Overload control on LM7809 output

Thread Starter

kiltro

Joined Oct 24, 2011
50
Hello everyone,
I've a symple PS circuit using a LM7809.
I'd like to put a limit to the load current so that if it goes above 100mA the output of the ldo is disconnected from the output connection (a simple barrel jack), and possibly and possibly signaled with a red LED.

Can you suggest some possible ways of doing this?

Thanks
 

crutschow

Joined Mar 14, 2008
34,454
Do you just want to limit the current, or have it go to zero and stay there until the power is removed to reset the supply?

Providing a limit circuit is simpler.
 
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Thread Starter

kiltro

Joined Oct 24, 2011
50
Do you just want to limit the current, or have it go to zero and stay there until the power is removed to reset the supply?

Providing a limit circuit is simpler.
I was thinking about to the second option you mention, but now I'm thinking about a foldback limiter with a led indicator for overcurrent
 

LowQCab

Joined Nov 6, 2012
4,075
The Spec-Sheet contains a recommended Circuit for Current-Limiting.
But, it calls for an outdated Transistor.
This Transistor should do the job, BD912, You will need 2.

As a Bonus, this Circuit will easily handle ~10-Amps of Current with an adequate Heat-Sink.
A small Cast-Aluminum-Box will handle the heat-Sinking for your present project requirements.
Resistor "Rsc" determines the Current-Limit.
Resistor "Rsc" will probably be around ~7-Ohms for a 100mA Current-Limit.
Higher Resistance-Values will lower the Current-Limit.
Your Power-Supply will have to provide at least ~2-Volts more than the Regulated-Output-Voltage.
This means that, at 100mA, and 2-Volts of "Drop-Out-Voltage",
this Circuit will dissipate between .2 and .4-Watts, so You shouldn't have to have a Heat-Sink,
but some sort of a Heat-Sink is a good idea in any case.
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Over-Current Protection .png
 

Attachments

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Thread Starter

kiltro

Joined Oct 24, 2011
50
The Spec-Sheet contains a recommended Circuit for Current-Limiting.
But, it calls for an outdated Transistor.
This Transistor should do the job, BD912, You will need 2.

As a Bonus, this Circuit will easily handle ~10-Amps of Current with an adequate Heat-Sink.
A small Cast-Aluminum-Box will handle the heat-Sinking for your present project requirements.
Resistor "Rsc" determines the Current-Limit.
Resistor "Rsc" will probably be around ~7-Ohms for a 100mA Current-Limit.
Higher Resistance-Values will lower the Current-Limit.
Your Power-Supply will have to provide at least ~2-Volts more than the Regulated-Output-Voltage.
This means that, at 100mA, and 2-Volts of "Drop-Out-Voltage",
this Circuit will dissipate between .2 and .4-Watts, so You shouldn't have to have a Heat-Sink,
but some sort of a Heat-Sink is a good idea in any case.
.
.
.
View attachment 288060
Thanks I have a bunch of BD140, I guess they are ok too
 

LowQCab

Joined Nov 6, 2012
4,075
Just keep in mind that the BD140 is a much smaller Transistor than the BD912.
The BD912 can easily handle 10-Amps,
but the BD140 can only deal with around ~1-Amp ( at less than ~10-Volts Input ).

But since you're limiting the Current to ~100mA, there shouldn't be any problems.
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Thread Starter

kiltro

Joined Oct 24, 2011
50
Just keep in mind that the BD140 is a much smaller Transistor than the BD912.
The BD912 can easily handle 10-Amps,
but the BD140 can only deal with around ~1-Amp ( at less than ~10-Volts Input ).

But since you're limiting the Current to ~100mA, there shouldn't be any problems.
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Ok, thanks

Was thinking about something like this
Schermata 2023-02-22 alle 18.21.03.png

Will do some test though
 

LowQCab

Joined Nov 6, 2012
4,075
At first glance it looks like it should work,
but 10nF Bypass-Capacitors are probably too small, and could cause problems.
I would go with 330nF and 100nF as shown in the previous Schematic.

If you're going to go to all that trouble, there are better options ..........
See PDF below .......
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crutschow

Joined Mar 14, 2008
34,454
Secondary is 12vac, then there's a diode bridge
So the filtered bridge output will be a little over 15Vdc.
That means the regulator will be dissipating about 0.6W at 100mA output, and Q2 will dissipate up to 1.5W under 100mA short-circuit conditions.
 

Thread Starter

kiltro

Joined Oct 24, 2011
50
BTW how would you go about testing overload/shortcircuit conditions? With a current sink?
 
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Thread Starter

kiltro

Joined Oct 24, 2011
50
I ended up making it much simpler by removing the current limiting circuit, after all LM has short circuit protection and I just place a fuse on transformer secondaries to protect it.

Now, I made this layout in kicad

Pedalboard PS_WOCL.jpg

Pedalboard PS_WOCL2.jpeg

Any suggestions?
 

BobTPH

Joined Jun 5, 2013
8,985
When did all regulators start being LDOs? LDO (low drop out) is not even a noun it is an adjective. And 7809 is not an LDO regulator. It's like people started calling all cars "fasts" whether they are fast or not.
 
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