Voltage Regulator - surprise

The output stage of the regulator is several transistors connected as a very high current follower that, when as on as it can be, drops between 1.5 and 2 volts, depending on the current. As long as the output voltage is below the regulated target voltage, this circuit stays fully on. However, the circuit also includes a temperature sensor that is supposed to turn the pass circuit off if the chip temperature exceeds something like 150 C. This is supposed to prevent rapid destruction of the regulator from overheating. But if overheated repeatedly, I have seen the regulators eventually fail.

If your fan draws only 250 mA, you need a heat sink on the regulator, to keep its temperature down. Another thing that makes these regulators run hot is if they become unstable and turn into megahertz oscillators. Do you have 0.1 uF of ceramic or film capacitance close to the regulator, input to ground? Do you have any capacitance between the output and ground?

I don't know any easy way to add a low voltage cut out to the 78xx series of linear regulators.

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Regards,

John Popelish

A regulator is an amplifier that operates on an output/reference voltage difference. It saturates HIGH when the output is low for any reason, including because of inadequate input voltage. Failure, though, shouldn't happen because of simple input undervoltage. I'd look at the input/output capacitors, it sounds like the regulator may be oscillating. The motor might be a problematic load, too (put a reverse-biased quench diode on the motor terminals, to be safe).

You can add a cutout circuit to the input to the regulator:

+------------------o o---+ | | | | | S1 |>| | | __ | +--o o---+--------+-----+-----+---[7815]----+--> + | | | | | | | | [10K] [Rly] [D1] +-[C1]-+-[C2]-+ |+ | | | | B | +-----+ | A P | k | T 10K O Gnd

S1 (momentary) turns the circuit on. When the battery drops below the set point established by the pot, the relay drops out and the circuit draws no power, even if the battery voltage rises (because current is no longer being drawn from it). The circuit will protect both your battery (from being discharged too far) and the 7815. Although the TL431 can sink up to 100 mA, it is best to use a low current relay, such as Allelectronics CAT# RLY-622, which draws about 16.3 mA. That is a 12 volt relay with a 740 ohm coil, so you would add a 330 ohm resistor in series. The low current relay reduces the current the circuit draws from the battery. You can add a high efficiency LED between the 10K and the pot as an on indicator, if you want.

Ed

Three terminal regulators would cook if the battery is connected to the output and the input falls below the battery voltage - it is in the application notes - they recommend an isolation diode on the output or diode from the output to feed voltage to the input (which is reversed biased during normal operation and keeps the regulator happy when the input voltage is missing).

Having a battery on the output and no input is just like connecting the regulator backwards in the circuit. A large cap on the output can kill a regulator when power is removed.

The diode makes a cutout circuit unnecessary unless you are concerned about the small power (quiescent current) the regulator wastes when sitting idle

Hi again, To eshjr I can see where a relay configured to 'cutout' the battery would avoid problems. Thanks.

To default The battery is connected to the VR only. There aren't any direct connections from the battery to the load. Nor would I know how to. But now due to a curious nature - I'll be looking into that! So thanks a lot - I think.

A big Thank you to all JP, eshjr, & default- much appreciated. The fan is rated 6 to 15 volts. The rechargeable battery power supply was 18Volts. My 'too simple' remedy was to use a 15V Regulator. Once again I learned how true the adage "a liitle learning is a dangerous thing". After your explanation, I replaced the Voltage regulator with a 33ohm

5W resistor in series with the load (DC Fan 60ohms). I tried a 5.1Volt 5Watt Zener (in series with the Fan),but without any calculations it just felt warmer to the touch. I'm hoping that this is a safer solution. . .? I'm still curious to know why the Voltage Regulator 'cooks' when the input voltage drops to (around) 7Volts and the regulator's voltage to way below regulation - (around) 5Volts. Shouldn't there be much less heat generated at this stage - or at least not enough to heat the VR to failure? I did have (recommended) caps at the input and the output. I'm guessing that at the Fan's dropout Voltage (5Volts), a 'switching' on- off oscillation occurred, and the momentary 'on current' was much higher than 250mAmps, enough to burn the VR. Any explanation would be apprecated, BobP

