DSO138 USB Power: No Boost Converter!

Introduction: DSO138 USB Power: No Boost Converter!

The JYE DSO138 is an excellent little oscilloscope for audio work and would make a great portable signal tracer. The problem is, it's not really portable because it needs a 9V power adapter. It would be better if it could be supplied from a standard USB power bank or any USB source. For some reason JYE did not take the opportunity to make the original DSO138 fully USB powered, which is strange as it is very easy to do. The PCB even includes a USB connector, but it does nothing! (There is an updated DSO138 MINI that is USB powered, but most people seem to own the original version).

There are instructions on the internet showing you how to add a boost regulator to convert USB power into 9V, but that is inefficient and unnecessary. In this instructable I will show you how to use the on-board USB connector directly for power input. I've also included an optional mod to get rid of the annoying waveform glitches that happen on some gain settings.

I've drawn the necessary mods on the original JYE schematic together with photos of my modded PCB.


Some hook-up wire
100uH 100mA (or more) inductor
2.2k resistor
470pF to 1nF capacitor
3k resistor (optional)
1.8k resistor (optional)
1.2k resistor (optional)

Step 1:

Remove D2. This is a safety precaution to disable the old DC power jack. This will ensure nothing goes wrong if you accidentally plug the old DC power supply in after you've done the USB power mod.

Solder a wire from the pad labelled VBUS to the pad labelled +5V. (These are labelled TP33 and TP21 on the schematic). This connects the USB power pin to the circuit's 5V power rail. The +3V rail is derived from this voltage by U3 which needs no changes.

Now remove U5 and solder a wire between the two outer pads, to jumper across where it used to be.

This takes care of the positive power rails, in the next step we'll mod the negative rail so it works off 5V USB too.

Step 2:

The DSO138 circuit uses a simple switching inverter to generate a rough negative voltage which is then regulated down to -5V by U4. Even though there is a feedback network to the CPU through R41/42, it appears JYE never implemented it in the firmware, the CPU simply produces a continuous 17.6kHz signal at R40. This means we have to modify the circuit to work from the 5V USB supply.

Replace the L2 with a 100uH inductor (rated for 100mA or more). I had one in my junk box. I had to bend the legs a little to make it fit in the same spot.

Add a snubbing network across D1 consisting of a capacitor in series with a 2.2k ohm resistor. It doesn't matter what type of capacitor you use, but the value should be between 470pF and 1nF. I used a 1nF plastic cap because that's what I had. This will clean up the switching waveform.

You're done! You can now plug in a USB cable and measure the voltages on the PCB test pads which should still be -5V, +5V and 3.3V. The next step is optional.

Step 3:

When viewing large signals you may have noticed glitches on the waveform. This is caused by excessive loading of U2B by the potential divider R6/7/8. The solution is easy:

Replace R6, R7 and R8 with resistors ten times greater in value, i.e. R6 = 3k, R7 = 1.8k, R8 = 1.2k.

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    1 year ago

    There's only one big issue with using USB power: Not all USB power sources are created equal.

    Hub power can vary over wide limits, (as great as +4 to +6 depending on the specific hub and the devices attached to it), and depending on the supply used it can be quite noisy.

    USB "phone chargers" can have significant ripple.

    Laptops, depending on how well made they are can also have noisy, drifting power.

    The best source is a supply that is specially designed for electronics, like a Raspberry Pi supply, or a power bank, though be careful as cheap power banks have poor filtering and can be noisy.


    Reply 10 months ago

    That doesn't really matter since all the voltages are regulated locally and Vcc is non critical.


    1 year ago

    It didn't work for me when I replaced L2 with a 100uH one. I used a 210uH inductor and the -AV is reaching only -4.9V and a voltage of -6.3V at TP25 (the U4 datasheet says it needs -7V minimum). The oscilloscope turns on but It doesn't seem to work correctly. I'll have to try using a 330uH inductor and maybe that value would be good enough.


    2 years ago

    thanks for the correction!
    I'm trying to implement it, but i a slightly different way (i.e. maintaining the original 9V supply, possibly by battery, switchable to 5V with a slide switch) and, doing so, I've discovered a problem in the unmodified board:
    when I power the board with 9V, the +AV is at 4.96V, while -AV is -4.3 or even less: investigating a bit, it comes out that the voltage at TP25 is -5.1V, that of course is not enough to be stabilized at -5 by the 7905.
    The effect of this undervoltage is an asymmetric waveform, and a wrong voltage measurement
    If I increase the power supply at 10.5 V or more, also -AV reaches -4.98V, and everything works fine.
    Looking at the schematic, I think that a problem could be a L2 inductance with a value lower that the 1mH stated in the schematic.. what do you think?

    Furthermore (again, I'm trying to understand how it works, so excuse me if I'm asking silly questions): I assumed that, in order to have that same switching working with a lower input voltage (5 instead of 9V), the L2 should have been bigger, to drain the same current from D1 when Q1 is switched off).. but you chose a 100uH, instead of a 1mH, so ten times smaller.... where am I going wrong?

    thanks for your patience



    Reply 2 years ago

    Hi, I'm no expert in switchmode converters, but I believe it must be operating in discontinuous mode, which is why a smaller inductor leads to more voltage gain: https://en.wikipedia.org/wiki/Buck_boost_converter
    I suggest you try different inductors until it works on 5V.


    Question 2 years ago

    I've already hacked the DSO138 by installing the mod firmware DLO138 (https://github.com/ardyesp/DLO-138), and creating the analog section for the second channel.. and now I would like to have it 5V powered.. so I'm trying to implement you solution.. I have a couple of questions about step2:

    a) can you explain a bit better how it -5V section of the circuit works originally, and how it is intended to work now? I'm not very familiar with the switching schematic.. in particular, I can not understand how the circuit can work if the V+ bus (downstream D2) is not supplied (or, in other words, where the +5V supplied by the USB is connected to the negative supply section).. perhaps it is a silly question.. sorry

    b) in case I do not remove the D2, apart the risk of supplying noth )V and 5V at the same time, is there any interaction across the +V bus?



    Answer 2 years ago

    Oops, you're right, I missed a step out! You need to remove U5 and jumper across it. Have a look at this ghetto sketch for a clearer picture. I have fixed the instructable.