raw beginner: heatsink without a datasheet

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.

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?

Click to expand...

I don't see that formula in the circuit, but I think the output voltage is set by the divider made up of R4, R5, and R2. The output voltage of the regulator will be 1.25 volts higher than the adjust pin (Reference).

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?

Click to expand...

I think in most cases you can look at it that way.

Q3.
Could I substitute an LM358 for the LM301A?

Click to expand...

No

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.

Click to expand...

If your ruler is CM then I think you could run at full power with the heatsink on the right. I use a rule of thumb that Thermal resistance = 50/sq root of the area in CM. So if that heatsink has 100 sq. cm it would be about 5C/Watt or 55C rise over ambient. Pretty toasty but still under the maximum junction temperature of 125C.

Thank you very much.

Q4. I'll go by maximum junction temperature.
Q3. What characteristic governs the choice of the LM301A?

LM301's I think are finally obsolete after 45 years. A TL081 or TL071 should work though. You will not need C3 anymore, the 081/071 is internally compensated.

The LM350 develops a constant 1.25 volts across pins 1 and 2. With a 15 and 230 ohm resistance across it, you will have I = E/R = 1.25/(230+15) = ~5ma of current flowing through the 3k resistor on its way to gnd. With E = IR = .005*3000 = 15 volts, the output of the LM350 will be 15 + 1.25 = 16.25 volts (not including the small current contribution from the 350's ADJ pin. This should all be on the 350's datasheet.

Jaguarjoe,

TL071 or TL081 will be easy to find, thanks.

I calculated 16.25V as well. The circuit notes claim 14.5V, which is in line with gel and marine battery manufacturers specs. Oh well, another bit of magic which will just have to wait for the multimeter.

I'm still think of putting a couple of power diodes 6A 400V between V+ and and the LM350's IN pin. To reduce the voltage difference across the LM350.

Thanks very much for your help.

There's one minor hitch, what to do with the "start" button between pin 1 and gnd. The functionality of that pin is probably different between a 301 and a TL0xx. I've never seen anyone use a compensation pin for anything other than that. I suspect its just an elegant way to force the 301's output to go to gnd but I'm not sure.

This identical ckt was lifted from the National Semi datasheet for the LM350. You have to watch these guys, they make transistors with 3 collectors or 4 emitters! They're good.

Be careful with the input voltage to the LM350. Lower is better but don't go so low that you let the 350 drop out. It needs a couple of volts across it.

If the transistor is cut off then you should see 16.25 out of the 350. With it on there should be around 14 because there is now a 15k res in || with the 3k.

Jaguarjoe,

I was thinking of connecting the TL071's two offset pins to GND. You think it could be more devious? Or side-stepping the issue by using a TL072.

So no problem with my using power diodes to bring the 20V laptop adapter down to 18V. Thanks.

The 15K and the 3K are in parallel! Yes, so obvious once it's pointed out. You'll never know how many permutation I tried, to get that sum to come out. Thanks twice.

I wouldn't ground the offset pins, the 071 will not like that, just let them hang.

Take a look at this, I just found it, it may be more useful:

**broken link removed**

The diode he put across the 350 is cheap insurance.

Also, if you use an LM350K in the TO-3 can it might be easier to cool. It costs more than the model T version though.

I think you might be able to add a small amount of resistance in the lead going to the - terminal (100, 200 ohms) of the op amp then put the reset switch to the - terminal.
Don't forget diodes dissipate heat as well. Your diodes will probably go to over 100C without heatsinks.

ronv,

100-ohms resistor to the Vcc- (pin 4) of the TL071C. Okay. Is that because the op amp is being used as a sink?

For this diode, I was thinking MUR840 with TO-220 package, using a small heatsink akin to the Wakefield 273-AB-01. Is this enough heatsink?

Thanks very much for your help.

Jaguarjoe,

I looked at **broken link removed**. Thank you, I like it.

Mr. Hamer uses an LM1458 with no switch. And he uses the common-or-garden 1N4148 and BC558. I like the use of the 100 ohm pot. Does this pot mean I don't have to dig up special 1% metal film resistors for the project?

In my junk collection, there are scavenged heatsinks that measure 5 C/W, using ronv's simplified formula. But I have nothing (except the metal frame of an old FM tuner) that dissipates at 3.3 C/W. Is that a worry?

I read and (surprisingly) understood Jan Hamer's notes on the 2nd half of the LM1458 does.

Thanks very much.

100-ohms resistor to the Vcc- (pin 4) of the TL071C. Okay. Is that because the op amp is being used as a sink?

Click to expand...

No in series with the negative input terminal. On power on the OP Amp comes on in low voltage mode so reset just makes the output go high.

For this diode, I was thinking MUR840 with TO-220 package, using a small heat-sink akin to the Wakefield 273-AB-01. Is this enough heat-sink?

Click to expand...

Well, if you look through the data sheet for the diode you will see it drops about 1 volt (Vf) at 2 amps = 2 watts. The heat sink has a spec of 55C per watt (almost no help) so it would rise 110C above ambient.

ronv,

Thank you for the lore concerning an op amp when it has just come on.

I found this heatsink (defunct FM tuner) with 163 sq.cm surface area, counting both sides. Using your calculation, this will dissipate at 3.9 C/W. I guess I can forget about those diodes?

Thanks again.

Should work. It will be hot but should work ok.

In the end, I'll build **broken link removed**, where I started. However, it was good to be able to compare with the other van Roon circuit.

Two last doubts, though.

1. Can I put both the 500R bleeder resistor and also a backwards 1N4002 protective diode in parallel with the LM350?

2. I'm still trying to reduce that 20V before it hits the LM350. What about the obvious, a resistor? Could I put a 1 ohm 5W resistor right next to Vin? The circuit is kept at 2A. Naively, 2 amps x 2 amps x 1 ohm = 4 watts. Will it work, or will it catch fire?

Thanks for all your help.

Most power supplies feed passive circuits. The only storage there is in electrolytic capacitors that could back feed through the regulator when the input shorts. A small rectifier across the reg can easily handle the cap's current. OTOH, with a battery charger, there is a lot of current available from the battery that can back feed. Much more current than a small rectifier can handle. The diode will need to be able to survive the short circuit current from the battery which will mean a pretty hefty diode.
The 500Ω resistor offers no input short circuit protection.
A diode in series with the input will protect the regulator, or skip all of this and just make sure the input never shorts to ground. I looked at a 6a diode, a 6A4 400v job, its Rθja is 10 deg/w. At 2a and 1.5w, it would be at 45degrees.

Resistor wattage should always be derated by a factor of 2 to 2 1/2. A 5w resistor dissipating 4w will get mighty hot. A 10w unit would be better.

Jaguarjoe,
Thanks for pointing out the Rθja of 10 C/W on the 6A04. Consider it now part of the circuit.

Thank you, both.
You've both been very patient (and lucid) with a beginner. I'm very grateful for your time.