Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Voltage Regulator ICs?
i want to use a voltage regulator IC (7809) to give a circut a constant supply of 9 volts. if i was to use a 9 volt battery with a voltage that is constantly dropping would that work (even though the voltage is under 9 volts? thanks.
5 Answers
- MarkGLv 71 decade agoFavorite Answer
No.. Linear regulators burn off excess power so that the applied voltage is reduced to the regulated value. It does this by changing its resistance to cause a voltage drop. The current flowing through this varying resistance will generate heat.
As you increase load on the power supply the output voltage starts to droop. THe 7809 regulator quickly adjusts its resistance so as to allow some of the extra applied power to be passed on to the load there by keeping the output voltage relatively constant.
Inorder to provide the load with additional power upon demand the input to the regulator must be more than the regulated output voltage. Any short fall in output voltage is boosted by passing additional voltage from the input.
If you apply less voltage than 9 volts the regulator will not have any excess power to work with. Its resistance will drop to its minimum and you will have an unregulated output that is less than 9vdc
- Anonymous1 decade ago
That type of voltage regulator only lowers voltage and dissipates excess as heat. To keep 9 volts from a 9 volt battery you would want a boost converter. Unfortunately they are more complex.
- 1 decade ago
No. A voltage regulator requires a certain minimum output to sense voltage drop, with the lowest I've seen about 1.2v, so your supply would need to be at least 1.2v about the voltage that you wanted (since you want 9v, you would need to source 10.2v)
- Anonymous4 years ago
until ingredient is costly, very soft, has many pins, and/or is probable to be destroyed in up plenty in use/in the time of soldering, this is appropriate to solder right now. eg many ICs are fantastically soft to warmth, and with all their pins they relatively used to get broken by capability of instruction manual soldering. Fuses have a holder via fact a) soldering is probable to break them and b) they want replaced in the event that they do blow. voltage regulators shouldn't blow in a decently designed circuit, and that they are somewhat warmth tolerant- take care of them extremely like a transistor, and solder it in. As others have stated a cooling warmth exchanger is a powerful concept, until drawing tiny currents from them(and that generally messes up their voltage administration, yet this is yet another question...) volt regs can get fantastically warm as warmth output[W]= voltage drop[V] * present day[A] .-NB that works for all gadgets inc transistors. length/spec a warmth sink by capability of determining what temperature you may enable the area of attain and subtract ambient temp to get temperature difference. Ambient temp could be warmer than room temp if that is in a case. divide warmth output by capability of temperature difference to get Watts/in line with degree temp. any warmth sink of that or bigger fee might desire to be waiting to do away with warmth rapid adequate to shrink the temperature
- How do you think about the answers? You can sign in to vote the answer.
- 1 decade ago
Yes as long that the voltage amount of your source is range from 8.0 - 9.0 V. But if it is below 7.9 >, Voltage regulator is useless already. You need transformer for that matter.
I