乾電池1本から100Vを作る
Summary
TLDRこのビデオでは、1.5Vの乾電池から100Vの電圧を生み出すブーストコンバータの制作過程を紹介しています。有名なコッククロフト・ウォルトン回路の代わりに、シンプルなブーストコンバータを使用し、パワーMOSFET、ダイオード、コンデンサーを活用します。原理説明から実際のPCB製作、そして実験による検証まで、一連の流れが詳しく解説されています。また、乾電池の電圧を5Vに昇圧する村田製作所のモジュールの使用や、複数のブーストコンバータモジュールをカスケード接続して高電圧を得る工夫が紹介されています。最終的には、単一の乾電池で150Vを超える電圧を実現し、その危険性と電圧が負荷によってどのように変動するかについても触れられています。
Takeaways
- 😀 They will make a 100V dry cell battery using power electronics
- 👷♂️ They will use a boost converter circuit to step up the voltage
- 🔌 The circuit only needs a single 1.5V AA battery as input
- ⚡️ The battery voltage is stepped up using an inductor and capacitor
- 🤔 Driving the MOSFET switch is challenging with only 1.5V input
- 🙌 Found a Murata module to boost 1.5V to 5V for MOSFET control
- ⚙️ Connected multiple boost converter modules in series to multiply voltage
- 🔋 Made a 150V 'dry cell battery' from a single AA battery
- 📉 Voltage drops under load due to no feedback control
- 💡Possibility to boost voltage further with more batteries in parallel
Q & A
What is the goal of the project in the video?
-The goal is to create a 100V dry cell battery using only a single 1.5V AA battery.
What circuit configuration is used to boost the voltage?
-A boost converter circuit configuration consisting of an inductor, switch, diode and capacitor is used.
How does the boost converter work to increase the voltage?
-It works by storing energy in the inductor when the switch is closed, and then transferring that energy to the capacitor when the switch is opened, boosting the voltage.
Why was the Murata power module used?
-It was used to boost the 1.5V battery voltage to 5V to provide sufficient voltage to drive the MOSFET switches.
How was a 100V output achieved?
-By connecting multiple boost converter modules together vertically in cascade to achieve a high voltage gain.
Why did the output voltage drop under load?
-Because there was no feedback control, the boost converters moved into discontinuous mode, and the battery voltage sagged under the high load current.
What was the maximum output current capability?
-Only about 1A could be drawn before the output voltage dropped significantly due to battery limitations.
What improvements could be made?
-Add voltage feedback control, parallel more batteries, add more cascade stages, and optimize the circuits for higher load currents.
What was learned from this project?
-That while voltage multiplier circuits can generate high voltages easily, getting them to work properly under load with feedback and control is more difficult.
What future applications are possible?
-With improvements, this circuit technique could generate voltages up to 3kV for specialized applications.
Outlines
😀 はじめに、ブーストコンバーターの原理説明
ブーストコンバーターはインダクタとスイッチ、ダイオード、コンデンサーから成り、インダクタにエネルギーを蓄え、スイッチのON/OFFによってコンデンサーにエネルギーを移すことで電圧を上げる。基本回路の原理実証のため、試作回路を作成し、入力電圧を上げていくことで出力電圧も上がることを確認した。
😊 MOSFET駆動用に1.5Vを5Vに昇圧するモジュール
ブーストコンバーターのMOSFETを駆動するのに最低5Vが必要だが、単電池では1.5Vしかない。そこでマウス社のDC-DCコンバーターモジュールを利用し、単三電池の1.5Vを5Vに昇圧してMOSFETの駆動に利用する。
🤔 1個のブーストコンバーターモジュールだと100Vまで昇圧するのは難しい
1個のブーストコンバーターモジュールで1.5Vを100Vに大きく昇圧しようとすると、インダクターに非常に大きな電流が流れる必要があり、単三電池ではそこまでの大電流は流せない。モジュールを垂直接続してステージを重ねる「カスケード接続」したほうが容易。
😮 3段階のカスケード接続で1.5Vを150Vに昇圧
3段のブーストコンバーター・モジュールをカスケード接続し、1.5Vを最初に2倍、次に2倍、最後に5倍して150Vに大きく昇圧することに成功した。これを使って単三電池サイズの携帯できる「150V乾電池」を作成した。
Mindmap
Keywords
💡ブーストコンバータ
💡垂直接続
💡ドライセル電池
Highlights
Introduction to the concept of creating a 100V dry cell battery using power electronics technology.
