engineering and applied mathematics.

Using the above linearization, what does the frequency response of the half bridge circuit look like?

After linearization, we can treat the diode as a resistor. As a result, the diode and resistor act together as an RC filter, and the frequency response would be similar to an RC high-pass filter. This is because the diode and resistor together (R_D) forms an RC circuit with the capacitor.

Measurement

Using the Red Pitaya’s Bode Analyzer tool, measure the frequency response (|T(f)|) as described in the previous lab. Keep in mind that for this circuit, we stated that the amplitude must be small. Set the DC bias to > 0.6V to ensure the diode is forward biased while testing.

Show the plot of the measurement below:
Try making the amplitude larger and see what occurs. Find a point at which the behavior is no longer linear

Using the Red Pitaya’s Bode Oscilloscope & Spectrum analyzer tools, measure the large signal response to a sinusoid:
With DC Bias of 0.7V, and amplitude 0.1
In this scenario, the diode is forward-biased because of the DC Bias of 0.7V which is greater than the threshold voltage of the diode (typically around 0.6V for silicon diodes like the 1N914). The amplitude of 0.1 is relatively small, and hence you will mostly see a sinusoidal output but there may be some small distortion due to the diode’s non-linear characteristics.
With DC bias of 0V, and amplitude 1V

In this case, there is no DC bias and the signal’s amplitude is large (1V), exceeding the diode’s forward voltage. The diode will start conducting when the input voltage is positive (creating a positive half-wave), but will block current when the input is negative. The output will therefore be a half-wave rectified sinusoidal wave, showing only the positive half-cycles of the input.

Comment on the Spectral content of the output signal when compared to the input signal.

The spectral content of the output signal will include the fundamental frequency and harmonics when compared to the input signal, due to the non-linearity of the diode.

Show a plot of the both the time waveforms and frequency domain.

Comparison

Respond to the following questions:

Find the -3dB point in the circuit, and compare this value to the one you previously calculated.

The -3dB point can be found from the frequency response plot by finding the frequency at which the gain drops to approximately 70.7% of the maximum value.