i have a cordless phone that specifies 9v DC input 500mA.
i am currently using a 12v DC 500mA adaptor for this cordless phone.
this has been plugged in for about 12 hrs now and i haven’t yet seen or perceived any damage to the unit, yet.
the adaptor is admittedly hot, but not so hot that i should keep a fire extinguisher handy.
so, my question(s) is (are):
What are the consequences, both long term and short term, of using a 12v input on a device rated for a 9v input. And should i change the adaptor with immediate effect ?
Also, what if the adaptor was indeed a 9v DC, but a 250mA instead of a 500mA.
note: for those who like redundant details, the input AC to the adaptor is 220-240v.
What you are doing is not advisable over time. Damage to the battery is more likely than damage to the adapter. In some cases an overcharged battery can explode after it’s removed from the charger and spew dangerous and toxic materials all over the place or catch on fore while it’s charging. This is not specualtion I was a Radio Shack Mgr. in my mis-spent youth and I have seen this happen.
An overcharged battery is not a “good thing” to have pressed against your face. 9V adapters are easily obtainable from your local Radio Shack or even the bargain bin at your local Goodwill or Salvation Army for a dollar or so. Just make sure ther voltage and tip polarity is correct and (milli) amperage is fairly close to (preferably not much over) the phone’s original unit.
Usually running a semiconductor device on a voltage higher than intended wil cause near-instant failure. In this case your cordelss phone has an internal voltage regulator which steps down the input voltage into the lower voltages needed inside. As long as the input voltage is does not exceed the maximum allowed voltage of the regulator circuit, it won’t fail immediatly.
In a simple linear tyle regulator (which is probably what we’re talking about here), the more voltage you put into it, the hotter it will get. So running your phone off 12V rather than 9V will cause the regulator in your phone to get significantly hotter than the manufacturer intended. This may shorten the life of the regulator circuit. How much, I can’t say.
Simple rule for electronics. Anything that runs hot dies young.
The current rating is the maximum for the adapter. These things are voltage sources, which means they will keep the voltage constant and let the current vary to deliver the desired power. If your phone draws 300 mA of current then it will run fine off of the 500 mA adapter, but is pushing the 250 mA adapter past its design limit. Worst case it could catch fire and burn down your house and kill dozens of people. Most likely it will just get hot and stop working.
Match up the voltage exactly, and use an adapter with the same current rating or higher. If you can find a 9 volt 750 mA one it would work fine. If the phone specifies 500 mA you don’t want to go lower than 500.
ok… so i’m looking at battery and regulator circuit damage in the long run… i guess i’ll go get a 9v 500+ mA adaptor… i was hoping that the extra 3 volts wouldn’t really make a difference to any such internal voltage regulator of the phone.
it’s been on for some days now and the adaptor and the phone are both generally warm, not hot. When i once used an adaptor with a lower ampere rating than the phone the adaptor was significantly hotter, as engineer_comp_geek said…
thanks AndrewL and engineer_comp_geek…
astro, i think i’ll add you to my phone’s quick dial list… you always end up answering my questions…
also, i don’t have a radio shack within a thousand miles from my home… so i guess i’ll have to look elsewhere… basically i was wondering whether it was a better idea to stay with the extra 3 volts damage, or risk a sub-standard but correct voltage adaptor… 'coz it’s difficult to find a reliable good quality adaptor around here…
so i was wondering which damage would be less risky…?
The others are pretty much correct… the problem is not that you’ll burn up the 12V power supply; it’s that the voltage regulator inside the phone will run hotter.
To illustrate, here’s an example.
Let’s say there’s a 6V linear regulator inside the phone. Let’s also say the maximum current the phone requires is 500 mA (i.e. the load requires 6V at 500 mA).
If the input voltage is 9 VDC, the regulator will dissipate a maximum power of 1.5 watts. Most TO 220-packaged devices (which includes most through-hole regulators) have a junction-to-ambient thermal resistance of around 40 °C/W without a heat sink. Assuming the ambient temperature is 25 °C, the regulator’s junction will be 85 °C. This is below the 125 °C “safe” limit for such devices.
But let’s say you boost the input voltage to 12 VDC. Because the load still requires 6V at 500 mA, the regulator will now dissipate a maximum power of 3 watts, and the regulator’s junction will be 145 °C. This is above the 125 °C “safe” limit for such devices. (It could even worse than that. Though the 40 °C/W rating assumes no heat sink, it also assumes the package is standing “upright,” thereby exposing all surfaces to the air. If the back surface of the package is screwed to the PCB as is commonly done, the effective junction-to-ambient thermal resistance goes up.)
The saving grace is that most regulators have thermal overload protection, which means you probably won’t permanently damage the regulator, at least in the short run. But in order to shut down, the regulator must exceed 125 °C for a certain amount of time. Do this enough times, and you will decrease the life of the regulator.
Solution: Get a 9V power supply.
Mouser Electronics sells them very cheap. Plus they do not have a minimum $ amount for an order:
BTW: The above example was just that - an example. Chances are, the battery-charging circuit does not feed off the linear regulator, but instead has its own charging circuit. But it’s still not a good idea to use a 12V power supply, since it could also damage the charging circuit and/or batteries in a similar fashion.
xash, everyone else here appears to know electronic circuits in greater detail than me, so I can’t comment much beyond what they said.
But to make one thing clear, the 500mA rating on your power supply is simply its maximum capacity. The wall socket in your house likely is rated for 15 amps or more, enough to run an air conditioner… but obviously, you can still plug a night-light into the same circuit… you could get a 100-jilliamp power supply if you wanted to, as long as it’s at 9 volts.
I would base your decision on whether or not to purchase a new power supply on economics. How much is the phone worth, and how much will the new power supply cost you? If it’s a cheap phone, I’d take a chance.
P.S. At least around my house, those “wall-wart” power supplies just seem to accumulate out of nowhere; I must have a good dozen or more various types in my basement. I have no idea what’s involved in shipping one to India, but let me know if you have trouble getting hold of one for some reason.
thanks Crafter_Man… your answer makes it amply clear that i need to get a new adaptor…
Chris Luongo… welcome to the board
i appreciate your generous offer… very kind… however, i think the shipping and custom duty costs will far exceed the cost of purchasing a new adaptor… thanks.
the phone cost me about US$130 so i guess a few more dollars should be worth a new 9v adaptor…
A 9V adapter is the best solution. However, if you’re really in a tight spot, there is another solution, which is quick, easy, and ultra-cheap…
You can easily reduce the 12V to 9V using a “diode string.” A single, forward-biased diode has a voltage drop of around 0.7V. Four diodes in series would thus reduce the 12V to around 9V. At 500 mA each diode will dissipate about 0.35 W, so use diodes rated at least 0.5W. This is not elegant, and it wastes energy, but it would work…
