Diode circuits formulas

Exponential model of the junction below breakdown (series resistance neglected). $I_S$ = reverse saturation current, $n$ = ideality factor (1 to 2). $V_T=25.9$ mV at 300 K; these notes round it to 25 mV, and 26 mV is also common, so take the value the question gives. For $V_D\gg n{V}_T$, $I_D\approx I_S{e}^{V_D/(n{V}_T)}$; for $V_D\ll -V_T$, $I_D\approx -I_S$.

Valid when both currents are much larger than $I_S$. With $n=1$ and $V_T=25$ mV, a tenfold change in current changes $V_D$ by 57.6 mV (about 60 mV with 26 mV).

Slope resistance $d{V}_D/d{I}_D$ at the dc operating current $I_D$ (valid for $I_D\gg I_S$), used when a small ac signal rides on a dc bias. With $n=1$, $V_T=25$ mV and $I_D=1$ mA, $r_d=25\Omega$.

At a fixed current the forward drop falls as temperature rises (figures between about $-2$ and $-2.5$ mV/°C are quoted, so use the value given). Problems often assume the reverse saturation current doubles for every 10 °C rise (about 7% per °C). This is a rule of thumb: some treatments give a faster rate for the ideal-junction $I_S$, so if a question states a rate, use it.

Diode anode at the output node, cathode at potential $V_R$. The diode conducts, and the output is flat, for $v_i\ge V_R+V_{\gamma }$.

Diode cathode at the output node, anode at potential $V_R$. The diode conducts for $v_i\le V_R-V_{\gamma }$. With the anode at $-5$ V, for example, the output cannot fall below $-5.7$ V.

Two shunt branches: an upper-level diode with its cathode at potential $V_1$, and a lower-level diode with its anode at potential $V_2$, where $V_2-V_{\gamma }<V_1+V_{\gamma }$. The output is confined to $V_2-V_{\gamma }\le v_o\le V_1+V_{\gamma }$, and at most one diode conducts at any instant. Two identical Zener diodes connected back to back in series (anode to anode, or cathode to cathode) across the output, fed through a series resistor, clip at $\pm (V_Z+V_{\gamma })$: on each half-cycle one is in breakdown and the other is forward-biased.

Diode anode at the output, cathode at potential $V_R$. The capacitor settles at $V_C=V_{i,\max }-(V_R+V_{\gamma })$, input-side plate positive. Unbiased ideal case: $v_o=v_i-V_m$, so the output runs from $0$ to $-2V_m$.

Diode cathode at the output, anode at potential $V_R$. For a sinusoid, $V_{i,\min }=-V_m$. Unbiased ideal case: $v_o=v_i+V_m$, so the output runs from $0$ to $2V_m$.

$V_{\text{shift}}$ is found from the actual input peaks and $V_{i,\text{dc}}$ is the input's average. For a zero-mean input such as a sinusoid, $V_{o,\text{dc}}=V_{\text{shift}}$. For a square wave, the time constant must be much larger than the interval during which the diode is OFF.

The far peak is the output peak on the side opposite the pinned one; $V_R$ is the signed potential of the diode's other terminal.

These apply to any periodic output. $V_{r,\text{rms}}=\sqrt{V_{\text{rms}}^2-V_{\text{dc}}^2}$ is the rms value of the ac part alone.

Currents follow the same ratios, with $I_m=V_m/R_L$. The ripple fundamental is at $f$, and PIV $=V_m$.

The ripple fundamental is at $2f$. PIV $=2V_m$ for the centre-tapped circuit and $V_m$ for the bridge.

dc load power divided by the ac power fed to the rectifier. $R_f$ is the total diode forward resistance in the conduction path: $r_f$ for half-wave and centre-tapped, $2r_f$ for the bridge. Add winding resistance if it is given. The maxima (ideal diodes) are $4/{\pi }^2=40.5\%$ and $8/{\pi }^2=81.1\%$. The figures 40.6% and 81.2% also appear; treat them as the same result.

Diodes modelled as ideal switches with forward resistance $r_f$ (no cut-in voltage). $R_f=r_f$ for half-wave and centre-tapped, $2r_f$ for the bridge. $R_s$ is the secondary winding resistance (one half-winding for centre-tapped). $V_{\text{dc,NL}}=V_m/\pi$ (half-wave) or $2V_m/\pi$ (full-wave), and $I_{\text{dc}}=I_m/\pi$ or $2I_m/\pi$ respectively.

