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electfans posted on 2021-1-1 11:07 I downloaded a calculation tool online, but it is incomplete. It does not include the wire diameter, the transistor amplification factor and inductance, etc. I don’t quite understand it.

The primary is 0.9 turns, the secondary is 29.75 turns, how can it be wound in less than one turn? How is it related to transistors?

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The number of turns of the transformer must be an integer, and fractional turns cannot be wound. To keep the turns ratio unchanged, round the number of turns. For example, [primary 0.9 turns, secondary 29.75 turns] the turns ratio is 0.0302521, the primary is rounded to 1 turn, then the secondary is calculated to be 33.05 turns, take 33 turns.  Details Published on 2021-1-1 12:21
The number of turns of the transformer must be an integer, and fractional turns cannot be wound. To keep the turns ratio unchanged, round the number of turns. For example, [primary 0.9 turns, secondary 29.75 turns] the turns ratio is 0.0302521, the primary is rounded to 1 turn, then the secondary is calculated to be 33.05 turns, take 33 turns.  Details Published on 2021-1-1 11:44
The number of turns of the transformer must be an integer, and fractional turns cannot be wound. To keep the turns ratio unchanged, round the number of turns. For example, [primary 0.9 turns, secondary 29.75 turns] the turns ratio is 0.0302521, the primary is rounded to 1 turn, then the secondary is calculated to be 33.05 turns, take 33 turns.  Details Published on 2021-1-1 11:41
The number of turns of the transformer must be an integer, and fractional turns cannot be wound. To keep the turns ratio unchanged, round the number of turns. For example, [primary 0.9 turns, secondary 29.75 turns] the turns ratio is 0.0302521, the primary is rounded to 1 turn, then the secondary is calculated to be 33.05 turns, take 33 turns.  Details Published on 2021-1-1 11:38

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electfans posted on 2021-1-1 11:11 The primary is 0.9 turns, the secondary is 29.75 turns, how can it be wound with less than one turn? How is it related to transistors?

The number of turns of the transformer must be an integer; fractional turns cannot be produced.

To keep the turns ratio unchanged, just round up the turns. For example, [0.9 turns for the primary, 29.75 turns for the secondary], the turns ratio is 0.0302521, the primary is rounded to 1 turn, then the secondary is calculated to be 33.05 turns, and 33 turns are taken.

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electfans posted on 2021-1-1 11:11 The primary is 0.9 turns, the secondary is 29.75 turns, how can it be wound with less than one turn? How is it related to transistors?

[Primary 0.9 laps, secondary 29.75 laps]

This result is far from 0.7V:7V. It may be wrong.

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electfans posted on 2021-1-1 11:11 The primary is 0.9 turns, the secondary is 29.75 turns, how can it be wound with less than one turn? How is it related to transistors?

[I don't quite understand what the transistor's amplification factor and inductance are. ]

The self-excited converter is of course related to the transistor amplification factor.

But can ordinary silicon transistors work under such a low power supply voltage of 0.5V? I doubt it. The emitter junction of ordinary silicon transistors must be turned on before oscillation can occur. The lower limit of your power supply voltage has reached 0.5V, which certainly cannot make the emitter junction of the transistor turn on.

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maychang posted on 2021-1-1 11:11 [What else do you need the amplification factor and inductance of the transistor?] I guess you downloaded the calculation tool for the self-excited converter transformer.

Yes, is there a more comprehensive introduction to calculation tools?

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I have posted the design manual for you. That should be the most complete, at least I think it is.  Details Published on 2021-1-1 12:23
 
 
 
 

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electfans posted on 2021-1-1 11:11 The primary is 0.9 turns, the secondary is 29.75 turns, how can it be wound with less than one turn? How is it related to transistors?

You'd better explain clearly: where does this 0.7V voltage come from? Is it a square wave or a sine wave? Is it a peak value or a peak-to-peak value or an effective value?

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It is a sine wave. The primary side is connected to 220V mains electricity. The 0.7V voltage drop of the diode is intercepted. After the transformer boosts and rectifies, 5V is obtained to power the microcontroller.   Details Published on 2021-1-1 13:32
 
 
 
 

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electfans posted on 2021-1-1 11:49 Yes, is there a more complete introduction to calculation tools?

I have posted the design manual for you. That should be the most complete, at least I think it is.

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maychang posted on 2021-1-1 12:21 You'd better explain clearly: where does this 0.7V voltage come from? Is it a square wave or a sine wave? Is it a peak value or a peak-to-peak value or an effective value...

It is a sine wave. The primary side is connected to 220V mains electricity. The 0.7V voltage drop of the diode is intercepted. After the transformer boosts and rectifies, 5V is obtained to power the microcontroller.

image.png (7.48 KB, downloads: 0)

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This post is from Analog electronics

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Seeing this picture, I want to advise the original poster: Don't think about winding a transformer yourself. If you want to power a single-chip system, find an old mobile phone charger and use the charger output to power your single-chip system.  Details Published on 2021-1-1 15:57
Seeing this picture, I want to advise the original poster: Don't think about winding a transformer yourself. If you want to power a single-chip system, find an old mobile phone charger and use the charger output to power your single-chip system.  Details Published on 2021-1-1 15:02
 
 
 
 

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This post was last edited by maychang on 2021-1-1 14:31
electfans published on 2021-1-1 13:32 It is a sine wave. The primary is connected to 220V AC power, the 0.7V voltage drop of the diode is intercepted, and after the transformer boosts and rectifies, 5V is obtained to power the microcontroller...

This circuit will trip when the plug is plugged in, and the diode will definitely burn out.

Besides, AC220V is 50Hz, not 50kHz.

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electfans published on 2021-1-1 13:32 It is a sine wave. The primary is connected to 220V AC power. The 0.7V voltage drop of the diode is intercepted. After the transformer boosts and rectifies, 5V is obtained to power the microcontroller...

Seeing this picture, I want to advise the original poster: Don't think about winding a transformer yourself. If you want to power a single-chip system, find an old mobile phone charger and use the charger output to power your single-chip system.

This post is from Analog electronics
 
 
 
 

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electfans published on 2021-1-1 13:32 It is a sine wave. The primary is connected to 220V AC power. The 0.7V voltage drop of the diode is intercepted. After the transformer boosts and rectifies, 5V is obtained to power the microcontroller...

This is just a partial schematic diagram of the circuit. If you know, just tell me the relevant parameters and winding method of the transformer. Thank you!

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How can I tell you the relevant parameters and winding method of the transformer for a circuit that cannot work? If I tell you to wind it this way, then I am talking nonsense.  Details Published on 2021-1-2 07:59
 
 
 
 

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electfans posted on 2021-1-1 15:57 This is just a partial schematic diagram of the circuit. If you know, just tell me the relevant parameters and winding method of the transformer. Thank you!

I said on the 29th floor that the AC220V in the picture on the 28th floor is 50Hz, not 50kHz.

In addition, even if the 28th post is just a schematic, the two ends of the transformer primary are approximately square waves, not the sine wave mentioned in the 28th post.

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electfans posted on 2021-1-1 15:57 This is just a partial schematic diagram of the circuit. If you know, just tell me the relevant parameters and winding method of the transformer. Thank you!

How can I tell you the relevant parameters and winding method of the transformer for a circuit that cannot work? If I tell you to wind it this way, then I am talking nonsense.

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