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# Where can I get these circuit diagrams?

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#### semiconductor

##### Full Member level 4
thirystor

I've just done a power supply that can provide: 24VDC - 4A.

But at the moment, I face a problem:
- How can I protect my supply from short-circuit
- How can I limit the current output (I out <= 4A), I mean, If Iout > 4A caused by the decrease in load for example, the protection circuit will pull it back to 4A or 0.

I really need circuit diagrams for my above purpose!

thirystor circuit

Hi

Look at this project if it could help :

thirystor data sheet

I've just tested it. It works well but if someone can, upload here another circuit diagram? I need more solutions!

Get a three terminal 24Volt 5Amp voltage regulator ic, TO3 can on heatsink, and wire it in current limiting mode. You can find the circuit in the data sheet.

It is dificult to propose current limiting circuit for your stabilizer because the addition should be compatible with your circuit. If you want any reasonable suggestion it is neccessary to post circuit I believe.

anyway

look at:

I have down the file: FUSE.PDF from the link you gave above. But I do not understand the operation of this Fuse. Can some one explain to me how it works (It limits the current over 2A).

explaination

It goes this way:
When ps is swiched on, both BC182 and BD241 are fully conducting because 1K resistor connected to base of BC182. Transistors are darlington connected.
Load current is flowing through BD241 and 0,35 ohm resistor. At current limit (2A) the voltage drop on 0,35 ohm is sufficient to open second BC182 (most right). BC182 through current limiting resistor 1K
which is conn. to collector and thirystor gate, trigers it. Thirystor conducts and grounds base of first BC182. Fuse breaks, load is switched off. Key is meant for reset. When pressed it shorts thirystor which stop conducting. After key release load is powered again.
For greater load current limit it is neccessary to change 0,35 ohm resistor and to check alowed collector current for BD241 so as base drive of first BC182

fir the current limitation u can use series voltage controllers or u may use parallel voltage controllers. u can design one in a way taht for a short circuir if voltage drop across a test resistor is increased than a transistor will get on and will draw the current from the circuit. u can refer to such circuits in the book "electronic devices and circuit theory" by boylestad

u can also use simple hardware logic. design counter through simple d or jk flip flops nad then decode the time where u wanna make alaram to binary. say for 3 output should be 11. and the output of these counter such that when they meet the desired time they produce a logic zero. feed that to the reset of the IC's/flip flops

to borber:
I think that I understood your explanation. But I'm wondering and it is also one problem that I encounter in pratical:

First, I assign the load resistance of 24 Ohm. It leads to the current 1A through the load. When I decrease the load resistance, The current flows through the load increase (of cause). When the load resistance reaches 6 Ohm. In theory, the fuse must shut down to protect the circuit and the load. But in fact, it does not. If the current flows through the load increase (from 1A up to 4A), it makes the U (CE) of the darlington (BC182 - BD241) increase (not as zero as designed) and it affect the output voltage, and current through loads (I think that the working points of these transistor drop into the active region (not cut-off or saturation region)).

The second problem is that: the thyristor does not grounds the Base of DARLINGTON (I use the type: NEC 5P4M)
so even when the Transistor BC182 (most right) does not open fully (Uce=0), the led lights because of current going through it into base of the DARLINGTON.

I'm now worried. I can't find any solution to these problems. I have been trying but my work does not preceed!

(One think that: what does the role of this DARLINGTON here? is it used as electronic switch?)

Hi Semiconductor,

The best solution to solve your problem it's to try posting your schematics.
I don't want to believe that it contains only a transformer, a bridge rectifier and one big cap.
You wrote:
(not as zero as designed) and it affect the output voltage
But believe me it was not designed like that, not at all. You can't decrease to zero the voltage across the darlington BC182- BD241. This voltage will always be subtracted from your output.

You wrote:
One think that: what does the role of this DARLINGTON here? is it used as electronic switch?)
Why do it think that the U (CE) of the darlington (BC182 - BD241) increase and it affect the output voltage ?
I'm confident you're a smart guy and have seen already the voltage drop across the darlington as a sum of BD241 Ube + BC182 Ube + voltage across R1.
How the circuit would work at 4A without darlington ? Where the bias will come from ?

You wrote:
the thyristor does not grounds the Base of DARLINGTON
Obvious it can't. Because the thyristor it's not properly biased. Usual the colector of BC182 drawn on the right goes to the base of darlington in order to cut off the bias current. But border wants some kind of lock and fit the thyristor. But unfortunatelly, not well.

Regards,
Silvio

OK, on pdf file it's a small example of electronic fuse protection.
The circuit drawn surrounded by dotted lines it's added to the circuit to be protected.
The input voltage is applied at J3 and J4.
Q1 and Q3 make up a bistable circuit.
C1 helps the bistable going into default state (Q1 off, Q3 on) when power supply is applied first time.
When the voltage across R-sense equals the Ube of Q2, Q3 goes off and Q1 "sucks" the bias current delivered by error amplifier to the base of Q4.
Since the R-Sense is fited outside the voltage control-loop, the voltage across R-Sense it's not subtracted from stabilized output voltage.
By closing S1, the output voltage recovers after short circuit or over current.

fuse

to semiconductor
as I understand your problems you connected the fuse after the regulator instead in front of it.
But the best way you can get propper helt is to post your schematics.

Your circuit is on page 13. Circuit is voltage/current stabilizer. Current through load can not exceede value set with current adjust setting. With load resistances smaller then certan value output voltage will go lower mantaining max. adjusted current limit.
Puting fuse circuit in front of regulator/stabilizer it will break the load current to negliglible value when trip value of the fuse is reached.
If your load can withstand 4 A as you want you don't need fuse. Just set current adjust to 4 A. In that case load current will not exceed 4 A.

Yeah, I have tried your solution, but something goes wrong.

- The second BC182 (most right) is npn type and if it is connected as in FUSE.PDF, the potential at Emitter is lower than potential at Collector of the second BC182, the voltage of U (CE) allows current going from C to E (inverse, because potential at output teminal is higher than potential at ground).

- I do not understand this sentence of BORDER (My english is not good):
). BC182 through current limiting resistor 1K
which is conn. to collector and thirystor gate, trigers it.

You are right. Second BC182(right) is wrong. It should be PNP type (BC212) connected in such a way that base and emitter are interchanged and collector is connected where it is. Now, when voltage drop on 0,35 Ohm resistor (because of load current increase) reaches 0.65V (Vbe) transistor BC212 starts to conduct. Collector current of BC212 is limited by 1k resistor in collector circuit (connected between collector and thyristor gate). But collector current should be high enough to fire thyristor. It is now mostly limited by 1k resistor.
Other part of fuse circuitry is OK.
My english is no good enough so I appologize for missunderstandings.

Another tip.
If you want to set trip level more precise put trimer (1k) in parallel to 0.35Ohm resistor and connect base of BC212 to the slide.

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