$20 LCR ESR Transistor checker project

[MaxFrister]

I'm pretty sure it is defective although I was unable to quantify how defective.  I tried to measure the current using a series resistor and managed to melt the probe grabber on my scope.  I don't think I'll try that again.

As for shorting K and A on the tester, I didn't do that but isn't that a normal diode failure mode?

Thanks.

[indman]

As for shorting K and A on the tester, I didn't do that but isn't that a normal diode failure mode?

This is really not dangerous if a current stabilizer or its limiter is provided in the circuit. If it is not present or it is incorrectly calculated, this will lead to the failure of the measuring port of the controller.

[Feliciano]

To characterize the boost converter you have in front, with the aid of your DMM:
a)you could measure the open-circuit voltage, it should be around 30V
b)then you could put a resistor, like 1K (around 1W to be on the safe side), and measure the voltage again, calculate the current, and check whether the internal current limitiation is working as it should. For testing a Zener, a few mA should be enough

[indman]

Feliciano,this measurement operation is true for the classical tester scheme. But in the clone version with a fake processor, the LM358 operational amplifier is present in the zener diode measurement circuit.What function does it perform in the scheme? A clear and understandable schematic diagram of the clone is needed to definitively determine the cause of this error.

[Feliciano]

For sure an schematic (official or reverse-drawn) would be better, but as you know a boost converter should follow an inductive (classical) or capacitive (voltage multiplier) principle, and we could try to guess a rough Thevenin or Norton equivalents of it around a given operating point. If the zener gets too hot, the Voc is too high, and/or the internal resistance is too low. Knowing which is the case, a modification could be devised to the circuit, trying to make it safe.

Regarding the op-amp in this case, it could be part of a not-so-regulated (cheapo) boost converter, or it could be part of the V sensing for protecting the microcontroller. That would deserve analysis after the initial findings. On the other hand, an LM358 can only source a few tens of mA, bellow it's Vcc. That's why if I were MaxFrister, I would start by trying to understand how the circuit behaves and then think whether a mod can be made.

[indman]

Look closely at the PCB - pins A are shorted to ground through a 40 ohm resistor(59x)! The output voltage from the DC-DC switch (28V) is directly fed through the diode to pin K. It turns out that the current through the zener is limited only by this resistor and open diode (and it's ~ 0.7A!!!) That's why the diode D2 damage and the zener diode is very hot. If I saw the right track on the board.
For the experiment I suggest you solder a 5-10 kΩ resistor in series with the diode.It's not hard mod.

[madires]

Look closely at the PCB - pins A are shorted to ground through a 40 ohm resistor(59x)! The output voltage from the DC-DC switch (28V) is directly fed through the diode to pin K. It turns out that the current through the zener is limited only by this resistor and open diode (and it's ~ 0.7A!!!)

Ouch! A constant current source with a BJT and a few diodes and resistors is simple and inexpensive. Just a series resistor causes a too large variation of the test current, which in turn also impacts the measurement (Zener voltage). AFAIK, the TC-1 and T7 Plus have a 10k series resistor. We would have less trouble if those cloners would stick to the recommended design instead of trying to cut even more corners.

[indman]

A constant current source with a BJT and a few diodes and resistors is simple and inexpensive.

You're right of course, a simple current constant regulator is called for in this design. But this is only one obvious problem that was detected. And I'm sure there will be others.

[MaxFrister]

Look closely at the PCB - pins A are shorted to ground through a 40 ohm resistor(59x)! The output voltage from the DC-DC switch (28V) is directly fed through the diode to pin K. It turns out that the current through the zener is limited only by this resistor and open diode (and it's ~ 0.7A!!!) That's why the diode D2 damage and the zener diode is very hot. If I saw the right track on the board.
For the experiment I suggest you solder a 5-10 kΩ resistor in series with the diode.It's not hard mod.

I was able to install a 5.1k resistor for current limiting.  I chose to install it after the cap rather than before.  I initially tried a 8.2k resistor but the tester reported a zener of 29.5V with nothing in the test socket.

