# Diodes – applications

In this tutorial we cover the following topics

LED

Flyback diodes and voltage spike suppression

Convert AC voltage into fixed DC voltage with diodes

Transformer

First indispensable component is the transformer. With this passive electrical device a harmful voltage from wall socket (230V) can be reduced to safe value.

Full wave bridge rectifier

In this type of rectifier four individual rectifying (signal, power) diodes are used connected in a closed loop "bridge" configuration to produce the desired output.

Four diodes labelled $D_{1}$ to $D_{4}$ are arranged in "series pairs" with only two diodes conducting current during each half cycle. This diode bridge circuit was invented by Polish electrotechnician Karol Pollak and patented in December 1895 in Great Britain and in January 1896 in Germany. In 1897, the German physicist Leo Graetz independently invented and published a similar circuit. Today the circuit is still referred to as a Graetz circuit or Graetz bridge.

During the positive half cycle of the supply, diodes $D_{2}$ and $D_{4}$ conduct in series while diodes $D_{1}$ and $D_{3}$ are reverse biased and the current flows through the load as shown below.

During the negative half cycle of the supply, diodes $D_{1}$ and $D_{3}$ conduct in series while diodes $D_{2}$ and $D_{4}$ are now reverse biased. The current flowing through the load in the same direction as before.

In consequence the current flowing through the load is unidirectional, so the voltage developed across the load is also unidirectional and the average DC voltage across the load is 0.637 of the peak voltage ($0.637V_{max}$, assuming no losses) or 0.9 of the effective voltage ($0.9V_{RMS}$, see RMS Voltage Tutorial for RMS explanation).

Because during each half cycle the current flows through two diodes instead of just one so the amplitude of the output voltage is two voltage drops ( $2 \cdot 0.7 = 1.4V$ ) less than the input $V_{max}$ amplitude. The ripple frequency is now twice the supply frequency (e.g. 100Hz for a 50Hz supply).

In our case we use four 1N4004 rectifier diodes to get full wave bridge rectifier.

Although we can use four individual power diodes to make a full wave bridge rectifier, pre-made bridge rectifier components are available "off-the-shelf" in a range of different voltage and current sizes that can be used directly into a circuit.

The smoothing capacitor

The resulting voltage produced by bridge rectifier implemented as it is showned on the image bellow

is not realy constant -- it forms a wave. To convert the full-wave rippled output of the rectifier into a more smooth DC output voltage a capacitor can be added. In our case, with 1000uF capacitor and 1kO load, the following results were obtained Basic bridge circuit with capacitor

 AC with no load ($V_{1}$) 14V AC with load ($V_{1}$) 14V DC with no load ($V_{2}$)(image Basic bridge circuit) 12.1V DC with load ($V_{2}$)(image Basic bridge circuit) 11.34V DC with capacitorand no load ($V_{2}$)(image Basic bridge circuit with capacitor) 18.75V DC with capacitorand load ($V_{2}$)(image Basic bridge circuit with capacitor) 17.1V

Stabilise voltage output with the Zener diode

The main drawback of the solution we have so far is a lack of stable voltage output. This could be achieved with a voltage regulator which main function is to provide a constant output voltage to a load connected in parallel with it in spite of the ripples in the supply voltage or variations in the load current. On of the simplest solution use a Zener diode which will continue to regulate its voltage until the diodes holding current falls below the minimum $I_{Z(min)}$ value in the reverse breakdown region (typically 5-10mA).

The Zener diode behaves just like a normal general-purpose diode consisting of a silicon PN junction and when biased in the forward direction, that is Anode positive with respect to its Cathode, it behaves just like a normal signal diode passing the rated current.

However, unlike a conventional diode that blocks any flow of current through itself when reverse biased, that is the Cathode becomes more positive than the Anode, as soon as the reverse voltage reaches a pre-determined value $V_{Z}$, the zener diode begins to conduct in the reverse direction if only current is greater than $I_{Z(min)}$. That is why a zener diode is always operated in its reverse biased condition. As such a simple voltage regulator circuit can be designed using a zener diode to maintain a constant DC output voltage across the load in spite of variations in the input voltage or changes in the load current.

The zener voltage regulator circuit consists of a current limiting resistor $R_{S}$ connected in series with the input voltage $V_{S}$ with the zener diode connected in parallel with the load $R_{L}$ in this reverse biased condition. Because the load is connected in parallel with the zener diode, so the voltage across $R_{L}$ is always the same as the zener voltage $V_{Z}$ ($V_{R}$ = $V_{Z}$). Other words, the stabilised output voltage is always selected to be the same as the breakdown voltage $V_{Z}$ of the diode. The supply voltage $V_{S}$ must be greater than $V_{Z}$.

Resistor, $R_{S}$ is connected in series with the zener diode to limit the current flow through the diode. By passing a small current through the diode from a voltage source, the zener diode will conduct sufficient current to maintain a voltage drop of $V_{out}$. With no load connected to the circuit, the load current will be zero, and all the circuit current passes through the Zener diode which in turn dissipates its maximum power. Also a small value of the series resistor $R_{S}$ will result in a greater diode current when the load resistance $R_{L}$ is connected and large as this will increase the power dissipation requirement of the diode so care must be taken when selecting the appropriate value of series resistance so that the zener’s maximum power rating is not exceeded under this no-load or high-impedance condition.

There is a minimum zener current $I_{Z(min)}$ for which the stabilisation of the voltage is effective and the zener current must stay above this value operating under load within its breakdown region at all times. The upper limit of current is of course dependant upon the power rating of the device.

Let's do some calculations to make real working circuit.

