sghr220
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To achieve what?should i add an emitter resistor to the output transistor?
A brief explanation of "fluctuations" would be helpful. Are you talking about permanent oscillations? It may be caused by too much load capacitance.
To achieve what?
.. of multiple paralled output transistors. But according to your schematic you don't have it.To prevent thermal runaway
I meaned more than in the datasheet example design.How much is too much output capacitance?
.. of multiple paralled output transistors. But according to your schematic you don't have it.
I meaned more than in the datasheet example design.
But I compared both designs and found the problem, I think. You added the capacitors C4 and C12 which must not be there. They are introducing a second pole in the loop gain and make the control loop unstable.
Don't bump your thread with useless followups
Forum rules say
Did you already try to operate the regulator without the inappropriate low-pass?
If you want to understand the problem in control theory terms, read about loop gain and loop phase margin.
I see that the value of R7 (4700 ohms) is MUCH TOO HIGH and prevents the BD140 transistor from driving the 2SC5200 output power transistor.
Similar circuits in manufacturer's datasheets use an extremely low input current LM195 output device instead of a high input current 2SC5200 power transistor.
The 2SC5200 might work properly if the 4700 ohm resistor is replaced by a piece of wire.
Forum rules say
Did you already try to operate the regulator without the inappropriate low-pass?
At 10A, the minimum current gain of the 2SC5200 transistor is about 22 so its base current must be 10A/22= 455mA.
The minimum current gain of the BD140 transistor at 455mA is about 25 so its base current in the 4700 ohm series base resistor is 455ma/25= 18.2mA and the base resistor will have a voltage drop of 18.2mA x 4700 ohms= 85.5V which causes the circuit to fail.
So remove the 4700 ohm resistor and replace it with a piece of wire.
You need 9W 225Ω (220Ω) resistor in front of 7812 voltage regulator (60V to 15V).
I dont know what you trying to do with this, but design is not the best. There is LM317HV with max input voltage of 60V but in your case this is not enough. Maybe you should consider usage of newer regulators which allow higher voltages (linear and switcher regulators).
Hi Peter, thanks for your input but could you elaborate on how you calculated these values. :smile:
The lm317 used is floating so it sees only an input-output difference within its ratings (40V).
Calculation:
Input voltage 70V
Needed voltage 15V (at least 3V higher then 12V because 7812 regulator)
Current 0,2A (Raw without other currents of regulator IC,...)
R = 70V - 15V = 55V / 0,2A = 275Ω
P = I²R = (0,2A x 0,2A) x 275Ω = 11W
Power dissipation for 5V and 3A on output will be 50V-5V=45Vx3A=135W.
2SC5200 can allow maximum 150W at Tc=25C.
Theoretically on longer you can use 10A from this design only if output voltage is set around 40V.
For voltage bellow 40V current should be decreased, because power dissipation is increased on bypass transistor.
All depends what currents at what voltages you will need to use, and you need to know what you can expect from this design.
Always you can add one more transistor in parallel to increase dissipation power. Also its good to implement maximum power and over temperature protection and signalisation or smoke is inevitable in some delicate situations.
Best regards,
Peter
That's a very good description of the problem. What the max. output current expected at 18v? :grin:
I try to tell you from first post, this should be first item for discussion before starting to make design.
50V - 18V = 32V (voltage difference)
32V x 3A = 96W (power dissipation) - count up to 100W power dissipation for that case with good cooling.
If you have more then 50V of input voltage, this situation is worse.
This is situation when you should start to think about switching psu.
Best regards,
Peter
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