You can make this easier to set up by using two silicon diodes in series in place of Rx. In this case the PNP transistor will turn on as soon as the load current exceeds about 1uA (sufficient to cause a collector current of about 100uA to pull the EN pin to the active state). At this point almost all the load current will flow into the base of the transistor because the voltage across the two diodes will be only about 0.6v total (ie 0.3v per diode) and thus too small to cause significant conduction. Once the unit powers up the output voltage will be about 1.2v below the voltage on C2 (ie two diode pedestal voltages) and will have better load regulation than you will get using the resistor Rx. I hope that is of use.
Yes, good suggestion. A further improvement would be installing your with @John Conrad's one: between diode and R1 (from voltage feedback). In this later case, the output voltage (Turn-ON) would be constant and the "internal" / additional Joule dissipation would be limited to (0.6 x I).
I'm a bit new to this game so sorry if this is a dumb question but will any silicon diode suffice? Will something like a 1N400x, or maybe a 1N540x, or a 1N4148 do the job (just because I have all of those on hand), or is there more to it than that?
@@uniquepete9417 A very sensible question actually! I would use the 1N540x as they will carry the full 2A output of the module though if you were using it exclusively for a known lower current load you could use diodes of lower rating appropriate for that current. Probably only worth doing that if you were pushed for space to accommodate the bigger diodes.
For multimeter application i have a solution for it. The solution is by searching for the vcc pin of the multimeter, vcc pin is will output a 9V if the mutlimeter is on & 0V when off. But it tricky to find this pin, but luckly i find one on my aneng m1 multimeter, so i just solder a cable to the enable pin of the mt3608 ic and the circuit works. It measure 40uA when off & 5mA when on.
For the transistor to turn on you need a voltage drop of around at least 0.6V on the resistor. That's quite alot, especially if the output current is a few 100mA. The resistor also makes the "output" voltage current dependent, which means you will often need another regulator (with it's own losses) after the this converter.
Or an op amp with global feedback. I honestly think it is the easiest implementation, perhaps not the cheapest. Any old op amp will compensate enough for the cross over distortion without any biasing, unless we are talking audio, but that's another kettle of fish. Regulating DC voltage is easy in comparison.
@@johnconrad5487 how do you mean between the diode and R1?? What way would you you place it because I would like to do this hack the most efficient way. Could you explain for me on which side of the diode and R1 and how dose it connect back to the transistor base??? I would Appreciate your input on this lots thank you✊✊
@@johnconrad5487 I thought exactly that way too. Using the current limiting feature (to turn on) before voltage feedback. But the circuit board might need more invasive modifications...
Don't pry the IC pin up. It will bend and then break off. It did that twice for me. Use hot air rework to remove the chip, bend it up carefully and then put it back with flux and hot air.
Ever use a current mirror FET? It taps off a sample of the drain currrent into a separate pin, which can be used to measure the drain current. Seems like you could make a current sampler with no voltage drop and no quiescent power drain.
I have an idea. If you replace the resistor with two diodes in series and a 1Kohm (more or less) resistor in parallel with them you can obtain a more fixed output voltage. The resistor will put off the transistor, and when there is load, there will be a drop on the diodes. Actually it should work with a single diode too.
I keep a stock of these MT3608's to add to my own board designs. Although I haven't used them in a design yet in practice. The modules are too easy to play with, and I'm lazy ;) Anyways, adding a circuit like this would be easy and could prove useful. I'm not sure how off the top of my head, but it seems like we could make a better (but more complicated) circuit based on continuity instead of current using some kind of JFET, but I'm just typing my half baked ideas now... Thanks for the upload. -Jake
Upcycle Electronics , hey I was wondering if you can solve my dilemma regarding this converter. I have a beard trimmer that I turned to lithium using an 18650 battery, a tp4056 charging module and a this converter. It worked perfectly except that after a couple of days it doesn't work anymore. I have been messing with this on 3 seperate occasions with it working each time but for only a few days. Please tell me what you think. Thanks
0:03 Yes indeed, everyone knows why we need DC to DC boost converters. Why I just asked an old fellow on the way to the crack house how he increases his DC voltages and replied "Well, we use DC to DC boost converters old chap".
