Solved Problems On Zener Diode Pdf
Janne Desir
Q3. A 7.2 V zener is used in the circuit shown in Fig. 3 and the load current is to vary from 12 to 100 mA. Find the value of series resistance R to maintain a voltage of 7.2 V across the load. The input voltage is constant at 12V and the minimum zener current is 10 mA.
If R = 43.5 Ω is inserted in the circuit, the output voltage will remain constant over the regulating range. As the load current IL decreases, the zener current IZ will increase to such a value that IZ + IL = 110 mA.
Q4. The zener diode shown in Fig. 4 has VZ = 18 V. The voltage across the load stays at 18 V as long as IZ is maintained between 200 mA and 2 A. Find the value of series resistance R so that E0 remains 18 V while input voltage Ei is free to vary between 22 V to 28V.
Q5. A 10-V zener diode is used to regulate the voltage across a variable load resistor [See fig.5]. The input voltage varies between 13 V and 16 V and the load current varies between 10 mA and 85 mA. The minimum zener current is 15 mA. Calculate the value of series resistance R.
When the desired regulated output voltage is higher than the rated voltage of the zener, two or more zeners are connected in series as shown in Fig. 6. However, in such circuits, care must be taken to select those zeners that have the same current rating.
Q7. What value of series resistance is required when three 10-watt, 10-volt, 1000 mA zener diodes are connected in series to obtain a 30-volt regulated output from a 45 volt d.c. power source ?
Q8. Over what range of input voltage will the zener circuit shown in Fig. 8 maintain 30 V across 2000 Ω load, assuming that series resistance R = 200 Ω and zener current
rating is 25 mA ?
(ii) If the zener diode is short [See Fig. 10 (ii)], you will measure V0 as 0V. The same problem could also be caused by a shorted load resistor (= 5kΩ) or an opened source resistor (= 1 kΩ). The only way to tell which device has failed is to remove the resistors and check them with an ohmmeter. If the resistors are good, then zener diode is bad.
When the filter capacitor shorts, the primary fuse will blow. The reason for this is illustrated in Fig 11. When the filter capacitor shorts, it shorts out the load resistance RL. This has the same effect as wiring the two sides of the bridge together (See Fig. 11).
If you trace from the high side of the bridge to the low side, you will see that the only resistance across the secondary of the transformer is the forward resistance of the two ON diodes. This effectively shorts out the transformer secondary. The result is that excessive current flows in the secondary and hence in the primary. Consequently, the primary fuse will blow.
When the filter capacitor opens, it will cause the ripple in the power supply output to increase drastically. At the same time, the d.c. output voltage will show a significant drop. Since an open filter capacitor is the only fault that will cause both of these symptoms, no further testing is necessary. If both symptoms appear, replace the filter capacitor.
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I am newer to this CAD. Installed, footprint downloaded and activated.
when used zener as voltage regulator, it is not working as expected. not working in rev. direction and breakdown never happen. i have used .model Diode D(Bv=5.6).
what is the problem. is it footprint or script in model?
Make a simple circuit: Voltage source driving the Zener diode. Sweep the voltage from forward to reverse, plot the diode current. Check if it is o.k., or polarity is reversed or no current at all. If not o.k., you may post this circuit here, and we may check.
I'm developing an LED indoor display. I use matrix LED display driver type TLC59581. I use 4 drivers on one panel (connected in series).
The supply voltage of the RGB LEDs is 4.2V, and driver supply voltage is VCC = 5V (separate). I use a 4-layer PCB. My solution is very similar to TIDA-00161 ( -00161 ).
At small LED currents, everything works fine, but at high currents, some LEDs do not light up (check the picture). Iref=25mA, registers GS/BC/CC=max. The absolute maximum ratings parameters are not exceeded. VCC (5V) supply voltage checked with an oscilloscope - in no case does the voltage drop by more than 100 mV during operation.
If this problem occurs, then rewriting data and control registers doesn't help. I need to reset the power supply. Adding bulk capacitors for VCC/VLED doesn't help. LED open detection error flag is not set (read by loopback).
Another test. I connected VLED to an external laboratory power supply and changed the voltage from 6V to 4.2V. Decrasing value of VLED voltage causes more LEDs to light off. Important - increasing the VLED again (and send new data) does not solve the problem. I need to reset the TLC59581 supply voltage (VCC).