Discussion on the challenge of boosting voltage from a single 1.5V dry cell battery and exploring circuit configurations.
Explanation of why the Cockcroft-Walton circuit, often used for DC high voltage creation, is not suitable for this project.
Introduction to using a boost converter as a simple and effective solution for increasing voltage.
Detailed explanation of how a boost converter works, including the roles of its components.
Discussion on the importance of transferring energy from the inductor to the capacitor for boosting voltage.
Practical demonstration of a boost converter increasing the voltage and the challenges faced with MOSFET operation.
Solution for driving MOSFETs with low voltage using a Murata Manufacturing Co. module to boost voltage for the control circuit.
Step-by-step assembly of the boost converter circuit, highlighting the use of components from DigiKey.
Demonstration of the assembled boost converter successfully increasing voltage in a controlled experiment.
Discussion on the limitations of boosting voltage to 100V with a single module due to current and resistance constraints.
Proposal of a cascading approach with multiple modules to achieve higher voltage outputs effectively.
Successful demonstration of cascaded modules boosting voltage significantly, with safety precautions emphasized.
Creation of a 150V dry cell battery casing using 3D printing to house the cascaded boost converter modules.
Final thoughts on the project's success, limitations, and the potential for scaling up the voltage further with additional modules.
Mention of DigiKey's support and the availability of educational content on vector analysis for electronic design.
Transcripts
This video is brought to you courtesy of DigiKey
Good afternoon.
Today we're going to take advantage of power electronics technology.
We're going to make a 100V dry cell battery.
This is a dry cell power supply that can produce 100V here
We use only one 1.5V dry cell battery here.
Actually from this one small battery
How much voltage can you get out of it?
I'm sure many of you are wondering.
So that's what we're going to try.
First of all, we try to figure out what kind of circuit configuration we want to use.
If you search for DC high voltage or something like that, you'll find a lot of stuff.
The first one is the Cockcroft-Walton circuit.
This is a very famous circuit
It is often used to make DC high voltage
But if you look closely, you can see that an AC power supply is required.
This is a circuit that converts AC voltage to DC voltage
But this time, we use DC dry cell batteries.
So, with this circuit, it seems to be a little difficult to boost the voltage.
Converting AC voltage to AC voltage
The circuit that converts AC voltage to DC voltage is
There are various
Further boosting the DC voltage
The circuit that converts a DC voltage to a DC voltage is
It looks quite difficult.
For example, a flyback converter like this
There are some circuits like this
It's a little bit difficult to make this transformer part.
So this time, number 1 straight forward
And I'd like to proceed in a simple way
We're going to use a boost converter here.
Very simple circuit configuration
Boost converter is an inductor switch
This time we use a power MOSFET
And it can be composed of a diode and a capacitor
It's a very simple circuit
I'll explain the principle of operation in a moment.
For clarity, the MOSFET here is
Rewrite it as a switch
What's the most important thing about this circuit?
The energy that's accumulated in the inductor here.
If you accumulate it in the capacitor
From inductor to capacitor
It's about transferring energy
Let's close this switch first.
Then here's what happens
Current flows through this inductance and battery loop
The energy stored in the inductor at this time is
1/2 LI².
This L is the inductance
And I is the current through the inductor
And next when the current is flowing in the inductor
This switch is opened
Then there will be no current flow to this switch, so
So current flows through this switch.
And that current is
It flows into this capacitor here
If there is a load
Of course it will flow like this with load
Let's turn off the load for a moment this time, because it's in the way
The energy stored in the inductor
It is transferred to the capacitor
At this point, the energy in the capacitor is 1/2 CV².