Hi, Is there any way to modify it for low current usage. I measured about 1.4mA. Want to use it for a smoke detector that works on a 9V battery. Instead I want to use a 16500 lithium with protection circuit and DC booster. But the smoke detector only uses about 0.03mA (calculated) so the 1.4mA of the booster without load is way to high
Hi, Played a little with the original one. Putting a small resisor in line with the inductor changed it a little but not much and depends on which side. Bigger stops it from working at all. Second attempt. I removed the trimmer pot and 222 resistor. Replaced the values by multiply by 4. (didn't have a 1M trimmer pot) 390K for the pot. 220k for the runner. And replaced the 222 for a 10K resistor. Output 9.4Volt On current dropped 4x so instead of 1.3mA is went to 0.28mA no load. 400/500mA must be possible without a big voltage drop. (tested) When startup current is to high it falls back to 2.2V and can be reset by disconnecting the battery. My smoke detector draws about 0.12mA. With the booster it's 0.48mA total. Still too high make some benefits. Maybe you can do something with it. I can't see how the output behavior is with this mod. Combined with the mod in the video maybe it makes a bigger difference. Or try a 1M trimmer pot and 22K resistor instead tp see how it behaves. Maybe I test this later.
They seem to release the magic smoke if you input more than 6v. I guess your PCB service won't take hand drawn schematics as I don't have the money for a license for PCB software or the time to learn how to use it
This is because of faulty design. Note from the schematic in the video how the trimpot wiper is not configured as a voltage divider. If wound to the end of its travel it will put nearly the whole of the input voltage into the feedback pin. This input is rated at 6V maximum, and feeding more than that into it will produce the black smoke. I have noticed that those versions which do not have thru-holes for the input and output pads are of a different design and work properly up to their rated input voltage for any output voltage adjustment. You can also identify the correct version because the wiper of the trimpot is earthed.
By maintaining a constant voltage across a resistor. A very simple way to do it is with, for example, a high gain NPN transistor. Put your load between Vin and the Collector. Then put a suitable transistor of resistance R between the Emitter and ground. The voltage, V, at the Emitter is I × R Volts, where I is the current you want. So, for a normal NPN resistor with current flowing through it, the voltage at the Base of the transistor must be 0.6 + V. You can control the voltage at the Base with a voltage divider, and then the correct amount of current will flow. Make sure the current flowing through the voltage divider is much larger than the current flowing into the Base. If you get that wrong, the current flowing into the Base will affect the voltage at the Base, and it won't work as expected.
@@petermcarthur7450 any possibility you could email a quick sketch of adding an npn? I'd like to try this but only have one board and don't want to fry anything. I'm looking to run one of these at 3-4a (max on datasheet) and need current regulation. 🍺😀
@@petermcarthur7450 it would be a single li-ion 18650 and the load being 5v. I'd even be happy with 2.2A and have slightly larger inductors I can add.. Problem is the load drops resistance with an increase in temperature.
Has anyone figure out yet how circulation pump motor works? I know it uses gate driver in IC to drive gates in IGBT. It is 3 phase too. pump name Wilo made in sweden or something. Same shit is in those compressor heater/cooler, the fans are run by PMM thats driven by weird pwm using gate drivers and IGBT. I wan to make them run on 12 or 24v DC. Any smart person here? I could even pay if you know where to tap the pcb.
I liked your old voice more! I have a video request for you. I need to wind a 2000Volts transformer, that is small like 5cm×5cm×5cm. And that can deliver very high power. The most you can make possible. Im not good at winding transformers nor designing them but I know you are perfect for the job with all your knowledge, so if you have the time please make a video on it. 2000volts and atleast 80watts.
sir kasyan please hear me can i request a contineuty high resistance tester like using tracing the electrical wires please................................
And how much more power is wasted into heat on the resistor you added once you connect the load to the circuit? Isn't it easier and better to add a switch and turn the device off when it's not used rather then do this mod that wastes energy while load is connected?
I'm not following the purpose of this circuit as applied to a multimeter. Most multimeters worth owning have 9V batteries that last until they leak, which is often more than five years depending on use. Change your battery every year and you won't have that problem. No turn-on problems with super-light loads either. This circuit requires a power source, either battery or buck or P/S to work at all, so two parts are now needed when only one was before. Why is that better than a plain old 9V battery?
9v batteries are the worst on the planet at $4 each. Some meters last a week on 9v rechargable (1253b oled is one). The old fluke 73,75,77 were excellent on batteries, but the newer stuff eats them in 3,4 months of daily use at work.
