# Getting 15VDC from an unreliable 16-18V source

Discussion in 'General Electronics Chat' started by mbek, Dec 9, 2011.

1. ### mbek Thread Starter New Member

Oct 30, 2011
16
0
Hello,

what would be the course of action to get 15V from a 16-18V DC source? The source is some kind of an adapter and it is not reliable.
L7815 requires at least 17.5V if I understand correctly?
I tried putting smoothing capacitor and a HF filter, but I still get ~16-17VDC.

Would it be ok just to put a 15V zener diode and a suitable series resistor (suitable as in terms of wattage) ?

Thanks.

Last edited: Dec 9, 2011
2. ### tommydyhr Active Member

Feb 3, 2009
39
4
A zener diode (ZD) would work perfectly fine, as long as you keep within it's maximum ratings.

When considering a value for the series resistor, keep in mind that you need to keep a minimum current flowing through the ZD for it to keep maintaining it's zener-voltage (commonly referred to as it's knee current, $I_{ZK}$, or simply it's minimum current). You must also keep in mind that in addition to a ZD always having a rated PD (Power dissipation), it also has a maximum current, $I_{ZM}$ at which it can keep up regulation.

This range, $I_{ZK}$ to $I_{ZM}$, is known as the ZD's operating area.

Last edited: Dec 9, 2011
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3. ### mbek Thread Starter New Member

Oct 30, 2011
16
0
Yes, well, it seems zener is not an option since I need to draw 2.5A, and that would mean cca 50W diode.

If I would get a bit higher rating input (18V), I could use L7815 in a high current regime.
But I am at a loss when calculating R1.

Formulas from L78xx datasheet:
$
R_1=\frac{V_{BEQ1}}{I_{REQ} - \frac{I_{Q1}}{\beta_{Q1}}}
I_O=I_{REG}+Q_1(I_{REG}\frac{V_{BEQ1}}{R_1})
$

I suppose they messed up, and that $I_{REQ}$ in first formula is actually $I_{REG}$, right?
And what is $Q_1$ in the second formula??
And is $I_{REG}$ actually quiescent current of L7815? (6mA)

And from datasheet of BD536:
$\beta=25
V_{BE}=-1.5V
I_{Q1}=-2A
$

Or would maybe a LM150 be a better solution?

Last edited: Dec 9, 2011
4. ### tommydyhr Active Member

Feb 3, 2009
39
4
It would indeed seem as if they've made some kind of typo there.

Since you don't need that much more current than the 7815 can support (and you're dropping a very low voltage across of the regulator), hooking up two of the 7815's is actually a viable solution.

You just need to isolate their outputs from each other by adding a rectifier (1N4007 would work just fine) to each of their outputs. This will, of course, reduce the total output voltage to $V_{out}=V_{reg}-V_{D}=15-0.7=14.3V$, but you can compensate for this by also inserting a similar rectifier to the regulator's ground pin, effectively also increasing their reference by a diode drop. You would also have to add a very small resistor (~0.25 Ohms) to each output, in order to equalise the load between the regulators. If the resistors weren't added, the regulator with the lowest output voltage (even though the regulator is rated at 15V, it may output 15.05V or 14.95V) will attempt to carry all of the output current.

This solution would, with the L7815, give you a total output current capability of 2x1.5=3A.

I can provide a schematic of this solution later today, if need be.

Last edited: Dec 9, 2011
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5. ### SgtWookie Expert

Jul 17, 2007
22,183
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Just to make sure everyone's on the same page, here's the current ST Microelectronics datasheet for the L78xx:
http://www.st.com/internet/com/TECHNICAL_RESOURCES/TECHNICAL_LITERATURE/DATASHEET/CD00000444.pdf
Have a look at Figure 29, "Dropout voltage vs. junction temperature" on page 37, top left.

Note that the highest output current plotted is 1A. The 500mA output plot is ~500mV below the 1A plot. It really is not reasonable to assume that 1.25A output will be 250mV above the 1A plot since it's so close to the maximum for this regulator, and above the 1A max other manufacturers use for their 78xx series - but we can roll with it for the moment.

So, when the regulator is at room temp, at 1A the dropout is 2V, and we're slapping a somewhat arbitrary 250mV on top of that for the 1.25A output.

