I am wanting to do a 12 volt Joule thief circuit to show my son a little about electronics. I have a 1.5 volt D cell battery and I would like to boost it to 12 volts and maybe as high as 1 or 1.5 amps, if possible? Would someone be willing to help me get the specific components list for this and provide me with a picture of the configuration? I have done some research and understand that a capacitor helps to level the electricity flow? I have also tried using an AI to help generate this information, but it is definitely lacking. lol. I am very new to electronics just trying to introduce electronics to my son. I appreciate any assistance I can get.

 

This is one posting I have found that supposedly does what I am trying to do, if it actually is real. I am not sure the size of the toroid or how many turns of wire on the toroid and what size wire on the toroid? This does show the components in the configuration and a picture too. Is this real? IF it is real, can someone please tell me the type of toroid and how many turns I need? This configuration does not have a capacitor? Would a capacitor help? If it would help, please explain where I would need to put it in this configuration. Please help. https://www.facebook.com/reel/913142461289310

 

It looks like I initially misread this diagram. It looks like there are 5 and 7 turns on the toroid? Can someone please help me confirm this and tell me what size mag wire it uses? Can someone please advise if this is a real solution or just some sort of AI generation?

 

Welcome to AAC.

I am very new to electronics just trying to introduce electronics to my son.

A good place to start is the AAC EDUCATION tab. Here's a link:
https://www.allaboutcircuits.com/textbook

 

I have a 1.5 volt D cell battery and I would like to boost it to 12 volts and maybe as high as 1 or 1.5 amps, if possible?

Before starting with electronics, you should start with physics. 12 V output at 1 to 1.5 A means 12 to 18 W output power. Ignoring any losses of the joule thief, this means 8 to 12 A input current, which is much more than a D cell can deliver.

 

It will be hard to teach your son about electronics, if you don't understand the basics yourself.
That will likely lead to a lot of misunderstanding about electronics (as you have already demonstrated).

Start with basic DC circuit theory, including Ohm's law for the relation between voltage, resistance, and current, series and parallel circuits, and Watts law for the power in a circuit.

 

What is a Joule Thief?



This a simple 1-transistor oscillator to squeeze out the last bit of energy from an already wasted 1.5 V battery. It is intended to be a novelty and not a practical boost converter circuit. Usually the battery voltage has already dropped to 1 V and lower. The circuit can extract just enough energy to power an LED.

Assuming that the LED needs 2 mA @ 2 V, that is 4 mW.

Do the math. If you are trying to get 1 A @ 12 V is 12 W. That is 3000 times more power than the circuit is designed to deliver.

Let's assume that a fresh D-cell has a capacity of 20 Ah. Theoretically, it can deliver 20 A @ 1.5 V for 1 hour.
If you created an oscillator to boost the output voltage to 12 V, theoretically, you can output 2.5 A for 1 hour.
In practice, the battery will die a lot sooner.

Your circuit has to be designed to handle 30 W. You need to reduce your expectations and lower the output current requirements, for example, 20 mA.

If your output is 20 mA @ 12 V, which is 0.24 W, your 1.5 V battery has to supply in excess of 160 mA.

12 V is eight times higher than 1.5 V. Hence whatever current you demand at 12 V, the 1.5 V battery has to supply more than 8 times the output current. You can use 10 times as a starting point.

The other thing to note is that the Joule Thief circuit is an oscillator circuit. For practical demonstration purposes, it is used to flash an LED. The output is not a steady current or voltage. If the oscillation frequency is high enough (above 30 Hz), the LED will appear to be on continuously even though in reality it is being pulsed.

 

the single AAA , AA ... D cell powered Joule -Thief can easily deliver 40V to insignificant load - one of which might be a gas discharge Ne bulb https://en.wikipedia.org/wiki/Neon_lamp - it has 90 to 120V breakdown and 50 to 60V hold voltage ...
... so - you should make an auto-transformer with an output voltage of - a bit over tripling - your collector load coil which means instead of original 20 turns ~ it'd require over 3²x e.g. 9 times of that . . .
. . . however your can double your output voltage so you need to go bit over 1.5 times with voltage say 1.7²x making it apx. 3x output turns

 

What is a Joule Thief?