Many hobbyist not careful about this fact, until they learn about durability, efficiency, and reliability on electronics instrument. If they love their instruments, they will never turn on the instrument when the electricity not stable enough at start ON. Bad habit, and it was famous habit 🤨
Interesting video! Perhaps you fond this interesting too: While making a video about using this module to make a 9V battery I fried the MT3608 chip. On my channel you can find another video about how I repaired the module.
![BLACK BAG - Official Trailer [HD] - Only in Theaters March 14](http://i.ytimg.com/vi/Du0Xp8WX_7I/mqdefault.jpg)
You can make this easier to set up by using two silicon diodes in series in place of Rx. In this case the PNP transistor will turn on as soon as the load current exceeds about 1uA (sufficient to cause a collector current of about 100uA to pull the EN pin to the active state). At this point almost all the load current will flow into the base of the transistor because the voltage across the two diodes will be only about 0.6v total (ie 0.3v per diode) and thus too small to cause significant conduction. Once the unit powers up the output voltage will be about 1.2v below the voltage on C2 (ie two diode pedestal voltages) and will have better load regulation than you will get using the resistor Rx.
I hope that is of use.
Thanks
Yes, good suggestion. A further improvement would be installing your with @John Conrad's one: between diode and R1 (from voltage feedback). In this later case, the output voltage (Turn-ON) would be constant and the "internal" / additional Joule dissipation would be limited to (0.6 x I).
I'm a bit new to this game so sorry if this is a dumb question but will any silicon diode suffice? Will something like a 1N400x, or maybe a 1N540x, or a 1N4148 do the job (just because I have all of those on hand), or is there more to it than that?
@@uniquepete9417 A very sensible question actually! I would use the 1N540x as they will carry the full 2A output of the module though if you were using it exclusively for a known lower current load you could use diodes of lower rating appropriate for that current. Probably only worth doing that if you were pushed for space to accommodate the bigger diodes.
I have some SX1308 it draws 0.5mA in idle, but what is drawing those 0.5mA is the SDB628 ic or the diode?
I have a few of those converter and constant empty battery problem with the same purpose! This is brilliant! Thank you for the design!
For multimeter application i have a solution for it.
The solution is by searching for the vcc pin of the multimeter, vcc pin is will output a 9V if the mutlimeter is on & 0V when off.
But it tricky to find this pin, but luckly i find one on my aneng m1 multimeter, so i just solder a cable to the enable pin of the mt3608 ic and the circuit works.
It measure 40uA when off & 5mA when on.
Love your videos. As of now no questions. If needed I will post here for help. Good luck
I could listen to you all day
For the transistor to turn on you need a voltage drop of around at least 0.6V on the resistor. That's quite alot, especially if the output current is a few 100mA. The resistor also makes the "output" voltage current dependent, which means you will often need another regulator (with it's own losses) after the this converter.
Or an op amp with global feedback. I honestly think it is the easiest implementation, perhaps not the cheapest. Any old op amp will compensate enough for the cross over distortion without any biasing, unless we are talking audio, but that's another kettle of fish. Regulating DC voltage is easy in comparison.
actually moving Rx to between the Diode and R1will solve the voltage regulation problem.
@@johnconrad5487 how do you mean between the diode and R1?? What way would you you place it because I would like to do this hack the most efficient way. Could you explain for me on which side of the diode and R1 and how dose it connect back to the transistor base??? I would Appreciate your input on this lots thank you✊✊
@@johnconrad5487 I thought exactly that way too. Using the current limiting feature (to turn on) before voltage feedback. But the circuit board might need more invasive modifications...
Don't pry the IC pin up. It will bend and then break off. It did that twice for me. Use hot air rework to remove the chip, bend it up carefully and then put it back with flux and hot air.
Ever use a current mirror FET? It taps off a sample of the drain currrent into a separate pin, which can be used to measure the drain current. Seems like you could make a current sampler with no voltage drop and no quiescent power drain.
Why don't you put the sense (3) MT3608 over R1 not behind the RX to have a defined output voltage?
I have an idea. If you replace the resistor with two diodes in series and a 1Kohm (more or less) resistor in parallel with them you can obtain a more fixed output voltage. The resistor will put off the transistor, and when there is load, there will be a drop on the diodes. Actually it should work with a single diode too.
I keep a stock of these MT3608's to add to my own board designs. Although I haven't used them in a design yet in practice. The modules are too easy to play with, and I'm lazy ;)
Anyways, adding a circuit like this would be easy and could prove useful.
