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Mini , 07-24-2024, 01:03 PM

It would charge input capacitor indeed because you would apply to the input 5V-0.2V(diode voltage drop)= 4.8V. But it would keep input and output difference quite small... only 0.2V or so (diode voltage drop)

Mini , 07-24-2024, 01:04 PM

What you tried testing? Where you get 3.3V?

AncientNord , 07-24-2024, 01:04 PM

voltage drop 1.7

Mini , 07-24-2024, 01:05 PM

What's your current draw from 5V? I think your supply going into current limit? Or I don't see how you can get 3.3V by applying 5V to the same point....

AncientNord , 07-24-2024, 01:06 PM

Mini , 07-24-2024, 01:06 PM

I think we misunderstand each other. That's not what i meant. I know this regulator can supply 100mA.

Mini , 07-24-2024, 01:07 PM

Can you explain once more what did you do? Where did you apply 5V and how did you get 3.3V?

AncientNord , 07-24-2024, 01:12 PM

give me some time, I will switch to Windows and do a simulation of this circuit in Proteus

AncientNord , 07-24-2024, 02:02 PM

AncientNord , 07-24-2024, 02:02 PM

This is what I meant

AncientNord , 07-24-2024, 02:04 PM

but overall it's weird😅

Mini , 07-24-2024, 02:12 PM

why would you disconnect output? That's not what I meant. If you would do this how would you expect to use your 9V input?

Mini , 07-24-2024, 02:13 PM

This picture is correct, that's what i meant. And you have everything as expected. But you have to try this in real world schematic. You have to check how much current you will lose into regulator.

Mini , 07-24-2024, 02:14 PM

Mini , 07-24-2024, 02:15 PM

Measure current here in these places, but in real world schematic. This way you know how much current goes into regulator.

Mini , 07-24-2024, 02:17 PM

In this simulation all your 0.11A goes into L1?

Mini , 07-24-2024, 02:18 PM

Seems like it, but one current meter is missing to tell that.

AncientNord , 07-24-2024, 02:25 PM

Mini , 07-24-2024, 02:28 PM

Can you add more in series with the load?

Mini , 07-24-2024, 02:29 PM

And the one in the regulator input is duplicate to the one series with the D3, but it won't hurt though.

Mini , 07-24-2024, 02:30 PM

We can calculate now easily based on this simulation. Seems like 1.21 mA+2.92mA =4.13 mA going into regulator. And 108mA - 4.13 mA = 103.87mA going into load. It looks totally fine.

Mini , 07-24-2024, 02:31 PM

Now test this in real world circuit. Btw if you choose schottky diode your voltage drop will be much less than here having 0.7V. Lets say like 0.15V-0.2V or so.

AncientNord , 07-24-2024, 02:34 PM

AncientNord , 07-24-2024, 02:37 PM

I need some time or read a book) i don't understand this moment.. why when i connect these point i get 5V

AncientNord , 07-24-2024, 02:58 PM

In general everything should work) thanks a lot everyone for your help

Mini , 07-24-2024, 03:03 PM

Btw note that your load draws more than 100 mA meaning if you use your regulator to supply this current then your regulator is rated as 100 mA max. Not sure how much current your external 5V source can supply. I'm sure you set up random load, but just in case.

Mini , 07-24-2024, 03:04 PM

Yeah this circuit looks fine.

Mini , 07-24-2024, 03:06 PM

But how would you use your regulator output if you don't connect it? Think what would happen if you connect 9V to your regulator input. Your regulator regulates it down to 5V, but your following circuit is disconnected. How you get 5V to your circuit this way?

Mini , 07-24-2024, 03:08 PM

AncientNord , 07-24-2024, 03:10 PM

I understand that if I take a 9V power supply and through the regulator, I will not get anything at the output..

Mini , 07-24-2024, 03:10 PM

Mini , 07-24-2024, 03:11 PM

Now think how currents will flow. 5V comes from your external 5V supply NOT from the regulator output.