On my unit, "A" is connect to ground via a 5-ohm resistor.  I'm not really sure why. 

It works now and best of all you can no longer use the DUT as a soldering iron.

[indman]

I chose to install it after the cap rather than before.  I initially tried a 8.2k resistor but the tester reported a zener of 29.5V with nothing in the test socket. On my unit, "A" is connect to ground via a 5-ohm resistor.  I'm not really sure why. 

I'm not quite sure how you installed the extra resistor? Can you show a high quality photo of this section of your PCB with the resistor installed?

[vklimk]

I was able to install a 5.1k resistor for current limiting.  I chose to install it after the cap rather than before.  I initially tried a 8.2k resistor but the tester reported a zener of 29.5V with nothing in the test socket.

On my unit, "A" is connect to ground via a 5-ohm resistor.  I'm not really sure why. 

It works now and best of all you can no longer use the DUT as a soldering iron.

Try to play with that 5 ohm resistor instead of 5,1k. It seems that 5 ohm resistor is a shunt (current sensor) in current supply unit (similar to power supply for led). Put variable resistor in place of 5 ohm and try to change current using it.
It is just a theory, I don't have such a device to check. But if sdb628 is in use here then 120 ohm shunt should limit current to 5mA. 5ohm shunt should limit current to 120mA.

[mebel]

Sorry for obvious question.

The link doesn't work anymore. Where from can I download this file?

1. To download from the author's site at the link https://www.mikrocontroller.net/svnbrowser/transistortester/Software/trunk/  the current distribution kit of firmwares.
For this purpose to click in the bottom of the page on a line of ""Download GNU tarball"".

[Fuzzy Star]

Sorry for obvious question.

The link doesn't work anymore. Where from can I download this file?

1. To download from the author's site at the link https://www.mikrocontroller.net/svnbrowser/transistortester/Software/trunk/  the current distribution kit of firmwares.
For this purpose to click in the bottom of the page on a line of ""Download GNU tarball"".

https://github.com/kubi48/TransistorTester-source/

https://github.com/madires/Transistortester-Warehouse/tree/master/Firmware/k-firmware

[indman]

Try to play with that 5 ohm resistor instead of 5,1k. It seems that 5 ohm resistor is a shunt in current supply unit (similar to power supply for led). Put variable resistor in place of 5 ohm and try to change current using it.

MaxFrister If now you have nothing warmed up and check stabilitrons works fine, then do not touch anything else and listen to other silly advice, otherwise you risk to lose serviceable device! 

[vklimk]

Try to play with that 5 ohm resistor instead of 5,1k. It seems that 5 ohm resistor is a shunt in current supply unit (similar to power supply for led). Put variable resistor in place of 5 ohm and try to change current using it.

silly advice

Actually the advice can be not so silly.
Look. It is typical application of SDB628 in image1, it is voltage source:

But SDB628 can be used as a current source, see image2:

A current through D2 will be equal to 0.6/R2 (in case D2 rating is bigger than Vin).
If R2=5 Ohm, then the current will be 120mA.
If R2=40 Ohm, then the current will be 15mA.
Look at the image you posted more carefully and you will see that it is the case. So, there will be no 0,7A current there, just 15mA.

[indman]

If R2=5 Ohm, then the current will be 120mA.
If R2=40 Ohm, then the current will be 15mA.

I don't need to explain the theory of how a simple chip DC-DC works.I can give you a big lecture about her work myself. The current of 15mA, and even more 120mA, which you have "calculated" is many times higher than the maximum permissible for checking a low-current zener diode. I repeat once again - do not give silly advice if you are 100% unaware of the complete connection scheme of the zener diode from Chinese "engineers".

[vklimk]

Whole circuit is clearly visible in the photo you provided. You can see that it is the same as I provided in image2.
I believe that you have a big background in electronics, much bigger than mine.
But please don't be so aggressive, be more polite.

