Connect with us

Application of 7805

Discussion in 'General Electronics Discussion' started by rai.abhishek, Jul 5, 2012.

Scroll to continue with content
  1. rai.abhishek

    rai.abhishek

    6
    0
    Jul 5, 2012
    Hi,
    I have attached the circuit.
    In this I was trying to get high current across resistance R2. But as soon as the circuit is connected the voltage dips to 0.8 volt (across R2 and base-emitter junction).

    But if I remove the transistor and then ground the other terminal of R2. Then a large current flows through R2. I don't understand why by placing the transistor in the circuit the voltage dips.

    Please help me out.

    Every response is appreciated.:)
     

    Attached Files:

  2. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    1. The 7805 will limit output current to 1A so you will not get the current you anticipate.
    2. The 7805 tries to raise the output voltage 5V above the reference terminal. The reference terminal is connected to the output so the output voltage will rise to about 2V below the input.
    3. A high current into the base of the 2N2222 could fry it, check that it still works.
    4. The base voltage relative to ground will not rise above about 0.7V if the transistor is still intact.

    Tell us what you are trying to do so that we can help with the circuit.
     
  3. rai.abhishek

    rai.abhishek

    6
    0
    Jul 5, 2012
    Hello sir,
    I apologize, in place of 2N2222 there is 2N3055 power transistor.
    From this circuit I want to drive the transistor to high current.
    All I am trying is to make sure that base of transistor receives current of about 1A.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,418
    2,788
    Jan 21, 2010
    A 12 ohm resistor between +12V and the base will do that
     
  5. rai.abhishek

    rai.abhishek

    6
    0
    Jul 5, 2012
    Hi steve,
    The problem is that maximum base emitter voltage of 2N3055 is 7V. So I tried to remain in this limit by using 7805 ic.
     
  6. duke37

    duke37

    5,364
    769
    Jan 9, 2011
    The maximum base/emitter voltage of the 2N3055 is with the base negative to the emitter. In this circuit, there is no negative voltage. When the 2N3055 is turned on, the base will be approximately 0.7V positive (one diode voltage drop).

    If you have a supply more than this you will need a resistor as *steve* says to limit the current.
     
  7. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    648
    May 8, 2012
    In the event that you're trying to obtain 5V regulated at a current higher than the 7805's 1 Amp limit. This might be what you're looking for. If not, nothing lost.
     

    Attached Files:

  8. rai.abhishek

    rai.abhishek

    6
    0
    Jul 5, 2012
    Hello,

    @CDRIVE

    Can I connect base resistance and 2N3055 in place of the load?

    As I have shown in the figure.
     

    Attached Files:

  9. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    648
    May 8, 2012
    That makes no sense. What are you trying to design?
     
  10. rai.abhishek

    rai.abhishek

    6
    0
    Jul 5, 2012
    Hi to all,

    I was working on the attached circuit. My aim is to make sure that transistor Q2's base terminal gets about 1A current.
    If anyone can please let me know whether this will happen in the attached circuit then it will be highly appreciated.
     

    Attached Files:

  11. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,270
    Nov 28, 2011
    (*steve*) and duke37 have already tried to explain this.

    You have a transistor with its emitter grounded, and you want to feed about 1A into the base, from a positive power source. Is that right?

    All you need is a resistor between the positive power source and the base.

    When the transistor is forward-biased, the base will have about 0.7V on it, with respect to the emitter. (Actually at 1A base current it will be closer to 1V than 0.7V so let's assume 1V; the difference isn't that great).

    So, if your resistor is connected from a 12V power source, it will have 11V across it, because the bottom end of it will be 1V above ground.

    If you want 1A to flow in the resistor, the resistor needs to be 11 ohms. This is from Ohm's law, which says R = V / I. In this case V=11 and I=1, so R=11.

    This resistor will dissipate a lot of power. P = V I with V=11 and I=1 so dissipation will be 11 watts. You'll need a resistor rated for at least 11 watts and it will get hot. There are alternatives to the resistor, but if you're dropping 11V at 1A, SOMETHING is going to be dissipating 11W and getting hot, unless you use a switching power supply which actually steps the voltage down without wasting all of the power. You can use a 5V regulator and a 4 ohm resistor; in this case the resistor will dissipate 4W and the regulator will dissipate 7W.

    Your mention of 7V being the "maximum base-emitter voltage" for the transistor refers to when the base is NEGATIVE with respect to the emitter. (Polarities are reversed for PNP transistors.) In this situation the base-emitter junction behaves like a zener diode with a zener voltage around 7V and the transistor can be permanently damaged if a lot of current is fed into the base in this direction. When the base is POSITIVE with respect to the emitter, the forward voltage of the base-emitter junction is generally between 0.6V and 1.0V depending on the current. A 2N3055 will not be damaged by a base current of 1A in the FORWARD direction.

    People here would be much better able to help you if you explained what you want to do. Your posts so far have been very brief and have contained no "context". You might think it's better not to burden us with a lot of background detail, but in reality, it just looks like you're being cagey, and often ends up in threads that go on for pages with nothing useful being said.
     
    Last edited: Jul 23, 2012
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,418
    2,788
    Jan 21, 2010
  13. rai.abhishek

    rai.abhishek

    6
    0
    Jul 5, 2012
    Hello Kris sir,

    I would like to tell you the purpose of this circuit.

