If you have a bunch of salvaged Zener diodes you need a quick way to sort them. In this article I describe how to build a simple test equipment that can be used to measure the breakdown voltage of a Zener diode. It can also be used to test LEDs and their color, since sometime a color LED can look white when not lit. This Zener tester is powered from a 9V battery so it doesn't involve dangerous mains voltage and can measure diodes up to 90V with the help of a 555 timer boost converter.
I also use it for checking for burned LEDs in a mains light bulb.
Main Features
- Constant 5mA test current
- Supports diodes up to 90V
- Powered from 9V battery making it safe and portable
- Can measure SMD components
- Has quick connect terminals for voltage and current display
- Cheap to build
- Protection on shorted and open (no load) terminals
- Can test a string of LEDs even in a mains light bulb
How to build a zener and LED diode tester
*DUT = Device Under Test or Diode
Under Test
Starting from left to right we have a 9V battery as a power supply and an external power switch (the red one in the above image). When the power switch is on the red LED is on. At this point the circuit draws 4mA. The zener diode D2 was added in series with the LED power indicator so when the input voltage drops bellow 5 volts, the red LED turns off and the battery needs replacement.
S1 is a momentary push button and is used to power the rest of the circuit when pressed. This TEST button should be pushed after the DUT is connected and only for a few seconds until the voltage on the multimeter is stable. This is for a few reasons: the inductor L1 will get a bit warm and to prevent the battery consumption.
A 555 timer IC is used to boost the voltage. The role of R2, R3 and C2 are to set the output frequency. With this setup the input current when the TEST button is pressed is 110mA. By making the resistors or capacitor a higher value, the frequency will decrease and so the power consumption. With 2.2K resistors and 100n capacitor the input current was 270mA! Even 110mA is a bit higher for a 9V battery but it only draws this current when the test button is pressed. C3 and C4 are decoupling capacitors for removing voltage noise.
When the power mosfet Q1 is on, inductor L1 will store energy in the form of a magnetic field. When the mosfet is off, the magnetic field collapses producing a higher voltage that will charge C5 and C6 through D3. D3 should be a Schottky diode but I didn't had one. Capacitors C5 and C6 must be rated for at least 100V and have low ESR. I've used two in parallel for a higher capacitance and lower ESR.
R5 is used to discharge the capacitors.
D5 is a zener diode for clamping the voltage to 100V.
The clamping voltage must be lower than the rated voltage of the output
capacitors. I used 3 Zeners in series. I had 30V + 30V + 32V = 92V.
CON5 is a quick connect terminal with springs used mainly for speakers. The connector has two pairs of red and black connectors. The right pair is used to connect the leads of a voltmeter and on the left pair can be connected a ammeter for checking the test current. J1 in a two pin jumper. When a current meter is connected the jumper bar needs to be pulled out and mounted back when the current meter is not connected.
CON6 and CON7 are two thick pins used to connect two alligator clips to them. The DUT is connected across them. PAD1 is just a copper pad used to test SMD diodes. At the end of the video you can see a 0805 LED tested.
The current control circuit
Since all datasheets specifies a 5mA test current for Zener diodes, we need a
way to maintain a constant current on all voltage ranges. Or almost, since at
higher voltages the current will decrease bellow 5mA since the battery can't
supply much current.
I've used the LM358 dual op-amp because was all I had but any single op-amp
will do. U2.2 is the unused op-amp and according to what I've read online, an
unused op-amp should not be left floating since it can cause noise, high
consumption and even internal chip damage. Instead, the non-inverting input
should be connected at a voltage between GND and VCC and the inverting input
is connected to the output.
U2.1 is the op-amp that controls the transistor Q3 which is a general purpose
NPN transistor and acts like a variable resistor to keep the current at about
5mA. R7 is a 200 ohm resistor for monitoring the current. When 5mA passes
through R7 a voltage of 1V will be across it. This voltage is monitored by the
op-amp.using the inverting (-) input. The non-inverting (+) input
monitors the voltage drop of diode D4. R4 provides enough current to cause a
1V voltage drop across diode D4. And since an op-amp is trying to keep its
inputs at the same potential, it will keep the transistor Q3 in a linear
region changing its "resistance" and so we have a steady current through the
DUT.
C7 is just a decoupling capacitor for the op-amp.
The enclosure
The box is made out of a sheet of plastic named Guttagliss. The battery is held inside by electrical tape.
The PCB is rested against two plastic tabs glued on both sides with the
distance from the top equal with the PCB thickness |
The nuts for the scews have been inserted by pushing them with a
soldering iron |
If you have any comments leave them down below.
Hello, Your circuit looks simple but effective, I like it. I just have one question. The collector of Q2 is connected only to DVM+ and CON6? I thik is like that.
ReplyDeleteThanks again.
Hi. I'm glad you like it. Yes the collector goes to CON6 and DVM+. I have updated the schematic image since the wire was overlapping the connector in the diagram. The transistor can be replaced with few zener diodes in series since it's hard to find one with a specific collector breakdown.
DeleteDo you Have any Recommendation on the value of schottky (D3) for replacement.
ReplyDeleteGreat Project by the way, simple and effective.
Keep it up
Thank you. Any schottky diode that can block 100V will do. I would choose one with minimum 150V blocking voltage and at least 0.5A of forward current capability and with the lowest voltage drop that I could find given that the battery voltage will drop with time. Most of the time I buy the components from Farnell and with the parametric search engine is very easy to filter for this values. Have a nice day.
Deleteis it possile for it to output more than 200v instead of 100?
Deleteis it possible to draw more than 200v instead of 100v?
ReplyDeleteI haven't tried but it might work. The output caps C5 and C6 should be rated for that voltage and clamping diode D5 needs to be changed to a higher voltage but lower or equal to what the capacitors can handle. If the voltage can't go higher, change the inductor L1 with a larger value.
Delete