% AC to DC Converters % Diego Trapero
A AC/DC Converter, or rectifier, is an electrical device that converts alternating current (AC), which periodically reverses direction, to direct current (DC), which flows in only one direction. Physically, rectifiers take a number of forms, including vacuum tube diodes, mercury-arc valves, copper and selenium oxide rectifiers, semiconductor diodes, silicon-controlled rectifiers and other silicon-based semiconductor switches. Historically, even synchronous electromechanical switches and motors have been used. #
- Uncontrolled rectifiers use diodes as switching devices. They don't need a control circuit.
Diodes
$$ % Diode \begin{circuitikz}[scale = 1] \draw (0,0) todiode, i^>=$i_D$, v_<=$v_{AK}$ ; \end{circuitikz} $$
Common Cathode Diodes If two or more diodes are connected with a common cathode, the closed diode is the one with the most positive anode voltage. The rest of the diodes are open.
$$ % Common Cathode Diodes, by Diego Trapero \begin{circuitikz} \draw (0,0) tobattery1 (0,1) todiode (1,0) tobattery1 (1,1) todiode (2,0) toR (0,0) -- (2,0) (0,2) -- (2,2) ; \end{circuitikz} $$
Common Anode Diodes If two or more diodes are connected with a common anode, the closed diode is the one with the most negative anode voltage. The rest of the diodes are open.
$$ % Common Anode Diodes, by Diego Trapero \begin{circuitikz} \draw (0,0) todiode (0,1) tobattery1 (1,0) todiode (1,1) tobattery1 (2,0) toR (0,0) -- (2,0) (0,2) -- (2,2) ; \end{circuitikz} $$
\tabulinesep=1mm \begin{longtabu} to\linewidth{X[1,m,c]X[1,m,c]X[1,m,c]}
\hline
Configuration & Circuit diagram & Conducting Diode \\
\hline
Common Cathode
&
$$ % Common Cathode Diodes, by Diego Trapero
\begin{circuitikz}
\draw
(0,0) to[battery1](0,1)
(0,1) to[diode](0,2)
(1,0) to[battery1](1,1)
(1,1) to[diode](1,2)
(2,0) to[R](2,2)
(0,0) -- (2,0)
(0,2) -- (2,2)
;
\end{circuitikz}
$$
&
The diode with the most positive anode voltage
\\
\hline
Common Anode
&
$$ % Common Anode Diodes, by Diego Trapero
\begin{circuitikz}
\draw
(0,0) to[diode](0,1)
(0,1) to[battery1](0,2)
(1,0) to[diode](1,1)
(1,1) to[battery1](1,2)
(2,0) to[R](2,2)
(0,0) -- (2,0)
(0,2) -- (2,2)
;
\end{circuitikz}
$$
&
The diode with the most negative cathode voltage
\\
\hline
\end{longtabu}
General rule: The most polarized diode is the one conducting. The one with the biggest
Thyristors
(Ideal)
- Closing conditions
-
$v_AK > 0$ , the thyristor is forward polarized -
$i_G$ , current in the gate (thyristors are current-controlled)
-
- Opening conditions
-
$i_AK < 0$ , current is reversed
-
$$ % AC to DC Half Wave rectifier, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,4) todiode, i^>=$i_D$, v_<=$v_D$ (0,0) toshort (4,4) toR, l=$R$, v_<=$v_o$ ; \end{circuitikz} $$
-
$v_i > 0$ : D ON,$i > 0$ -
$v_i < 0$ : D OFF,$i = 0$
$$ % AC to DC Half Wave rectifier, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,4) toshort (0,0) toshort (4,4) toR, l=$R$, v_<=$v_o$ ; \end{circuitikz} $$
$$ % AC to DC Half Wave rectifier, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,4) toopen (0,0) toshort (4,4) toR, l=$R$, v_<=$v_o$ ; \end{circuitikz} $$
Output voltage,
The voltage in the resistive load,
$$ % AC to DC Full Wave rectifier, by Diego Trapero \begin{circuitikz}[scale = 1] \draw %Sinusoidal source and connections (0,2) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,2) toshort (4,2) node[circ] {} (0,4) toshort (2,4) node[circ] {} %Rectifier branch left (2,0) todiode, i^>=$i_D$, v_<=$v_D$ (2,2) toshort (2,4) todiode, i^>=$i_D$, v_<=$v_D$ %Rectifier branch right (4,0) todiode, i^>=$i_D$, v_<=$v_D$ (4,2) toshort (4,4) todiode, i^>=$i_D$, v_<=$v_D$ %Load and connections (2,6) toshort %upper node (2,0) toshort %lower node (8,6) toR %load ; \end{circuitikz} $$