I'm not sure how off the top of my head, but it seems like we could make a better (but more complicated) circuit based on continuity instead of current using some kind of JFET, but I'm just typing my half baked ideas now...
Thanks for the upload.
-Jake
Upcycle Electronics , hey I was wondering if you can solve my dilemma regarding this converter. I have a beard trimmer that I turned to lithium using an 18650 battery, a tp4056 charging module and a this converter. It worked perfectly except that after a couple of days it doesn't work anymore. I have been messing with this on 3 seperate occasions with it working each time but for only a few days. Please tell me what you think. Thanks
0:03 Yes indeed, everyone knows why we need DC to DC boost converters. Why I just asked an old fellow on the way to the crack house how he increases his DC voltages and replied "Well, we use DC to DC boost converters old chap".
Can you do a vid on LM 2596 ?
Hi,
Is there any way to modify it for low current usage. I measured about 1.4mA. Want to use it for a smoke detector that works on a 9V battery. Instead I want to use a 16500 lithium with protection circuit and DC booster. But the smoke detector only uses about 0.03mA (calculated) so the 1.4mA of the booster without load is way to high
Hi,
Played a little with the original one.
Putting a small resisor in line with the inductor changed it a little but not much and depends on which side. Bigger stops it from working at all.
Second attempt.
I removed the trimmer pot and 222 resistor.
Replaced the values by multiply by 4. (didn't have a 1M trimmer pot)
390K for the pot. 220k for the runner.
And replaced the 222 for a 10K resistor.
Output 9.4Volt
On current dropped 4x so instead of 1.3mA is went to 0.28mA no load.
400/500mA must be possible without a big voltage drop. (tested)
When startup current is to high it falls back to 2.2V and can be reset by disconnecting the battery.
My smoke detector draws about 0.12mA.
With the booster it's 0.48mA total.
Still too high make some benefits.
Maybe you can do something with it.
I can't see how the output behavior is with this mod.
Combined with the mod in the video maybe it makes a bigger difference.
Or try a 1M trimmer pot and 22K resistor instead tp see how it behaves. Maybe I test this later.
I never knew about this converter. Probably cheaper than THis version. Thanks!
Just add switch in the input and save your battery voltage
legends JUST add a switch before the module
thats exactly what I use to do lol
Exactly 🤣
Ikr
Pretty interesting mod, man... It can be really useful!!! Thanks a lot! 😃
Is their a way to add an LED to indicate it's reached a certain voltage, like in a battery full indicator ?
same question here. hope someone will notice.
what is the value of RX and Transistor pls?
You can see from the close up shot that the transistor used is a 2n5401 PNP transistor
Wouldn't it be possible to use the already existing voltage drop across the diode to trigger the transistor?
No because it's acting as a voltage divider for the base aswell!!
this little is amazinggg .... (from tech guy's point of view) ..
Do a video on induction heater!!
They actually have video about induction heater on their channel.
Just love this channel😍
Could you do a power supply 30v and 10A? With voltage e current regulator? Please
what is the transistor value of your own?
Actually you can see from the close ups that it is a 2n5401 PNP transistor used. For the resistors you can just read the value from the color codes.
They seem to release the magic smoke if you input more than 6v. I guess your PCB service won't take hand drawn schematics as I don't have the money for a license for PCB software or the time to learn how to use it
This is because of faulty design. Note from the schematic in the video how the trimpot wiper is not configured as a voltage divider. If wound to the end of its travel it will put nearly the whole of the input voltage into the feedback pin. This input is rated at 6V maximum, and feeding more than that into it will produce the black smoke. I have noticed that those versions which do not have thru-holes for the input and output pads are of a different design and work properly up to their rated input voltage for any output voltage adjustment. You can also identify the correct version because the wiper of the trimpot is earthed.
@@danielthackeray7798perfect, you are the only one I have seen get this correctly, I made a video about it 4 years ago on my channel.
Thank you! It's very useful for most applications! I'm sure that I will use it in the future!😀
Can anyone tell me what will be the value of R1 AND R2 if we want to get fixed 6.7V output not using rheostat to adjust.Input is 3.7-4.2v
I set it to the correct voltage then took the rhiostat out. Measured the impedence then selected the correct resister to put in the circuit.
Do a video on xl6009 reworking
I have any questions, how I add constant curent output?