[Feliciano]

Let me try to shed some light on the subject:
1.- Let's say the "designer" of this clone took the typical schematic from the boost converter datasheet, and forgot to add a current limitation (by mistake, or to save a few more cents)
2.- Let's say the boost can source 100mA and 30V (just a guess, I don't have measurements of the circuit with its chinese ICs):
2.1.- If you try to test a 30V zener, you are applying 3W to it (P=V*I). If it's an small zener (0.25W or 0.5W) you will fry it, or if supports it for short term, you would be testing it outside its intended operating window
2.2.- On the other hand, if the efficiency of the boost converter is around 85%, you will be asking close to 1A from the battery. For a small Li-ION battery, you will be exceeding the recommended C-rate
2.3.- If you put a small resistor in parallel to the zener, you will be lowering the maximum zener voltage you can measure, and for small-V zener (like 3.3V), you will still applying a lot of current to it, and drawing high current from the battery, whether you're testing a zener or not
2.4.- If you modify the IC voltage divider, you will be still lowering the open circuit output voltage, but not limiting the current
Therefore, as madires suggested, the best mod for a circuit like this it would be to add a current limiter, or if space is a problem, at least a series resistor, as indman suggester earlier (just make sure to add it before the measuring loop)

[indman]

A competent and simple solution, as indicated in the screenshot. Selection of the most optimal current and its stabilization for safe testing of high-power and middle(low)power zener diodes.

[MaxFrister]

I'm not quite sure how you installed the extra resistor? Can you show a high quality photo of this section of your PCB with the resistor installed?

Attached is a picture of my ugly fix as well as an attempt at reverse engineering of the LCR-T7 zener test circuit.

It looks to me like the resistor R16 is bad and that it is intended to control the feedback pin of U7 in constant current mode (0.6V/60R = 10mA).   

[indman]

On my unit, "A" is connect to ground via a 5-ohm resistor.  I'm not really sure why. 

Why do you think that R16 on your board has a resistance of 5 ohm? The internationally coded 59X stands for 40.20 ohms ± 1%.
Your board has exactly the same resistor rating as the one I showed in my example.

[MaxFrister]

Why do you think that R16 on your board has a resistance of 5 ohm?

That is approximately what it measures in circuit.  I can pull it and measure it accurately tonight.

Does anyone know what IC U7 is?  It is not the SDB628 mentioned above; the pin wiring does not match.

[indman]

That is approximately what it measure in circuit.  I can pull it an measure it accurately tonight.
Does anyone know what IC U7 is?  It is not the SDB628 mentioned above; the pin wiring does not match.

U7 is very similar to SY7208CABC in the location of the contacts, but this cannot be said with 100% certainty, because the HYDSI or HYDWE marking does not appear anywhere yet.

P.S Here is an even more likely interpretation of U7 - this is AN_SY7200A (marking HY6VE). If you believe the documentation for this chip, then the output current is calculated by the formula I=0.2/R1. If R1 = 40 ohms, then the test current I=0.2/40=0.005A, that is, 5mA.

Now it remains to solve the riddle from Chinese friends -why does this circuit not withstand a given R1 current and allows it to conduct a much higher current?

[vklimk]

Because of open circuit. There is no load to inductance if you do not measure zener diode and there is no current, there is no voltage drop at current sensor resistivity and IC tries to add more and more power, it works at its maximum, it loads a lot of power into inductance and inductance generates high voltage pulses. And these high voltage pulses from inductance go to both capacity and IC itself. Both of them get fried in some time. If there is a big current then it means that feedback part of IC has already been fried and resistance between pins 2 and 3 is very small (shorted), such fried IC continues to work but with no ability to adjust the current by mean of that 40 ohm resistor any more.

[indman]

It is possible, although according to the documentation, this chip should have good overload protection for mode of open circuit.
And here is also an important note from the datasheet, which should be taken carefully:
"6) If the system chip interfacing with the EN pin has a
high impedance state at shutdown mode and the IN pin
is connected directly to a power source such as a Li-Ion
battery, it is desirable to add a pull down 1Mohm
resistor between the EN and GND pins to prevent the
noise from falsely turning on the regulator at shutdown
mode."