The thermal overtemperature cutout for these regulators compares a reference voltage derived from an internal 6.3V zener diode to the cut-in Vbe of an internal temperature sensing transistor positioned to shunt the base drive away from the output transistors at a die temperature of 175oC. At your low input voltage of 7V or so, the zener bias current is too low which makes the zener reference voltage low, and this in effect shifts the resulting thermal cut-out threshold to a higher die temperature causing the 'VR' to fail. The output transistor SOA protection circuit will also be inoperative, leaving the current limiting protection as the only safety feature in place. You also risk damaging your battery by running it down to one third of its full charge cell voltage. You need to include a battery voltage monitor which latches the whole battery load off when it drops below a threshold of say 12V under load. Because this is a battery powered system, you would be better off using CMOS logic and a MOSFET series pass element for the monitoring and control.

Not plausible...

On Jan 16, 6:51 am, Fred Bloggs wrote:

BobP Wow! and Thanks! Your explanation may be just a 'little too clever for me', but there was a flash of Iight here and there, enough for me to at least "believe" if not "undestand". By the way just in passing, who did create the universe? It wasn't you or - was it? Please say it wasn't. Now I have a real value for 'deep cycling' batteries. That alone is great information. Another Question: The circuit for all this came from an old mercury 'heat' type thermostat. The contacts are open to a set 'low' temperature (e.g. 70oF), at which point the contacts close. The contacts stay closed as long as the temp is below 70oF and stay open as long as the temp is above 70oF. Problem was, I needed the Thermostat to operate as 'cooling' or air conditioning type. As such - it would have to turn on a small fan when the contacts were 'open' and turn off the fan when the contacts were 'closed' - the exact opposite to its normal usage. A kind of Solution: To do this, I used a PNP (Darlington) transistor configured as a switch. One Thermostat contact is tied to B+. The Other Thermostat contact is tied to the base of the Transistor and Ground (B-) through a resistor (10k). With the contacts 'open', the base of the transistor base is grounded through the resistor. i.e. Transistor on. With the contacts 'closed', the base of the transistor base is positive - i.e. Transistor off. The circuit works - but is somewhat inelegant.. For a start - it draws power even with the transistor off. i.e thru th 10k resistor - a mA or two. Its kept awake to find a better configuration (using the same 'heat' type thermostat). Any suggestions? BobP P.S. I could have and maybe should have just 'recyled' the thermostat

- but then again, look at all I've learned. (There was a previous "lesson" on discarded lithium-ion batteries. Enough just to say I'm using Nickel-Cadmium now.)

Well, the "elegant solution" that occurs to me is to use a regulated wall wart that provides the proper voltage and sufficient current for your relay circuit to start with. That eliminates the need for a 7815 regulator and battery and protection circuit for both of those. It also makes the 1 or 2 mA quiescent current unimportant. You can then make it "more elegant" (at the cost of more current) by doing this:

/ + 12---+---o o----+---[1.5K]---+ | | | +---[Rly]---+ | | Gnd ----------------------------+

It uses a MOSFET relay (Mouser #653-G3VM-61A1) for low current draw (< 7 mA). The relay output is open when the t-stat is closed, and vice versa. It is rated for 500 mA continuous load current.

Ed

Ed

BobP Thanks ehsjr. The location makes a "wall wart" impractical. I'm in Montreal ( = cold). I have a heating radiator in a closed cabinet under a sink (not my design). Needless to say, it gets quite warm in the cabinet maybe 15oF higher than the room temperature. The fan is supposed to blow out the warmer air into the room. Ergo the thermostat. I know there are (inexpensive) air conditioning thermostats that would solve the design problem. But . . . BobP P.S. Had to google 'wall wart' just to be sure. Very colorful description.