Ideal diodes, resistive load. Half-wave: $2\sqrt{2}/{\pi }^2\approx 0.287$. Bridge: $8/{\pi }^2\approx 0.81$. Centre-tapped: $\approx 0.57$ for the secondary and $0.81$ for the primary; the 0.69 in the table below is the average of the two. A lower TUF means a larger transformer for the same dc power. In the half-wave circuit the secondary also carries a dc current, which can saturate the core.

Use the peak output in place of $V_m$ in the $V_{\text{dc}}$ formula, e.g. $V_{\text{dc}}\approx (V_m-V_{\gamma })/\pi$ for half-wave. This approximation slightly overestimates $V_{\text{dc}}$, because the diode actually conducts for less than a half-cycle. To first order in $V_{\gamma }/V_m$, the half-wave average is $V_m/\pi -V_{\gamma }/2$. PIV becomes $V_m$ (half-wave), $2V_m-V_{\gamma }$ (centre-tapped) and $V_m-V_{\gamma }$ (bridge).

Assumes the capacitor discharges at a constant current $I_{\text{dc}}\approx V_m/R_L$ for the full period $T$ (half-wave) or $T/2$ (full-wave), neglecting the conduction interval. This standard approximation slightly overestimates the ripple.

Treats the ripple as a triangle between $V_m$ and $V_m-V_r$.

A triangular ripple of peak-to-peak $V_r$ has rms value $V_r/(2\sqrt{3})$. The last two forms take $V_{\text{dc}}\approx V_m$.

Angle of each charging pulse, measured from diode turn-on to the peak. Smaller ripple gives a shorter pulse and a larger peak diode current.

Ideal diodes. Here $V_m$ is the secondary peak (each half, for centre-tapped), which with ideal diodes equals the peak output. In the half-wave case the capacitor holds $+V_m$ on the cathode while the anode swings to $-V_m$, so the PIV is double the resistive-load value. With constant-drop diodes these become $2V_m-V_{\gamma }$ (half-wave and centre-tapped) and $V_m-V_{\gamma }$ (bridge).

Half-wave doubler: a clamper ($C_1$ charges to $V_m$) followed by a peak detector that captures the clamped waveform's $2V_m$ peak. Its ripple is at $f$, each diode sees a PIV of $2V_m$, and $C_1$ is rated $V_m$ while $C_2$ is rated $2V_m$. Full-wave doubler: two peak detectors that charge on alternate half-cycles, stacked in series, with ripple at $2f$. Cascading more clamper–detector sections gives $3V_m$, $4V_m$ and so on. Under load the output falls below these values.

Piecewise-linear model, with $r_z$ the Zener (dynamic) resistance. With $r_z=0$ the output is exactly $V_Z$.

Valid only while the Zener is in breakdown.

If this fails, the Zener is OFF and $V_o=V_S{R}_L/(R+R_L)$.

A heavier load (smaller $R_L$) takes current away from the Zener, and below $R_{L,\min }$ the Zener drops out of breakdown. At the other end, $I_{L,\min }=I_R-I_{Z,\max }$. If that is zero or negative, running with no load is safe.

Upper limit: at minimum input and maximum load, the Zener must still get $I_{Z,\min }$. Lower limit: at maximum input and minimum load, the Zener must not exceed $I_{Z,\max }$. If the lower limit comes out above the upper one, no value of $R$ meets the specification.

Occurs at maximum input with minimum load (usually no load, $I_{L,\min }=0$).

Small-change results from the $V_{Z0}+r_z{I}_Z$ model. A change in load current shifts the output by $\Delta V_o=-R_o\Delta I_L$. Line regulation is also quoted as a percentage or in mV/V. With $r_z=0$ both regulations are perfect.

Photon energy is approximately the bandgap. For example, $E_g=1.42$ eV gives $\lambda \approx 0.87\mu$m (infrared). The same relation gives the longest wavelength a photodiode with bandgap $E_g$ can detect.

$V_R$ = reverse bias, $V_{bi}$ = built-in potential. $m=1/2$ for an abrupt junction and $1/3$ for a linearly graded one; hyperabrupt junctions have $m>1/2$. Raising $V_R$ lowers $C_j$ and raises the tuned frequency (when $C_j$ is the only tuning capacitance).