Now, tommydyhr suggested a 1N4007 be used on the outputs. I am not sure why, as these diodes are rated for 1A as an absolute maximum for continuous current.
Example datasheet: http://www.fairchildsemi.com/ds/1N/1N4001.pdf
Note the Vf specification under "Electrical Specifications" on page 1: Forward Voltage @ 1A = 1.1V

So, depending on temp, your total dropout will be somewhere around 2.25v + 1.1v = 3.35v; and even this 3.35v dropout is a risky calculation - it will likely be more.

Since the desired Vout is 15v, the absolute minimum input voltage would be 15v+3.35 = 18.35v, greater than the highest voltage specified by mbek. Clearly, this solution is not viable.

Getting a 15v output from a 16v to 18v input is going to be pretty tough with a linear regulator.

One solution might be a National Semiconductor LM1084 adjustable:
http://www.national.com/ds/LM/LM1084.pdf
The dropout voltage could be acceptable once the regulator has warmed up; that shouldn't take long with a 2.5A current flow through it.

Last edited: Dec 9, 2011
6. ### tommydyhr Active Member

Feb 3, 2009
39
4
The drop-out voltage will indeed become an issue, but if you look at the aforementioned L78xx-datasheet, you have an output current capability of >1.5A, even at 125*C, with a dropout voltage of 3-4V (I don't get how you get the highest plotted current to be 1 Amp).

Yet again, this dropout voltage will indeed pose a problem (you can subtract the rectifier voltage from the total dropout, though, as if you read my previous post you'll see I suggested adding a rectifier to the regulators' ground terminal so as to compensate for the rectifiers drop).

My bad on the 1N4007's. A BY299 will do the job, though.

I will actually try and build a circuit very much like this, under the same conditions, tomorrow, just to satisfy my curiousity.

7. ### SgtWookie Expert

Jul 17, 2007
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Take a look at ST Microelectronics' datasheet that I posted the link to in my last reply.
The 1.5A is only guaranteed when Tj=25°C. Since the thermal coupling from the junction to the substrate and the case is something like 4°C/W, it'll be pretty tough to keep the junctions at 25°C - even if you DO have a very large forced-air cooled heat sink.

As far as the highest current plotted, look at Figure 29, "Dropout voltage vs. junction temperature". I'd already mentioned this. I don't see where you are getting your numbers from?

Even with a 1A output, when Tj=25°C and Iout=1A, dropout is 2v; which already disqualifies it. Our OP needs a maximum of 1v dropout.

No, you need to ADD the rectifier Vf to the total dropout, as it will be in series with the regulator. Adding a diode between ground and the regulator ground does nothing to raise the input voltage, or decrease the regulator's dropout.

A BY299 does have a suitable current rating. With 1A flowing through it, the Vf is a tad over 0.8v. This leaves 0.2v for the regulator when the input is 16v. That's not going to work either.

8. ### tommydyhr Active Member

Feb 3, 2009
39
4
The attached picture should clear things up.

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9. ### SgtWookie Expert

Jul 17, 2007
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Notice that the title says "Peak output current vs...", the operative word being "Peak", which to me, means a very short period of time.

You need to look at figure 29 instead.

10. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
LM2941 only needs about 1V headroom to regulate. It will do it for you.

11. ### SgtWookie Expert

Jul 17, 2007
22,183
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Bountyhunter,
mbek did not state the current requirement in their original post; it is 2.5 Amperes, where the LM2941 is only rated for 1A.

The LM1085 might do the job at a bit less cost than the LM1084; however 2.5A is just a bit too close to the maximum 3A rating of the LM1085. I usually like to see at least 20% overcapacity to spare; which means 3,125mA for the stated requirements.

In light of our OP's complete requirements:
Out: 15v @ 2.5A
In: 16v-18v
and your inside knowledge of these regulators, what would you suggest?

12. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
If the 1085 devices are rated at 3A then they are good for 3A guaranteed. Current limit is usually set at least 50% higher. They are a really good product but I recall their dropout is one VBE + VSAT so it's actually about 1.2V maximum (not 1V) but would probably work in this app.

Micrel used to make some true LDO's with dropout under a Volt that do 5A and 7.5A as I recall. Probably pricey.

Last edited: Dec 9, 2011
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13. ### bountyhunter Well-Known Member

Sep 7, 2009
2,498
507
You know I recall I developed a cheap LDO controller IC called the LP2975 that is good for this (building an LDO) with an external transistor. Could do it cheap and easy. handles input voltages up to 26V.

http://www.national.com/ds/LP/LP2975.pdf

I can help with component values so you don't have to read the whole 20 page data sheet.

Last edited: Dec 9, 2011
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