Sorry my bad English
By maintaining a constant voltage across a resistor.
A very simple way to do it is with, for example, a high gain NPN transistor. Put your load between Vin and the Collector. Then put a suitable transistor of resistance R between the Emitter and ground.
The voltage, V, at the Emitter is I × R Volts, where I is the current you want. So, for a normal NPN resistor with current flowing through it, the voltage at the Base of the transistor must be 0.6 + V. You can control the voltage at the Base with a voltage divider, and then the correct amount of current will flow.
Make sure the current flowing through the voltage divider is much larger than the current flowing into the Base. If you get that wrong, the current flowing into the Base will affect the voltage at the Base, and it won't work as expected.
@@petermcarthur7450 any possibility you could email a quick sketch of adding an npn?
I'd like to try this but only have one board and don't want to fry anything.
I'm looking to run one of these at 3-4a (max on datasheet) and need current regulation. 🍺😀
@@ZeroMass Is your supply voltage stable?
@@petermcarthur7450 it would be a single li-ion 18650 and the load being 5v. I'd even be happy with 2.2A and have slightly larger inductors I can add..
Problem is the load drops resistance with an increase in temperature.
@@ZeroMass Is your supply voltage stable?
Creative video, thanks :)
Jlcb dosent have that library of components for which I was designing a ckt
Best boost converter in the low price
Has anyone figure out yet how circulation pump motor works? I know it uses gate driver in IC to drive gates in IGBT. It is 3 phase too. pump name Wilo made in sweden or something. Same shit is in those compressor heater/cooler, the fans are run by PMM thats driven by weird pwm using gate drivers and IGBT. I wan to make them run on 12 or 24v DC.
Any smart person here? I could even pay if you know where to tap the pcb.
Excellent project
I liked your old voice more! I have a video request for you. I need to wind a 2000Volts transformer, that is small like 5cm×5cm×5cm. And that can deliver very high power. The most you can make possible. Im not good at winding transformers nor designing them but I know you are perfect for the job with all your knowledge, so if you have the time please make a video on it. 2000volts and atleast 80watts.
160watts
some questions...
can you varry op voltage without load ?
why dont you use small slide switch before converter to save battery ?
sir kasyan please hear me can i request a contineuty high resistance tester like using tracing the electrical wires
please................................
i love this voice😍😍😍😍
is so relaxing
i love
Dudes projects told by women makes me feel weird tbh.
And how much more power is wasted into heat on the resistor you added once you connect the load to the circuit?
Isn't it easier and better to add a switch and turn the device off when it's not used rather then do this mod that wastes energy while load is connected?
Dont you notic the error in your schematics? See my video on how to fix it.
Nice mod. Thanks 😊
use L1 instead of RX.
woa good idea!
Could you connect it to a 5050 LED strip and check if it flickers on camera please?
Very useful.
Mam why battery charge half amp then the battery amp
Thank
Very smart
I'm not following the purpose of this circuit as applied to a multimeter. Most multimeters worth owning have 9V batteries that last until they leak, which is often more than five years depending on use. Change your battery every year and you won't have that problem. No turn-on problems with super-light loads either. This circuit requires a power source, either battery or buck or P/S to work at all, so two parts are now needed when only one was before. Why is that better than a plain old 9V battery?
9v batteries are the worst on the planet at $4 each. Some meters last a week on 9v rechargable (1253b oled is one).
The old fluke 73,75,77 were excellent on batteries, but the newer stuff eats them in 3,4 months of daily use at work.
Bad habit, never turn on mains already connected power supply to the load... let it stabilize, then connect load.
Many hobbyist not careful about this fact, until they learn about durability, efficiency, and reliability on electronics instrument. If they love their instruments, they will never turn on the instrument when the electricity not stable enough at start ON.
Bad habit, and it was famous habit 🤨
good, i like this hack
Interesting video! Perhaps you fond this interesting too: While making a video about using this module to make a 9V battery I fried the MT3608 chip. On my channel you can find another video about how I repaired the module.
It fried because the Vfb pin got more than 6v because the pot is missing a connection, see my video on how to fix it.
Not bad, but usually not needed.
Nice voice
Thumbs up
Grate
Take a look at the light i made. Make a better one.
Wow
🤔🤔🤔
Почему то ничего не понял)
I cooked all mine...lol
I could not under stand plz talk slower and make an other video