where
-
$v_i$ is a pure sinusoidal wave$v_i = V_i \sqrt{2} \sin{2 \pi f}$ - RMS value is
$V_i$ - Peak value, or
$V_{ip}$ is$V_i \sqrt{2}$ - Frecuency is
$f$ -
$\varphi = 0$ beacause it is the reference for phase shifts
- RMS value is
-
$R$ is resitance that models the load
Full Wave Rectifier Symbol
Diode table
| D1 | D2 | D3 | D4 | |
|---|---|---|---|---|
| ON | OFF | OFF | ON | |
| OFF | ON | ON | OFF |
$$ % AC to DC Full Wave rectifier, by Diego Trapero \begin{circuitikz}[scale = 1] \draw %Sinusoidal source and connections (0,2) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,2) toshort (4,2) node[circ] {} (0,4) toshort (2,4) node[circ] {} %Rectifier branch left (2,0) toopen (2,4) toshort %Rectifier branch right (4,0) toshort (4,4) toopen %Load and connections (2,6) toshort %upper node (4,0) toshort %lower node (8,6) toR %load ; \end{circuitikz} $$
$$ % AC to DC Full Wave rectifier, by Diego Trapero \begin{circuitikz}[scale = 1] \draw %Sinusoidal source and connections (0,2) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,2) toshort (4,2) node[circ] {} (0,4) toshort (2,4) node[circ] {} %Rectifier branch left (2,0) toshort (2,2) toshort (2,4) toopen %Rectifier branch right (4,0) toopen (4,2) toshort (4,4) toshort %Load and connections (4,6) toshort %upper node (2,0) toshort %lower node (8,6) toR %load ; \end{circuitikz} $$
Output voltage,
The voltage in the resistive load,
Note: Although this
Reduced model of the rectifier If
- a
$v_o$ voltage source, that represents the bridge output voltage - a diode that represents that current cannot enter in the rectifier
$$ % Rectifier model, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) tovoltage source, l=$v_o$ (0,2) todiode (2,2) togeneric, l=$\text{load}$ (0,0) toshort (0,0) nodeground{} ; \end{circuitikz} $$
$$ % AC to DC Full Wave Rectifier with R load, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) nodefullWaveRectifier{} (-19mm, -5mm) to[sinusoidal voltage source](-19mm, 5mm) (9mm, 5mm) toshort (19mm, 5mm) toopen (29mm, 5mm) toR (-19mm, 5mm) -- (FWR.1) (-19mm, -5mm) -- (FWR.2) (19mm,5mm) -- (29mm, 5mm) (FWR.3) -- (29mm,-5mm) ; \end{circuitikz} $$
A condenser can be added in parallel with the R load to smooth the voltage across it. In this case, the circuit is
$$ % AC to DC Full Wave Rectifier with RC load, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) nodefullWaveRectifier{} (-19mm, -5mm) to[sinusoidal voltage source](-19mm, 5mm) (9mm, 5mm) toshort (19mm, 5mm) toC (29mm, 5mm) toR (-19mm, 5mm) -- (FWR.1) (-19mm, -5mm) -- (FWR.2) (19mm,5mm) -- (29mm, 5mm) (FWR.3) -- (29mm,-5mm) ; \end{circuitikz} $$
Output voltage,
- Mean value
$$ % Rectifier model, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) tovoltage source (0,2) todiode (2,2) toC (2,2) toshort (3,0) toR (0,0) toshort (0,0) nodeground{} ; \end{circuitikz} $$
Not exactly like this, it's more complex.
- When the diode conducts,
$v_o = v_R$ and the capacitor is charged with the current from the source.
$$ % Rectifier model, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) tovoltage source (0,2) toshort (2,2) toC (2,2) toshort (3,0) toR (0,0) toshort (0,0) nodeground{} ; \end{circuitikz} $$
The charge is sinusoidal:
- When the diode is not conducting, the capacitor provides the current to the load. It would continue to discharge until
$v_o = v_i$ .