C is the capacitance of the capacitor
By the way, the voltage of an ultracapacitor is
The integral of the current through the capacitor
This current charges the capacitor
The switch was turned off
Turned on again
When current flows through the inductor
And if you turn this switch off
Now we're going to put the current through the capacitor again.
Repeat this operation to increase the voltage
The current flowing in the inductor boosts the capacitor to
Let's estimate how much the capacitor can be charged
For example, this circuit
1mH inductor and 100µF capacitor.
1A flows through this inductor
Then the energy stored in the inductor is
1/2・LI²
Calculation yields 0.5 mJ
The energy of this inductor is
Assume 100% of the energy is used to charge the capacitor
Capacitor energy is 1/2 CV².
Actual calculation shows that the voltage of the capacitor is 3.16V
As explained earlier
By turning the switch here on and off
The energy stored in the inductor
The energy is transferred to the capacitor
By the way, when the voltage is low, it means
The voltage of the capacitor can easily go up.
When the voltage of an ultracapacitor is high, it means
The voltage increase becomes a little more difficult
because the energy of the capacitor is
because it is proportional to the square of the voltage
For example, let's say the initial voltage of a capacitor is 100V
If the same energy is stored in a 100V capacitor as before
The voltage of the capacitor only rises to 100.05V
So the voltage is increased like this
By the way, you can watch the following video about boost converters.
I made it before.
If you watch that video as well
You'll learn more about boost converters, and you'll be able to see how to use them.
Please take a look.
There is a link in the summary section
We are going to make a boost converter
Whether the boost converter actually works
We'll go ahead and verify the principle first.
This is Ichiken's original PCB and
Combination of inductors, capacitors, power semiconductors, etc.
Making a boost converter
Electrolytic capacitor
Inductors
And power semiconductors.
This is a diode
This is a capacitor in the output section
Step-up converter for verification of principle is completed
We will run this one.
The front side looks like this
The reverse side looks like this
By the way, here is a breadboard type printed circuit board
Sold on Amazon
If you would like to use it
Please purchase from the link in the summary column.
Then we'll do a proof-of-principle test.
Here's how it's set up
There's a voltmeter.
The top is the input voltage
The bottom is the output voltage of the boost converter
Now let's add the voltage
First, we set the input voltage to 5V
This time, the voltage is set to be boosted by a factor of 2.
Since the output voltage is 9.5V
Roughly doubled
Let's increase the voltage further
Now
10V
That gives 19.7V
Looks about right
Here we increase the voltage further
It's on fire
Thus the boost converter is
It was found to be working without any problems
There's just one problem
What it is is the voltage to run the MOSFETs.
This is a MOSFET
To run a MOSFET
Between this GATE-SOURCE
You will need at least about 5V
But this time we can use it as a power supply.
Only one of these batteries can be used as a power source.
The voltage of the battery is 1.5V
between gate and source here
Not quite the minimum required voltage of 5V
So without these power supplies
So driving MOSFETs without these power supplies?
I'm thinking, "This is going to be difficult.
I'm in a bit of a pickle.
I was looking for parts on DigiKey.
What an amazing part I found!
This is a product of Murata Manufacturing Co.
This is a module that boosts the voltage of 1.5V of a dry cell battery to about 5V.
This time, I will use this for the control circuit of MOSFET.
This is the circuit we are going to make.
With the Murata power supply module that we just used
Boosts the battery's 1.5V voltage to 5V
This voltage is used to control the MOSFET
This is used for the control circuit part.
Now we use the MOSFET as a switch
It can now be turned on and off
Then you can use a boost converter to turn the voltage of this battery on and off
We can now boost the voltage.
Now we will actually build the circuit
As usual, the printed circuit board is made from scratch.
And the components that we use there are
purchased from our sponsor DigiKey.
This time, I used surface mount components for weak electric circuits.