This is a simple 1-transistor oscillator to squeeze out the last bit of energy from an already wasted 1.5 V battery.
Usually the battery voltage has already dropped to 1 V and lower. The circuit can extract just enough energy to power an LED.

Assuming that the LED needs 2 mA @ 2 V, that is 4 mW.

Do the math. If you are trying to get 1 A @ 12 V is 12 W. That is 3000 times more power than the circuit is designed to deliver.

Let's assume that a fresh D-cell has a capacity of 20 Ah. Theoretically, it can deliver 20 A @ 1.5 V for 1 hour.
If you created an oscillator to boost the output voltage to 12 V, theoretically, you can output 2.5 A for 1 hour.
In practice, the battery will die a lot sooner.

Your circuit has to be designed to handle 30 W. You need to reduce your expectations and lower the output current requirements, for example, 20 mA.

If your output is 20 mA @ 12 V, which is 0.24 W, your 1.5 V battery has to supply in excess of 160 mA.

12 V is eight times higher than 1.5 V. Hence whatever current you demand at 12 V, the 1.5 V battery has to supply more than 8 times the output current. You can use 10 times as a starting point.

the single AAA , AA ... D cell powered Joule -Thief can easily deliver 40V to insignificant load - one of which might be a gas discharge Ne bulb https://en.wikipedia.org/wiki/Neon_lamp - it has 90 to 120V breakdown and 50 to 60V hold voltage ...
... so - you should make an auto-transformer with an output voltage of - a bit over tripling - your collector load coil which means instead of original 20 turns ~ it'd require over 3²x e.g. 9 times of that . . .
. . . however your can double your output voltage so you need to go bit over 1.5 times with voltage say 1.7²x making it apx. 3x output turns
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Is an auto transformer a buck converter? I have looked at Buck converters online and I haven’t found one online that will use a starting voltage of 1.5 volts. A lot of them start out at around 3volts? I appreciate your help a lot!!! Please help me figure out what to do next.

 

Is an auto transformer a buck converter? I have looked at Buck converters online and I haven’t found one online that will use a starting voltage of 1.5 volts. A lot of them start out at around 3volts? I appreciate your help a lot!!! Please help me figure out what to do next.

AN Auto Transformer is not a converter in the same sense BECAUSE it has no active components. THAT is a very big difference, and why I try to defend the correct use of the term "TRANSFORMER" It is quite different from a converter.

 

If you and your son want to explore the vast and exciting field of electronics, starting with buck and boost converters is a huge first step.

Start off with learning how to power an LED with a 9 V battery and a resistor. Move on to making the LED flash using a 555-timer IC.

Experiment with small solar panels, toy DC motors, LDR (light dependent resistors). For a goal, build a small line or light following robot.

 

What to do next? Provide realistic and achievable requirements and feasible expectations.
As thoroughly explained earlier, there is no way you will obtain that much power from a single alkaline cell and a simplistic circuit as the Joule Thief.
The closest I can think of, is a Royer converter. Google it. But still, your original power requirements are not possible to be met with a single cell.

 

Prior to any discussion of actual ELECTRONICS, your son needs to understand basic electrical circuits, and basic circuit theory. Those are fundamental requirements.
The actual functioning of the classic "Joule Thief" is not trivial, it requires understanding of a transformer , not trivial in itself.

 

Going from 1.5V to 12V at 1.5 amps is asking quite a lot from a single D cell, and I don't think a basic Joule thief would manage that. A capacitor can help smooth out the output, but it won't give you extra power. I'd probably get a cheap boost converter module for the 12V experiment and build a simple Joule thief separately. That way your son gets to see how both work without making the first project too complicated

 

I appreciate your help and suggestion! I have a joule thief circuit from Engineering Mindset, https://theengineeringmindset.com/joule-thief-kit/ Engineering Mindset has directions and learning tools about how and why the circuit works. The joule thief works perfectly to boost the 1.5 volts to make a 3 volt LED lamp light. I have already done this using the joule thief circuit. And I just ordered a Mini Boost Converter from Drok. https://www.droking.com/Mini-Boost-...ge-Regulator-DC-12V-24V-Adapter-Driver-Module I think these two circuit boards will do exactly what I need them to do. Again, I appreciate your help. Have a great day!

 

Welcome to AAC.

it is not at all clear to me that you understand the fundamental nature of the Joule Thief or boot converter. Each is constrained by fundamental electrical laws and must trade off between voltage and current (Amps).