    I want to build voltage source of 5V capable enough to supply more than rated current of 7805 ic. I searched a lot and then found the circuit that I have posted. Now to make use of this large current I used it to trigger the base of the transistor Q2.

    In the circuit I have connected 7805 ic to ensure 5V supply. This means if I connect 4.3 ohms across the o/p of 7805 ic and base of Q2 then I would be able to get 1A current.One more thing I want to clarify is that in which mode Q1 is operating?

    I just want to take advice on this circuit before I go and purchase the items.
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,418
    2,788
    Jan 21, 2010
    The first thing to do if you want to do something like this is to get hold of the datasheet.

    Toward the end they typically include circuits you might find useful (less so for simple components, but very common for things like this).

    At the bottom of page 24 you'll find 2 circuits, one with short circuit protection and one without.
     
  15. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,270
    Nov 28, 2011
    Is that all you want to do? You want a 5V supply that can provide more than 1A current? That circuit won't do that; all it does is feed current into the base of the transistor, which makes the transistor conduct, but nothing more.

    You need to look at an earlier post on this thread by CDRIVE. He wrote:
    That post includes a schematic that will do what you want. It uses a power transistor as an emitter follower (aka common collector), to boost the amount of current available. This circuit does not provide current limiting, so if you short out the output, the transistor will dissipate a LOT of heat. It will also dissipate heat during normal operation. You can calculate its power dissipation from the formula P = V * I where P is power in watts, V is the voltage ACROSS the transistor (i.e. the input supply voltage minus the output voltage; the voltage that is being DROPPED by the transistor) in volts, and I is the output current in amps.

    For example, if your input voltage is 12V and your output current is 3A, power dissipation in the transistor will be 21 watts. This is calculated from 7 volts multiplied by 3 amps (7 volts is the voltage dropped across the transistor). The transistor will definitely need a heatsink, otherwise it will overheat and be damaged. Transistors don't have any kind of protection feature; they're "dumb".

    This circuit also has poor output accuracy and regulation, but this shouldn't be a problem unless you're powering something that needs a very exact voltage.

    The circuit you show with a second transistor, a load resistor, and a second supply doesn't do anything useful. You just need the 7805, the diode, and the power transistor.
    That's right, if you connect a 4.3 ohm resistor from the output of a 7805 to the base of a transistor, with its emitter grounded, you will get about 1A of current flowing into the base-emitter junction. This will make the transistor conduct. So current will flow in the collector circuit. It doesn't really do anything useful; it doesn't give you a regulated voltage source. The transistor would be operating in common emitter mode.

    As Steve suggests, you should download the data sheet for the 7805 regulator. It will include circuits for boosting the regulator output voltage (try to get data sheets from several manufacturers; the 7805 is made by several companies). This is probably where the emitter follower design came from originally.
     
  16. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    648
    May 8, 2012
    I already showed you how to do that here.

    [​IMG]
    You then ignored my suggestion, modified my circuit (for reasons unknown) and posted it here?

    [​IMG]
     
  17. CocaCola

    CocaCola

    3,635
    5
    Apr 7, 2012
    I guess I have to asked, why use the 7805 in the first place, why not just get a higher rated 5V regulator to start with? You can easily find 2A drop in replacements, and if you are not locked in to the same package 3A and 5A ones are easy enough to source...
     
  18. lramirev

    lramirev

    2
    0
    Mar 4, 2013
    CDRIVE:
    Thank you for your so helpfull suggestion. I'm doing something like that. But really required less current as capable of providing th 2N3055, so I'm switching to a TIP31C which support 3A, enough for my project. I do have a question about your circuit. Your diagram shows RL on the emitter of 2N3055. I'm assuming this RL is the load I'm attaching to the power source. My question is: Do I need to provide this RL in the circuit or you just place it there to represent the load of the power source? What would be the concecuences of removing RL from the circuit and sending the current directly though the emitter to my power feed? How would I calculate the value of RL?

    I have attached a schematic of the power source I'm planning.

    I know I may use a equivalent 5V regulator that supports 3A, the thing here is that I don't have the, in my local store and I need to order them plus delivery and time. To much trouble.
     

    Attached Files:

  19. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,270
    Nov 28, 2011
    The emitter-to-ground resistor in CDRIVE's diagram is actually labelled "Load", not RL. It is intended to represent the resistance of the device or circuit that is being powered from the regulator.

    But! You should use a resistor from the transistor's emitter to ground, unless the circuit you're supplying will always draw a significant amount of current (e.g. at least 10 mA).

    Without any current being drawn from the transistor's emitter, the output voltage will rise due to leakage current in the transistor. Also the transistor's base-emitter forward voltage varies somewhat with the collector/emitter current, so ensuring a minimum load current will improve regulation, which is not very good with this design.

    If your load doesn't draw significant current at all times, I suggest using a resistor from the emitter to ground. A 470 ohm resistor will draw about 10 mA and dissipate about 50 mW.

    P.S. Have you changed your username from rai.abhishek to lramirev or are you a new user joining this thread?
     
    Last edited: Mar 4, 2013
  20. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    648
    May 8, 2012
    Good advice Kris. ;)

    Chris
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-