And for the power circuit part
We use a lot of through-hole components
Assembled as soon as possible
There's a printed circuit board, and then we're going to put a stencil on top of it.
There is this stencil hole
We put solder paste on top of the stencil
And then you can use a credit card or something like that.
And then you rub it here.
Solder paste in the stencil holes.
It is an image of pushing it into the stencil
It's done.
This is the difference between before and after applying solder paste
Below is before applying solder paste
Above is after applying solder paste
The area where solder paste is applied
Surface mount components are placed
Now that all the components have been placed
Next, we will bake them on this reflow plate.
We're going to put it on this one.
Put the board on top of this
Now we will burn it.
Now it is heated
So the temperature is going up and up and up.
I'm going to leave it like this for a while.
The solder seems to have melted.
Let's take it out.
I was able to solder quite nicely
First, let's see if it works with just the control circuit.
Put a single AA battery in
Voltage is 1.5V
Now let's look at the control circuit voltage
Now there is no battery connected to the control circuit
When this jumper is connected
Power is supplied to the control circuit and
5V voltage is supplied
Next, let's look at the voltage between the gate and source of the MOSFET
Here we have NE555
Pulse wave is emitted from there to drive MOSFETs
When you turn on the power
Now it looks like this
The amplitude of the pulse wave here is 4.5V
To drive a MOSFET
I found that I needed about 5V
4.5V is enough to drive the device.
There is no problem with the voltage level.
This control part sets the boost ratio of the boost converter
There is a trimmer resistor here, but
Adjust this part
Like this
In this state, the pressure increase is not that high
Conversely, in this state
The boost ratio becomes high
That duty ratio.
The longer the duration of ON
The higher the output voltage of the boost converter
Now that we know something is going to work
We're going to solder the power circuit components as well.
Electrolytic capacitors of 450V/100μF are used
Diode, 650V withstand voltage
N-channel MOSFET, also 650V
Inductor 100μH
The circuit is completed as shown here
Let's try an actual experiment
When you put a battery in
First of all, here is the input voltage. Here is the battery voltage.
1.5V
And right side output voltage
It is lower than the input voltage
This is because the control circuit is not powered yet
It's just the voltage drop across the diode
Drop from input voltage
Now turn on the control circuit
The voltage is boosted
It is working properly
The variable resistor here can be used to change the boost ratio.
Like this
Higher or lower voltage
Pretty good
You can increase the voltage like this
I know it works fine and looks good, but...
So, if I try to make 100V from here
There is only one problem
What is that?
Just one module here
If we try to boost the voltage from 1.5V to 100V
A very large current must flow through the inductor
And with only one dry cell battery
The current cannot be that high.
It is possible, but...
It's not like the voltage is 1.5 or something.
It drops considerably to 0.7V or something like that
This is because there is an internal resistance in the battery
And also because of the high current
Inductor here
Worst case scenario, it burns up
Other than that, power semiconductors are
must withstand high currents
So, with one module
Boosting voltage from 1.5V to 100V is
I guess I better rethink this.
I've been trying to think of different ways to boost the pressure.
I thought that this is the best way to do it.
I came to a conclusion
This boost circuit as a single module
If we connect the beads like this
I think it would be easy to create a high voltage
This kind of connection is called a vertical connection
Cascade connection.
For example, if you had a circuit that doubled the voltage in one module
where the voltage is doubled
Double here, too
And here also doubled
Connecting three of them gives a total multiplier of x8
So, for example, if you have a 1.5V input
This is calculated to be 12V.
Besides that, for example, here is 5 times
Here is also 5 times
And if we multiply the last step by 5
That's 5 to the power of 3, so we get 125 times the voltage.
Then it is 125 times 1.5V, so
That gives 187V
So to make a vertical connection
We will make several of these modules
Three modules are created and connected vertically
Boosting the voltage at the first stage
Boost the pressure in the second stage as well
And the third stage also boosts the pressure
We actually turn on the power supply
I think we're going to get a pretty high voltage, so
I'm going to put on insulated gloves as well.
Now we turn on the power.
