Hi.
Starting from a 20V laptop adapter and some pre-owned heatsinks, I'm gathering parts for one of Tony van Roon's marvellous small circuits, **broken link removed**. This circuit has been endlessly copied; there must be a way to cool it without a fan.
I have four raw-beginner questions.
Q1.
Somewhere in van Roon's circuit are resistors that answer the equation Vout = 14.5V = Vref * (1 + R2/R1). They're hiding. Please could someone point it out?
Q2.
The battery must be an active player in this circuit. I want to think of it as a resistor whose resistance increases as the battery gets fuller. Is that valid?
Q3.
Could I substitute an LM358 for the LM301A?
Q4.
And finally, could someone please give me the benefit of your experience, concerning power dissipation and heatsinks?
(20V-14.5V)*2A = 11W. Yikes! That's a lot.
Could I put 3 1N4004 diodes between +Vin and the LM350T, to get the voltage down to 17.9V?
And then -- wishing I didn't have to -- limit the current to 850 mA, by increasing the value of R7 and probably changing R4 and R5 (with your help)?
(17.9V - 14.5V) * .85A = 2.9W. This is less wattage than a nightlight; I feel safer.
See the picture of heatsinks scavenged from old computer PSUs. These don't come with a datasheet or even a part number, naturally, so I'm trying to back into what I need.
To keep the circuit temperature below 60 C (LM317T Heatsinking - Electric Circuit among others), seems mostly a function of the heatsink surface area ( Heat Sink Design, Calculators, Thermal Analysis Software , he counts both sides).
Tmax = 60 C
Ta = 30 C (Harry Lythall's recommendation)
Pd = 2.9W
Rthj-c = 4.0 C/W, junction to case, from datasheet
Rthc-hs = 1.6 C/W, case to heatsink, a google
Rth-hs = 5.17 C/W, required thermal resistance
Area = 9441 sq.mm., of aluminium heatsink needed.
The heatsink in the middle of the picture is my candidate for this project. It measures 51 mm high x 47 mm wide x 5 mm thick. There are fourteen fins, each measuring 4 mm x 47 mm. That's an area of 10058 sq.mm., counting both sides. So, > 9441, and it seems we're good. However, it's just math to me, and might bear no relation to reality. Could someone who has seen a few heatsinks please let me know his/her thoughts?
Well, that's probably five questions.
Thanks very much for your help.
Starting from a 20V laptop adapter and some pre-owned heatsinks, I'm gathering parts for one of Tony van Roon's marvellous small circuits, **broken link removed**. This circuit has been endlessly copied; there must be a way to cool it without a fan.
I have four raw-beginner questions.
Q1.
Somewhere in van Roon's circuit are resistors that answer the equation Vout = 14.5V = Vref * (1 + R2/R1). They're hiding. Please could someone point it out?
Q2.
The battery must be an active player in this circuit. I want to think of it as a resistor whose resistance increases as the battery gets fuller. Is that valid?
Q3.
Could I substitute an LM358 for the LM301A?
Q4.
And finally, could someone please give me the benefit of your experience, concerning power dissipation and heatsinks?
(20V-14.5V)*2A = 11W. Yikes! That's a lot.
Could I put 3 1N4004 diodes between +Vin and the LM350T, to get the voltage down to 17.9V?
And then -- wishing I didn't have to -- limit the current to 850 mA, by increasing the value of R7 and probably changing R4 and R5 (with your help)?
(17.9V - 14.5V) * .85A = 2.9W. This is less wattage than a nightlight; I feel safer.
See the picture of heatsinks scavenged from old computer PSUs. These don't come with a datasheet or even a part number, naturally, so I'm trying to back into what I need.
To keep the circuit temperature below 60 C (LM317T Heatsinking - Electric Circuit among others), seems mostly a function of the heatsink surface area ( Heat Sink Design, Calculators, Thermal Analysis Software , he counts both sides).
Tmax = 60 C
Ta = 30 C (Harry Lythall's recommendation)
Pd = 2.9W
Rthj-c = 4.0 C/W, junction to case, from datasheet
Rthc-hs = 1.6 C/W, case to heatsink, a google
Rth-hs = 5.17 C/W, required thermal resistance
Area = 9441 sq.mm., of aluminium heatsink needed.
The heatsink in the middle of the picture is my candidate for this project. It measures 51 mm high x 47 mm wide x 5 mm thick. There are fourteen fins, each measuring 4 mm x 47 mm. That's an area of 10058 sq.mm., counting both sides. So, > 9441, and it seems we're good. However, it's just math to me, and might bear no relation to reality. Could someone who has seen a few heatsinks please let me know his/her thoughts?
Well, that's probably five questions.
Thanks very much for your help.