I have a feeling that the problem is in the internal structure of the TLC59581. For some reason, some analog part of this IC latches and stops working. Re-uploading the data/control registers does not help. Only on/off power supply. The problem is that there are separate supply voltages VCC and VLED.
There are two problems.
1) Too low voltage at the OUTx output. The TLC59581 needs about 0.5 V for the correct operation of the output current source. In my circuit, this voltage temporary drops to about 0.4 V, which causes incorrect operation. Adding bulk capacitors, using better MOSFETs, slightly increasing the VLED supply voltage solved the problem (+improving the PCB). Problem solved.
2) When the LED matrix power supply (VLED) is too low, the output OUTx of the TLC59581 is permanently turned off (for a row). I understand that for too low voltage, the output current source does not work properly. But why after increasing the supply voltage (VLED), the outputs still do not work? Why is a power supply reset needed?
It is because you turn on LOD_Removal function, and LOD detection threshold default value is 0.4V. So the LEDs are disabled after detecting the open failure, even you set higher Vled it still cannot light up.
If the Zener voltage is set too high, the upper side ghosting would not be eliminated, if the Zener voltage is set too low, the LED will stand too high reverse voltage, which may decrease the life span of LEDs.
A Zener diode is a type of semiconductor device that allows current to flow in both directions and is designed to have a specific breakdown voltage, known as the Zener voltage. It is commonly used as a voltage regulator, protecting circuits from voltage spikes and maintaining a constant voltage output.
A Zener diode works by using a process called the Zener effect, which occurs when a reverse-biased diode reaches its Zener voltage. This causes a large number of electrons to cross the junction, creating a small amount of current that is used to regulate the voltage. The Zener diode also has a sharp breakdown voltage, allowing it to maintain a constant voltage output even with changes in input voltage.
Some common problems that can occur with Zener diodes include overheating, voltage spikes, and incorrect voltage regulation. These problems can be caused by improper circuit design or incorrect usage of the diode, such as exceeding the maximum current or voltage ratings.
If you are experiencing issues with your Zener diode, the first step is to check the circuit design and make sure it is within the diode's specifications. You can also use a multimeter to measure the voltage and current across the diode to ensure it is within the correct ranges. If the diode is overheating, you may need to add a heat sink to dissipate the excess heat.
Some tips for using Zener diodes effectively include choosing the correct diode for your circuit, following the manufacturer's specifications, and using a heat sink if necessary. It is also important to properly connect the diode in the circuit, ensuring the correct polarity and avoiding any short circuits. Additionally, regularly checking the diode's voltage and current values can help prevent potential problems.
So my question is: Is my solution even correct (if not then where did I go wrong?) and secondly, is there a solution that I can do in "auto-pilot" so to speak. Like node-voltage analysis or another way that doesn't need clever thinking.
Since the inverting and non-inverting pins are shorted, you have 0 V at one end and -3 V at another end of resistor \$R_1\$. So the current of 3 mA flows from inverting pin towards the supply voltage. But where does this current come from? This is where the feedback resistor comes in action. The feedback resistor allow 3mA of current to pass from \$V_o\$ pin to inverting pin. But this creates a voltage drop of 3V across the resistor and Zener diode.
As you have told, the Zener voltage is 2 V which implies that the voltage across the resistor can only be 2 V. So, the output can no longer increase the voltage. Then how is the current balanced? Well, it is the duty of Zener diode to supply the current when it has crossed its reverse breakdown voltage.
You're correct. A useful concept here is a virtual ground...when connected in a negative feedback situation, the output does whatever it can to make the negative input look exactly like the positive one. Since the positive input is grounded (and I'm assuming that's within its common mode voltage range) the output will increase with decreasing Vin until it reaches the zener breakdown, after which it can supply all the current it needs with a voltage of 2V.
To protect against short overvoltage cases, where the user moves the motor in turning direction, I am using a zener diode (BZG04-22-M) connected to Motor_VCC and Ground. With Software (simple PID Controller) I can detect the user moving the motor faster than expected and reversing the polarity of the motor, which slows down the movement until the actual position is again equal to the expected position. Therefor the application has no problem with overvoltages during normal operation.
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