$$ % Rectifier model, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) tovoltage source (0,2) toopen (2,2) toC (2,2) toshort (3,0) toR (0,0) toshort (0,0) nodeground{} ; \end{circuitikz} $$
The discharge is exponential:
The voltage in the load is:
Triangular approximation
- The charge is considered instantaneous:
$t_{\text{charge}} = 0 \rightarrow t_{\text{discharge}} = \frac{T}{2}$ - The discharge is linear, with constant current:
$I_R = V_m / R$ - The diode opens in the maximum of
$v_o$ :$I_R = V_i / R$ (in the slides it says "if ripple is small enough").
$$ % Constant current condenser discharge, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) toC (0,2) toshort (2,2) toI, l=$\frac{V_i}{R}$ (0,0) toshort (0,0) nodeground{} ; \end{circuitikz} $$
The ripple of the wave using the triangular approximation is
Mean value Can be calculated from the triangular approximation
Advantages of a smoothing condenser
- Small size
- Cheap
- Robust
Disadvantages of a smoothing condenser
- High peaks of current through the diodes.
- Harmonics in the source current.
- Need of a big capacitor if the required ripple is small.
$$ % AC to DC Full Wave Rectifier with RLC load, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) nodefullWaveRectifier{} (-19mm, -5mm) to[sinusoidal voltage source](-19mm, 5mm) (9mm, 5mm) toL (19mm, 5mm) toC (29mm, 5mm) toR (-19mm, 5mm) -- (FWR.1) (-19mm, -5mm) -- (FWR.2) (19mm,5mm) -- (29mm, 5mm) (FWR.3) -- (29mm,-5mm) ; \end{circuitikz} $$
The mean value of
$$ % Rectifier model, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) tovoltage source, l=$v_o$ (0,2) todiode (2,2) toL (4,2) toC (4,2) toshort (5,0) toR (0,0) toshort (0,0) nodeground{} ; \end{circuitikz} $$
For the mean values, the resulting DC circuit is:
$$ % Rectifier model, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) tovoltage source, l=$\bar{v_o}$ (0,2) todiode (2,2) toshort (4,2) toopen (4,2) toshort (5,0) toR (0,0) toshort (0,0) nodeground{} ; \end{circuitikz} $$
The resulting
RLC circuit transfer function
Ripple is due to the first harmonic:
Inductor current,
Inductor current can be approximated with its mean value and ripple.
- Mean value. Using superposition
$$ % Rectifier model, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) tovoltage source, l=$\bar{v_o}$ (0,2) todiode (2,2) toshort (4,2) toopen (4,2) toshort (5,0) toR (0,0) toshort (0,0) nodeground{} ; \end{circuitikz} $$
-
Ripple. As in the
$v_R$ ripple, maximum variation of the current is due to the 1st harmonic of$v_o$ . Thus, ripple can be calculated:
$$ % Rectifier model, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) tovoltage source, l=$v_o$ (0,2) todiode (2,2) toL (4,2) toC (4,2) toshort (5,0) toR (0,0) toshort (0,0) nodeground{} ; \end{circuitikz} $$
If
The rectifier is in CCM if the inductor is always conducting some current:
Power factor in the source
- Only first harmonic delivers power:
$$P = V_1 I_1 \cos{\varphi}$$ $$P = \frac{V_{ip}}{\sqrt{2}} \frac{4}{\pi} \frac{I_o}{\sqrt{2}} \cos{\alpha}$$ - Apparent power
$$S = V_{\text{RMS}} I_{\text{RMS}}$$ $$S = \frac{V_{ip}}{\sqrt{2}} I_o$$
Advantages of LC filtering
- No current peaks in the diodes.
- Less condenser ripple.
- Less capacity and current capacitor required.