Mini , 07-24-2024, 03:11 PM

And you get 5V because they are same point. They are at the same potential.

AncientNord , 07-24-2024, 03:14 PM

but before connecting I have another potential here.. what happens to it?

Mini , 07-24-2024, 03:16 PM

Mini , 07-24-2024, 03:17 PM

I made a quick drawing about the current flow. Do you understand now?

Mini , 07-24-2024, 03:18 PM

AncientNord , 07-24-2024, 03:22 PM

And now it’s completely different) now I understand .. so I need to read something so that I can also independently determine how exactly the current flows in the circuit

Mini , 07-24-2024, 03:22 PM

You have there some voltage because you give 4.3V to the regulator input. Remember regulator is big circuit. It's made of many transistors. So you get some leakage voltage on the output. Now how that voltage gets there is more complicated. I don't know on 7805 case what would happen if you put load on the output now. Voltage might drop to 0V. Remember there is a whole circuit inside 7805. And I'm not sure how accurate this simulation is as well.

Mini , 07-24-2024, 03:24 PM

This is from random LM7805 datasheet. You should analyze this if you want proper answer.

QDrives , 07-24-2024, 07:28 PM

The datasheet capture is from another regulator. As you can see for the adjustable version, there is a limit of output voltage compared to input.
That is why you may want to add the anti-parallel diode.

As both @Mini and I mentioned, if your load is 100mA, pick a regulator that can do at least 150mA. Otherwise you might get one that limits the current or even shuts down.

AncientNord , 07-25-2024, 10:46 PM

Hello. one more small question. if you have time can you look at the diode? is this the best choice or did i choose wrong?

Mini , 07-26-2024, 02:10 PM

It depends on what input voltage you want to use on 9V input. With 9V totally fine, but since you mentioned up to 35V then it isn't very good choice. 40V reverse voltage is kind of close, I would choose something with bigger margin, but it will work. Generally next thing to look at is how big voltage drop this diode would have at max current you need and also calculate how much power it will dissipate. But since your current in this configuration will be very low like few mA perhaps no need to worry about power dissipation. And I checked it's datasheet about voltage drop - looks like under 0.2V at 10 mA and at 25 degrees. So in this way it would be fine just depends on what input voltage you want to use.

Mini , 07-26-2024, 02:10 PM

Mini , 07-26-2024, 02:13 PM

Also note that if you want to use it for example in series with the input/regulator then you need to look at this graph at 100mA(your max current) and so on.

QDrives , 07-26-2024, 08:10 PM

As I mentioned: it depends.
Forward voltage (@ current), max (continues/peak) current, reverse voltage, reverse current, capacitance, reverse recovery time, temperature, thermal conductivity, etc.
What is important for your application?

AncientNord , 07-27-2024, 12:38 AM

Forward voltage (@ current), - The lower the better
max (continues/peak) current, - Equal to or Higher than 100mA
reverse voltage, - Equal to or Higher 40V

Mini , 07-27-2024, 10:02 AM

Not sure why you go on so much about simple diode. If your input voltage will be up to 35V then you need better diode. When it comes to current in this configuration you don't need current capability since it will draw less than 10 mA. If you want to put it in series then you will have close to 0.3V at 100 mA at 25 degrees. Told you everything you need to know already.

AncientNord , 07-27-2024, 05:11 PM

I was answering his question) from your answer I understood the selection of the diode... I paid so much attention to it because there are a lot of diodes when choosing) and if the first prototype 100 boards will be successfully tested. then i would need 10k so careful selection diode makes sense)

QDrives , 07-27-2024, 07:45 PM

Ok, but what application?
Are you still referring to your original 'solution'?
Are you referring to the anti-parallel one?

For the original solution you need to make sure that the current drawn (load) is (much) lower than the diode can handle. Thermal conductivity may be important too.
For the anti-parallel solution, as @Mini already mentioned, you do not need much except for the voltage and perhaps reverse current (but that depends on the load current).