W will use 1N4733A Zener diode with the 1.3W maximum power rating $P_{Z}$ with the Zener voltage $V_{Z}$ ranges from 4.845V to 5.355V with typical value 5.1V. With this parameters the maximum current flowing through the Zener diode can be calculated as follow
$$I_{max} = \frac{Watts}{Voltage} = \frac{1.3}{5} = 0.26A = 260mA$$

Having the maximum current flowing through the Zener diode we can calculate minimum value of resistor $R_{S}$
$$R_{S} = \frac{V_{S}-V_{Z}}{I_{Z}} = \frac{18V-5V}{260mA} = \frac{13V}{0.26A} = 50 \Omega$$

For tests we will use one $51 \Omega$$resistor (green-brown-black) with the 0.25W maximum power rating. Applying all the components • Transformer. • Graetz bridge G made with four D_{1} to D_{4} rectifying diodes 1N4004. • The smoothing capacitor 1000\mu F. • Zener diode: 1N4733A with P_{Z}=1.3W and V_{Z}=5.1V. • The resisitor R_{S}=51\Omega limit the current flow through the diode with P_{max}=0.25W. • The load resisitor R_{L}=1k\Omega. to the above circuit we got a working circut. Unfortunately when turn on, within just a second, a smell can be detected and resistor R_{S} became VERY HOT. Something has gone wrong. Simple calculatons explains what. Power rating for R_{S} in this configuration can be calculated as follow$$ P_{R_{S}} = I_{R_{S}} \cdot V_{R_{S}} = 0.26A \cdot 13V = 3.38W $$3.38W is much greater value than P_{max} for R_{S} which is 0.25W. So either resistor with higher power rating or resistor with higher resistance should be choosen. Considering second option, we calculate$$ I_{R_{S}} = \frac{P_{R_{S}}}{V_{R_{S}}} = \frac{0.25W}{13V} = 0.019A = 19mA $$Now we know that if voltage is 13V and current 19mA than power raiting should reach P_{max}=0.25W for resistor R_{S}. The resistance for R_{S} in that case is equal to$$ R_{S} = \frac{V_{S}-V_{Z}}{I_{R_{S}}} = \frac{13V}{0.019A} = 684.2\Omega $$To have some safety margin we choose R_{S}=820\Omega (gray-red-violet) and in that case$$ I_{R_{S}} = \frac{V_{S}-V_{Z}}{R_{S}} = \frac{13V}{820\Omega} = 0,0158A = 15.8mA $$Use LED for load Before we check if the way we think expressed in a previous section is correct, let's make a target circuit, that is a AC to DC fixed voltage convert with diodes and LED load as it is showned below  Left: Scheme of AC to DC fixed voltage convert with diodes and LED load. Right: Close to real view of a schema from left image Resistor R_{LED} may be calculated as follow$$ R_{LED} = \frac{V_{Z}-V_{LED}}{I_{LED}} = \frac{5V-2.5V}{0.015} = 166,6\Omega$$We took$R_{LED} = 180 \Omega$(brown-gray-brown). To verify all our previous calculations a voltmeter and ammeter (from Ampere Meter) can be attached as it is showned below In this configuration measured values were as follow $V_{1} = V_{S} = 17.27VA_{1} = A_{R_{S}} = 0.0157A = 15.7mAA_{2} = I_{D_{Z}} = 0.8mAV_{2} = V_{LED} = 4.77VA_{3} = I_{LED} = 0.148A = 14.8mA$When turn on, everything seems to be working correctly but 1. Resistor$R_{S}$become hot. It is not as hot as previously for$R_{S}=50\Omega$, we can say that it is stable hot, but anyway noticeably hot. Even with$R_{S}=1k\Omega$resulting with$13.4mA$current$R_{S}$become hot as it is hot for$R_{S}=820\Omega$. 2. Zener diode should keep voltage constant but when AC voltage is off, DC voltage is decreasing and along with it$V_{LED}$is decreasing -- other words, despite the fact that decreasing$V_{S}$voltage is (for a few seconds) grater than$V_{Z}$,$V_{LED}$is also decreasing instead stays on the same$V_{Z}$level as long as$V_{S} > V_{Z}$. And this happen regardless of the capacitance of the capacitors used (even if$4 \times 1000\mu F4 are used).

Problem number 2 could be explained in the following way. Zener diode keeps voltage stable on (more or less) $V_{Z}$ level as long as

• $V_{S}$ voltage is grater than $V_{Z}$ and
• current flowing through it is greater then minimum $I_{Z(min)}$ (typically 5-10mA).

The second condition is not fulfilled in our case.

The questions are

• How to select $R_{S}$ so it wouldn't be hot?
• What should be changed to have (for a few seconds) stable $V_{Z}$ voltage when AC voltage is off.

One of the possible solution is reducing LED ($A_{3} = I_{LED}$) current. The following configurations were tested

 $R_{LED}$ [$\Omega$] $A_{3} = I_{LED}$ [mA] $A_{2} = I_{D_{Z}}$ [mA] Comment 270red-violet-brown 10.9 4.58 Voltage $V_{2} = = V_{LED}$ drops rapidly when AC is offeven when $4 \times 1000 \mu F$ capacitors are used 300orange-black-brown 10.1 5.35 330orange-orange-brown 9.38 6.1 360orange-blue-brown 8.58 6.9 430yellow-orange-brown 7.38 7.95 Voltage $V_{2} = V_{LED} > 4.9V$ for 3 secondswhen $4 \times 1000 \mu F$ capacitors are used.Voltage $V_{2} = V_{LED} > 4.9V$ for 1.5 secondswhen $2 \times 1000 \mu F$ capacitors are used.

Convert AC voltage into fixed DC voltage with diodes and transistor

Convert AC voltage into fixed DC voltage with linear voltage regulator or switching converter