Disadvantages of LC filtering
- Size and weight of the inductor
Flyback Diode
$$ % AC to DC Full Wave Controlled rectifier, by Diego Trapero \begin{circuitikz}[scale = 1] \draw %Sinusoidal source and connections (0,2) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,2) toshort (4,2) node[circ] {} (0,4) toshort (2,4) node[circ] {} %Rectifier branch left (2,0) tothyristor, i^>=$i_D$, v_<=$v_D$ (2,2) toshort (2,4) tothyristor, i^>=$i_D$, v_<=$v_D$ %Rectifier branch right (4,0) tothyristor, i^>=$i_D$, v_<=$v_D$ (4,2) toshort (4,4) tothyristor, i^>=$i_D$, v_<=$v_D$ %Load and connections (2,6) toshort %upper node (2,0) toshort %lower node (8,6) toR %load ; \end{circuitikz} $$
where
-
$v_i$ is a pure sinusoidal wave$v_i = V_i \sqrt{2} \sin{2 \pi f}$ - RMS value is
$V_i$ - Peak value, or
$V_{ip}$ is$V_i \sqrt{2}$ - Frecuency is
$f$ -
$\varphi = 0$ beacause it is the reference for phase shifts
- RMS value is
-
$R$ is resitance that models the load -
$\alpha$ is the delay angle for thyristors
\begin{align*} \bar{v_o} & = {1 \over T} \int_{\theta_0}^{\theta_0 + T} { v_o(\theta) d\theta}\ & = {1 \over \pi} \int_{0}^{\pi} { v_o(\theta) d\theta} \ & = \cancelto{0}{{1 \over \pi} \int_{0}^{\alpha} { 0 d\theta}} + {1 \over \pi} \int_{\alpha}^{\pi} { V_{ip} \sin{\theta} d\theta} \ & = {V_{ip} \over \pi} [ -\cos{\theta} ]{\alpha}^{\pi} \ & = {V{ip} \over \pi} [ -\cos{\pi} + \cos{\alpha} ] \ & = {V_{ip} \over \pi} ( \cos{\alpha} + 1 ) \end{align*}
When it is loaded with a RL load, current cannot be suddenly cut and thyristos continue to be closed even when they're polarized backwards, until the other bramch is triggered. The
$$ % AC to DC Full Wave Controlled rectifier, by Diego Trapero \begin{circuitikz}[scale = 1] \draw %Sinusoidal source and connections (0,2) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,2) toshort (4,2) node[circ] {} (0,4) toshort (2,4) node[circ] {} %Rectifier branch left (2,0) tothyristor, i^>=$i_D$, v_<=$v_D$ (2,2) toshort (2,4) tothyristor, i^>=$i_D$, v_<=$v_D$ %Rectifier branch right (4,0) tothyristor, i^>=$i_D$, v_<=$v_D$ (4,2) toshort (4,4) tothyristor, i^>=$i_D$, v_<=$v_D$ %Load and connections (2,6) toshort %upper node (4,6) toL (8,6) toR %load (2,0) toshort %lower node ; \end{circuitikz} $$
\begin{align*} \bar{v_o} & = {1 \over T} \int_{\theta_0}^{\theta_0 + T} { v_o(\theta) d\theta}\ & = {1 \over \pi} \int_{\alpha}^{\pi+\alpha} { V_{ip} \sin{\theta} d\theta} \ & = {V_{ip} \over \pi} [ -\cos{\theta} ]{\alpha}^{\alpha + \pi} \ & = {V{ip} \over \pi} [ -\cos{\alpha + \pi} + \cos{\alpha} ] \ & = {2 V_{ip} \over \pi} \cos{\alpha} \end{align*}
\begin{align*} \bar{v_o} & = {1 \over T} \int_{\theta_0}^{\theta_0 + T} { v_o(\theta) d\theta}\ & = {1 \over \pi} \int_{0}^{\pi} { v_o(\theta) d\theta} \ & = \cancelto{0}{{1 \over \pi} \int_{0}^{\alpha} { 0 d\theta}} + {1 \over \pi} \int_{\alpha}^{\pi} { V_{ip} \sin{\theta} d\theta} \ & = {V_{ip} \over \pi} [ -\cos{\theta} ]{\alpha}^{\pi} \ & = {V{ip} \over \pi} [ -\cos{\pi} + \cos{\alpha} ] \ & = {V_{ip} \over \pi} ( \cos{\alpha} + 1 ) \end{align*}
Phases hacer mejor, poner ciruiticos, explicar por que se extingue la corriente a traves de los tiristores
-
$v_i > 0$ , T1 and T4 triggered- T2 and T3 are backwards polarized
- T1 and T4 have been triggered and are conducting
$i_i > 0$
-
$v_i < 0$ , T2 and T3 haven't been triggered yet- T2 and T3 are forward polarized but haven't been triggered yet
- T1 and T4 are still conduction because current through the inductor cannot be discontinued
$i_i > 0$
-
$v_i < 0$ , T2 and T3 triggered- T1 and T4 are triggered and start conducting.
