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L200 Linear PSU with adjustable output Voltage and Current

ChrisP58

Well-Known Member
Most Helpful Member
And the heatsinks need to be on all power semiconductors.

Depending on your settings and load conditions, the current control mosfet could easily dissipate more than your voltage control transistors.

Figure 23 of this datasheet shows a better way of doing a variable current limit. Instead of adding a separate current limit function, it uses the opamp to generate a current error signal that's feed back into the regulating engine of the L200. So no additional power devices are needed.

It does use the ancient LM741 opamp. If you change that, make sure that the replacement opamps inputs can operate near the positive rail. (Many can't)

https://www.mouser.com/Semiconducto...-Regulators/Datasheets/_/N-5cg9g?keyword=L200
 
Last edited:

GiacomoPascon

New Member
And the heatsinks need to be on all power semiconductors.

Depending on your settings and load conditions, the current control mosfet could easily dissipate more than your voltage control transistors.

Figure 23 of this datasheet shows a better way of doing a variable current limit. Instead of adding a separate current limit function, it uses the opamp to generate a current error signal that's feed back into the regulating engine of the L200. So no additional power devices are needed.

It does use the ancient LM741 opamp. If you change that, make sure that the replacement opamps inputs can operate near the positive rail. (Many can't)

https://www.mouser.com/Semiconducto...-Regulators/Datasheets/_/N-5cg9g?keyword=L200
"Connecting point A to a negative voltage (for example - 3V/10 mA) it is possible to extend the output voltage range down to 0 V and obtain the current limiting down to this level (output short-circuit condition)."

I chose to use the Mosfet because I need short circuit protection and this circuit doesn't provide it. At least, this is what I understood....
 

MacIntoshCZ

Active Member
"Connecting point A to a negative voltage (for example - 3V/10 mA) it is possible to extend the output voltage range down to 0 V and obtain the current limiting down to this level (output short-circuit condition)."

I chose to use the Mosfet because I need short circuit protection and this circuit doesn't provide it. At least, this is what I understood....
Yes and no.
Your current will not rise to HIGH value becouse is limited to certain value.
Just add resistor for highest accetable reference voltage to adjust peak current...
 

GiacomoPascon

New Member
Yes and no.
Your current will not rise to HIGH value becouse is limited to certain value.
Just add resistor for highest accetable reference voltage to adjust peak current...
I watched again EEVBlog's video about opamps but I can't understand the circuit on fig.23.
I think it's a kind of Gain adjusting with P1, but I don't know why they put 1K resitor from pin 2 to pin 5...
1618386993924.png
 
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GiacomoPascon

New Member
Hi to all, I'm back after making and testing the l200 board.
The voltage regulation works fine until I put a load on the output.
When the voltage is set to 2.66V, the minimum, and a light bulb is connected the voltage raises to about 5/6V (without touching the potentiometer!).

I initially tought that the problem was the current regulation circuit, but disconnecting it didn't changed anything.
Does anyone know the cause of this behaviour?

Thanks in advace.
Giacomo.
 

MacIntoshCZ

Active Member
weak power supply?
 

crutschow

Well-Known Member
Most Helpful Member
So with no heatsink, the MOSFET became a nice, high temperature heating element.
A MOSFET can generally tolerate no more than about 1W of dissipation without a heatsink.
 

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