There's a little beeping sound.
but it seems to be working fine.
I'm going to measure the voltage.
Here it comes!
156V
So the voltage of the dry cell here is 1.5V
100 times more than that.
The pressure has been boosted.
I made this kind of battery type case with a 3D printer
There are three modules inside.
If you put one AA battery for power supply here
Here's the positive pole metal
And the negative pole is
It is pasted here
There is an ON/OFF switch on the back side of the unit
Close the lid
It's something like this, pretty much like a battery.
This is quite interesting.
This is a huge 150V dry cell battery.
Just like a normal battery
If you always put out 150V
Very dangerous Dangerous Dangerous Itiken specification
So just to be safe
I put an ON/OFF switch here
But this one is also turned on
I don't have any LED indicators or anything on.
The switch ON/OFF is the only thing that determines whether the switch is ON or OFF.
It is a rather dangerous specification.
Let's measure the voltage of this giant dry cell battery again
The needle is inserted properly, but
Is it 144V?
The voltage is firm.
Let's do a little experiment
Normal dry cell battery, low voltage
Even if both ends of this terminal are short-circuited
No sparks will be generated.
But what happens with these 150-volt dry cells?
I'll try an experiment.
Here is a mechanical pencil lead
I'll try to short-circuit it.
It crackles pretty solidly
As expected from the high voltage
Voltage comes down a little bit
This is just one dry cell.
It's not as powerful as it should be.
I think the voltage drops a lot when you take a load.
I think it will drop to about 50V.
That's quite a jump, now.
Can you hear it crackling a lot?
Once it shorts out and discharges
Until the voltage rises like this
Since it seems to take quite a long time
Sparks don't fly easily when discharged continuously
If you give it a little time
There's a big spark, but
If you leave it only for a short time
Sparks will be such that they will hardly fly
This is because it takes time to charge the voltage booster circuit inside
By the way, what happens when you take the load off
I'll try it with this resistance
If you look at the voltmeter display
It is below 50V
With no load, 150V was being produced, but
When the load is taken like this
The voltage drops as soon as it reaches 50V
When a load is connected to the step-up circuit
The voltage is now lower
There are many reasons for this
There's about three reasons this time.
The first one is
is that we don't have voltage feedback.
If the circuit converts the voltage in this way
Usually there is a voltage divider resistor that monitors the voltage
This is how the voltage is input to the control circuit
The calculated pulse width is then input to the MOSFET
And so the feedback works nicely
The output voltage of the load is always constant
but this time there is no feedback control
This signal is input by fixed input
So the voltage of the load goes down.
The second one...
The current flowing through the inductor here is
Very important in a step-up circuit
It's called current continuous mode, and
There is a mode called current discontinuous mode
The boost ratio changes depending on the mode of operation.
In this case, the operating mode is not calculated.
So the mode changes depending on the load current
It means that the voltage will not be increased as expected.
And here's reason number three.
These dry cell batteries here
Actually, the current is always about 1A in the dry cell
This current of 1A is
This is a very large load for an AA battery.
If you put this much current into a battery
1.5V will drop to about 1V
The voltage at the input drops, which means
The output voltage drops as well
For this reason, we have changed to this 100V dry cell battery.
When I connected a load, the voltage dropped
So I'm going to summarize
Dry cell here
Voltage can be output up to about 150V
But if we just take the load
I found that the voltage drops accordingly.
We are using only one AA battery in this case.
If you connect some of these AA batteries in parallel
There is a possibility to boost the voltage a little more.
We're going to need a more robust battery that we're using for the power supply.
And if you increase the number of vertical connections in the module
If you go up to about 3kV
This circuit configuration seems to work.
But what I found out by doing it this time is that
The fact that the voltage multiplier circuit itself is somehow working is a good thing.
It's easy, but
If you try to make it work properly
It turns out to be quite difficult.
So here's today's sponsor, DigiKey
Here you go.
I hope you all know that DigiKey has a Japanese YouTube channel.
Do you know about DigiKey Japanese YouTube channel?
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