In a very simplified version of things, each one uses an inductor (coil) and takes advantage of a particular behavior of an inductor in a circuit. The inductor acts as an electromagnet and develops a magnetic field when the current passes through it. When that current is removed the field collapses and induces a current of a higher voltage and the opposite polarity into the circuit.

This is called back EMF and is usually an undesirable side effect that must be suppressed but in the case of the Joule Thief and boost converter it is used by the circuit to provide a voltage that is higher than the input. However while the voltage is higher the current must be proportionally smaller—this is a demand of the laws of physics!

The controlling equation is a simple one: W=V*A. W is Watts—the total power that the source (D cell) can deliver, V is voltage and A is current. Since W is fixed we can only manipulate V and A. In addition all real circuits have parasitic losses, that is, they all turn some of the power applied into heat rather than electrical output.

The Joule Thief "cheats" by using time. The pulsing is the trick and is caused by accumulating power in the circuit's capacitor and then periodically releasing it into the load. It still provides less power to the LED than the cell is providing t the circuit!

This is why a boost converter has limits on how much higher the voltage can be than the input and why the current will be much lower than the input current. There is no magical escape from W=V*A, it will always be the bottom line.

This is why I am skeptical that what you think is the solution to getting your 12V@1.5A is viable. If you intend to use the output of the Joule Thief to power the boost converter you will be disappointed.

You never mentioned the load that you hope to power, what is it? Just for the record 18W is a lot. If you tried to get that from a fresh D cell you'd be asking for 12A! (remember W=V*A so A=W/V or A=18/1.5). Even if it could provide it, that cell would be getting very hot and be drained in a very short time.

Ultimately, the requirement doesn't change—it just shifts to the capacitor in the Joule Thief or boost converter circuit. If you managed to build a Joule Thief with an output of 12V@1.5A using a D cell you would be waiting a very long time for a very short pulse of current at those specs.

This brings us back to what you hope to power. Perhaps there is an alternative that can make this practical. Would you explain what this will be doing? What is your application?

Whatever you do, good luck and don't become discouraged if this doesn't work. Electricity and electronics are fascinating, useful, and very satisfying areas of study. Even if you and your son only pursue them lightly they will give you mental tools that no other area of study offers. Electricity isn't magic, but it feels like it and when you start to understand how it behaves it can make the whole world seem different.

 

I would like to light a 12 volt 450 lm light bulb for as long a possible. I know it won't last for very long. I want to see how long it will last. I have done it using two AA batteries in series using a small circuit board, but the board gets really hot. Here is the circuit board I used to do it for a short period of time. https://www.amazon.com/dp/B089JYBF25?ref_=ppx_hzsearch_conn_dt_b_fed_asin_title_9&th=1 I bought a different circuit board that has a heat sink on it, I am hoping it will help with the heat issue. This site would not let me upload a video of the lighted 12 volt 450 lm light bulb using two AA 1.5 volt batteries using the circuit I listed above. This site (All about Circuits) told me the video was too long even thought it was only 13.9 mb

 

I would like to light a 12 volt 450 lm light bulb for as long a possible. I know it won't last for very long. I want to see how long it will last. I have done it using two AA batteries in series using a small circuit board, but the board gets really hot. Here is the circuit board I used to do it for a short period of time. https://www.amazon.com/dp/B089JYBF25?ref_=ppx_hzsearch_conn_dt_b_fed_asin_title_9&th=1 I bought a different circuit board that has a heat sink on it, I am hoping it will help with the heat issue. This site would not let me upload a video of the lighted 12 volt 450 lm light bulb using two AA 1.5 volt batteries using the circuit I listed above. This site (All about Circuits) told me the video was too long even thought it was only 13.9 mb

Yes, a 3V battery made of 2 AA cells can produce quite a bit of current—for a very short time. But you would get more from the battery if you powered a 3V bulb directly than adding in the losses of a boost converter.

And yes, things are going to get hot, and a battery of D cells will certainly last longer. Since this is an arbitrary thing without a reason other than "I want to do this", then make a 3V D cell battery and make sure the boost converter has good cooling; a fan would be a good idea.

The Joule Thief circuit is superfluous—it will do nothing for you.