- Current extingueshes through T2 and T3, that cur and become backwards polarized
$i_i < 0$
-
$v_i > 0$ , T2 and T3 haven't been triggered yet$i_i < 0$
Source current,
Tip: For drawing the current, when the thyristors are triggered during the positive semicycle of the
Power
Calculated in the source:
Only first harmonic produces power:
Power factor in the source
RLE loads can be used to transfer power to the net, for example in solar powers. Using a rectifier to inject power in the net is cheaper than using an inverter because it uses thyristors instead of MOSFETs/IGBTs, which are more expensive.
$$ % AC to DC Full Wave Controlled rectifier, by Diego Trapero \begin{circuitikz}[scale = 1] \draw %Sinusoidal source and connections (0,2) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,2) toshort (4,2) node[circ] {} (0,4) toshort (2,4) node[circ] {} %Rectifier branch left (2,0) tothyristor, i^>=$i_D$, v_<=$v_D$ (2,2) toshort (2,4) tothyristor, i^>=$i_D$, v_<=$v_D$ %Rectifier branch right (4,0) tothyristor, i^>=$i_D$, v_<=$v_D$ (4,2) toshort (4,4) tothyristor, i^>=$i_D$, v_<=$v_D$ %Load and connections (2,6) toshort %upper node (4,6) toL (8,6) toR (8,3) tobattery1, l=$E$ (2,0) toshort %lower node ; \end{circuitikz} $$
\begin{align*} \bar{v_o} & = {1 \over T} \int_{\theta_0}^{\theta_0 + T} { v_o(\theta) d\theta}\ & = {1 \over \pi} \int_{\alpha}^{\pi+\alpha} { V_{ip} \sin{\theta} d\theta} \ & = {V_{ip} \over \pi} [ -\cos{\theta} ]{\alpha}^{\alpha + \pi} \ & = {V{ip} \over \pi} [ -\cos{\alpha + \pi} + \cos{\alpha} ] \ & = {2 V_{ip} \over \pi} \cos{\alpha} \end{align*}
Power flow Three operanting regions, depending of the value of
$$ % Rectifier model, by Diego Trapero \begin{circuitikz}[scale = 1] \draw (0,0) tovoltage source, l=$\bar{v_o}$ (0,2) todiode (2,2) toR (2,1) tobattery1, l=$E$ (0,0) toshort (0,0) nodeground{} ; \end{circuitikz} $$
- if
$i>0$ - if
$\bar{v_o} > 0$ - The rectifier produces power
- The E source produces power
- R consumes all the power
- Bad operating regime
- if
$\bar{v_o} < 0$ - The rectifier consumes power
- The E source produces power
- R consumes some power
- Good operating regime if you want to inject power in the net
- if
- if
$i=0$ the diode blocks the current and no power is consumed or generated
So, if you want to inject power from a DC source into the net with a rectifier, the output voltage of the rectifier should be negative, but current must be flowing from it.
The limit angle is calculated from the condition
$$ \begin{circuitikz}[scale = 1] \draw %Sinusoidal source and connections (0,5) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,4) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,3) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (2,5) todiode (2,4) todiode (2,3) todiode (0,5) toshort (4,5) toshort (4,4) toshort (5,4) toR (0,0) toshort (0,0) nodeground{} ; \end{circuitikz} $$
$$ \begin{circuitikz}[scale = 1] \draw %Sinusoidal source and connections (0,2) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,3) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,4) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (1,4) toshort, -* (1,3) toshort, -* (1,2) toshort, -* (0,2) toshort %Rectifier branch A (2,0) todiode, l=$D4$ (2,2) toshort (2,4) todiode, l=$D1$ %Rectifier branch B (3,0) todiode, l=$D3$ (3,2) toshort (3,4) todiode, l=$D2$ %Rectifier branch C (4,0) todiode, l=$D6$ (4,2) toshort (4,4) todiode, l=$D3$ %Load and connections (2,6) toshort %upper node (2,0) toshort %lower node (6,6) toR %load ; \end{circuitikz} $$
$$ \begin{circuitikz}[scale = 1] \draw %Sinusoidal source and connections (0,2) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,3) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (0,4) tosinusoidal voltage source, v^>=$v_i$, i^>=$i_i$ (1,4) toshort, -* (1,3) toshort, -* (1,2) toshort, -* (0,2) toshort %Rectifier branch A (2,0) tothyristor, l=$T4$ (2,2) toshort (2,4) tothyristor, l=$T1$ %Rectifier branch B (3,0) tothyristor, l=$T5$ (3,2) toshort (3,4) tothyristor, l=$T2$ %Rectifier branch C (4,0) tothyristor, l=$T6$ (4,2) toshort (4,4) tothyristor, l=$T3$ %Load and connections (2,6) toshort %upper node (2,0) toshort %lower node (6,6) toR %load ; \end{circuitikz} $$