I'm working on a project using kit stm32 blue pill. I designed a power suply circuit, but the kit was damaged. Please tell me what was wrong in my circuit?
I used Adapter 15V 2A.

 

I can't see any reason that circuit would not work if built as shown. You were taking the output across C3 weren't you?

 

Yeah, I put voltage 3V3 for the kit stm32. Do you know what the 5W resistor stands for?

 

5W resistor is to limit the current, ideally it should be 10W, you don't have any 100uF electrolytic capacitors in your circuit on the outputs of the regulators...

 

But the current already depends on the load (kit stm32). It's about several miliampes.

 

It's just like I plug directly the phone charger (5V, 2A) into the kit without any resistor. Is that right?

 

It's just like I plug directly the phone charger (5V, 2A) into the kit without any resistor. Is that right?

Yeah, if you’re lucky. Any failure of downstream components will cause a chain reaction. So you design to protect that from happening.

The recommended circuit for the 7805 from the datasheet, calls for much larger capacitors. Without the, the regulator will fail, and well, NOT regulate.

 

Check the polarity of your input from the adapter. You may have the positive voltage applied to ground at JDC3.

 

Yeah, if you’re lucky. Any failure of downstream components will cause a chain reaction. So you design to protect that from happening.

The recommended circuit for the 7805 from the datasheet, calls for much larger capacitors. Without the, the regulator will fail, and well, NOT regulate.

I'll find it out, thanks for your suggestion.

 

Check the polarity of your input from the adapter. You may have the positive voltage applied to ground at JDC3.

That didn't happen, because my circuit run well and get hotter then end up with the kit damaged

 

Yeah, if you’re lucky. Any failure of downstream components will cause a chain reaction. So you design to protect that from happening.

The recommended circuit for the 7805 from the datasheet, calls for much larger capacitors. Without the, the regulator will fail, and well, NOT regulate.

I checked the datasheet and didn't see any resistor limiting the current between source and input of 7805. Do you think the 5W resistor here is necessary. Because I know that the input current depends on the output current.

 

Since we don't have pic, we also have to assume stuff. What is the output voltage?

We don;t know what package your using either. The tabs of a TO-220 generally are kept electrically insulated from each other with a to-tto mounting kit. Some kits need a silicone grease, others do not.

For the 7805 in a to-220 pkg, the tab is ground and for the LM1117 in a TO-220 pkg, the output is Vout. Connect them together bad things happen. Chassis with no mounting kit.

 

Since we don't have pic, we also have to assume stuff. What is the output voltage?

We don;t know what package your using either. The tabs of a TO-220 generally are kept electrically insulated from each other with a to-tto mounting kit. Some kits need a silicone grease, others do not.

For the 7805 in a to-220 pkg, the tab is ground and for the LM1117 in a TO-220 pkg, the output is Vout. Connect them together bad things happen. Chassis with no mounting kit.

Sorry if there's something not clear. From this power suply circuit, I take 15V voltage for LM324 and a gate mosfet driver IC (IR2125); 3V3 voltage for kit STM32F103C8T6 (blue pill).
And also I use TO-220 for both LM7805 and LM1117. then what bad things happen when I connect them like that? Can you expain it to me? I'm newbie, so sorry if my question is awkward. Because I can't research these things on Google.

 

R3 was used for reducing the input voltage, normally the 7805 can be input from 8~12V, when the power input is 15V then you can count it as input 8V and the other 7V (15V-8V) using the R3 to take over the power dissispation, if the max output current is 500mA then W = V*I = 7V*0.5A = 3.5W, use it at least 3 times of calculation values in the real world, if the draw current almost less 300mA then 10W is enough, if the draw current always close to the max current then you will need to use 15W.

 

You have the output of the 3.3 V regulator connected to ground.

Take the regulators off the heatsink and measure the voltages.

You need two of these https://www.digikey.com/product-det...ivision-of-boyd-corporation/4880SG/4880SG-ND/ kits. This uses a silpad, so no grease.

 

You might consider a switching regulator 7805 replacement. https://www.digikey.com/product-detail/en/murata-power-solutions-inc/7805SR-C/811-1116-ND/

 

Thank you all. Now I know how it works.
For more protection of the kit stm32, I intend to add a resettable fuse (500mA) before 3V3 pin of the kit.
Do you think that's useful?

 

If you make R3 a metal oxide resistor, it will help with overloads. They tend to vaporize and act as fuses,
A diode would help.
If you can afford the voltage drop of a full wave bridge, that would make the adapter not care.

Watch out for unregulated adapters. A transorb might help. They come in directional and bi-directional.

 

If you make R3 a metal oxide resistor, it will help with overloads. They tend to vaporize and act as fuses,
A diode would help.
If you can afford the voltage drop of a full wave bridge, that would make the adapter not care.

Watch out for unregulated adapters. A transorb might help. They come in directional and bi-directional.

I use a ceramic resistor for R3. Is adding a fuse better?

 

A fuse would not hurt, but the regulators already have over-current protection.

R3 is there to lower the temperature of the 7805 by reducing the voltage on its input. That's a different function compared to a fuse.