LM7805 and TIP2955 not working

Discussion in 'General Electronics Chat' started by Domskis, May 21, 2015.

  1. Domskis

    Thread Starter New Member

    May 21, 2015
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    [​IMG]
    I made this circuit shown above. Seems very simple but it does not work. I get 0.5 volts on output and my LM7805 heats up a lot. Some info:
    TIP2955 is made by ST, looking at the front of the transistor going from left to right:
    PIN 1 - BASE
    PIN 2 - COLLECTOR
    PIN 3 - EMITTER

    I am using 1000 uF 25 V instead of 3300uF 50V.
    1 OHM 5 Watt Resistor
    I did not have 0.33 uF ceramic capacitor so I connected two ceramic capacitors in parallel: 224K and 104K. From what I gather that will equal to 0.33uF.
    I also did not use the 10uF capacitor at all since it is just a filter. I am using 19 Volt DC power supply to test the circuit.
    I tested TIP 2955 using simple multimeter. Checking resistance:
    PIN 1 to PIN 2 - 480 OHM
    PIN 1 to PIN 3 - 480 OHM
    Any other connection is infinite. If I remove the TIP 2955 from the circuit I get nice 5 Volts. What am I doing wrong ? Is there any way to test this one by one ?
    Please help....
     
  2. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    The tab on the TIP and the tab on the 7805 are at different potentials. If they are mounted on a common heatsink without insulation, the circuit will not work.

    Also, some capacitance on the 7805 output improves stability. Probably not your problem here, but bad practice not to use.

    ak
     
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  3. Domskis

    Thread Starter New Member

    May 21, 2015
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    OMG - without me even saying that you already knew. I removed the common heat sink and it started working right away !!!!!!!!!!! Thank you so much.
     
  4. Domskis

    Thread Starter New Member

    May 21, 2015
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    I have another issue. This circuit is suppose to provide 5 Amps. I am running just one hard drive from it and instead of TIP 2955 getting hot - it is the LM7805 that gets hot. Seems like the current is not running thru TIP 2955. Why would that be ?
     
  5. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Read...Think...
    Read...Think...
    Read...Think...
    Write.

    ak
     
  6. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    With a 1 ohm sense resistor, anything below 0.7 A will not wake up the TIP.

    ak
     
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  7. Domskis

    Thread Starter New Member

    May 21, 2015
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    Thanks. I am sorry but I am very amateur at this. I was just playing around but got interested and it turned into a project. I was trying to read but there is no good explanation of this circuit. I don't really know what to look for either. I hope I am not abusing this thread by asking most likely fundamental questions.
    Anyways, you say that 1 OHM resistor in between Base and Emitter requires more than 0.7 Amp to wake up the TIP. I believe that the hard drive must be consuming more than 1 Amp. I know from reading that those LM7805 can handle up to 1 AMP but get hot. I have it sitting on the heat sink and the heat sink gets very hot but the TIP 2955 is cold. Could you suggest changing the resistor to a smaller one or bigger one ?

    Thank you
     
  8. Domskis

    Thread Starter New Member

    May 21, 2015
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    Found a solution. I don't know why this works but it does. I changed the power supply to 12 Volts and set up the resistors like this:
    1 Ohm in series with two 1 Ohm resistors in parallel. Now running under load both IC get hot which means that the load is being spread in between them which is exactly what I wanted.
    Thanks for all your help.
     
  9. Bordodynov

    Active Member

    May 20, 2015
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    I made a simulation of stabilizer using LTspice.From the simulation, I made the following changes:1. increased capacity of the condenser up to 10000µF(3300µF not sufficient value).
    2.Ceramic capacitor 0.33µF moved.Now it's parallel to the condenser filter rectifier.
    Ifound that thescheme is verypoorlyif you replace theoutput capacitor at theceramic (a parasitic generation).
    Bordodynov.
     
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  10. atferrari

    AAC Fanatic!

    Jan 6, 2004
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    The circuit or the basic idea of adding a xtor to increment the current handling capability of linear regulator is in most (almost all ) datasheets. LM78XX LM317 LM338 LM337
    Go for one from NatioNational Semiconductors (now TI)
    Read anyone of them in full.
     
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  11. Domskis

    Thread Starter New Member

    May 21, 2015
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    Thank you for your suggestion. I went thru few of them and found this one that actually shows the circuit I am working with: https://www.fairchildsemi.com/datasheets/lm/lm7805.pdf

    The problem is I do not understand certain things and the datasheet will not explain this.
    What is the purpose of that 0.33 uF capacitor ? Is it just for filtering ? How does the value affect the circuit ?
    I found formula for I=V/R which they use for calculating the current. This is my dilemma. See the TIP2955 needs 0.7 V across the 1 Ohm resistor to start working. This voltage depends on the current flowing thru that resistor and my final output load has variable current needs, so when I connect just one load (hard drive) than I only get 0.4 Volts across that resistor which makes my LM7805 super hot. When I connect two 1 Ohm resistors in series I get the TIP 2955 super hot. When I connected the 1 Ohm resistor in series with another two 1 ohm resistors (connected in parallel which I believe gives me 0.5 Ohm) both IC are working since I get 0.7V across the whole resistor set ( the hard drive is working hard though ). When the hard drive goes to idle it heats up the LM2955 again.

    The goal for me is to make it so that the LM7805 gets maximum 0.6 - 0.7 Amps thru it and that's it (I have a heat sink on it - not very big but at 0.7 Amps I believe I should be able to touch the thing and not be super hot). After that all extra current should be handled by TIP2955. It seems though that variable load makes this circuit unstable. I do not know how to calculate this or what to use to make it how I want it. I am asking these questions here since I would have to spend few days reading up on Ohms Law, the Kirchoff law and any other - that is assuming that I would get all of this in the first try. I am learning a lot more by practical tryouts and can read up on things I already understand which brings me to this forum looking for hints.

    Thank you guys
     
  12. MikeML

    AAC Fanatic!

    Oct 2, 2009
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  13. atferrari

    AAC Fanatic!

    Jan 6, 2004
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    My grandaughter also uses it all the time. Seems very useful.

    What is the consumption of your HD?

    Use paragraphs from time to time. A mass of text is not appealing for the reader to be.
     
  14. dl324

    Distinguished Member

    Mar 30, 2015
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    You should consider this in your design criteria. The 7805 has built-in safe area protection. If you let it run up to it's maximum current, it will protect your external pass transistor under overcurrent and overtemp conditions. There is a wealth of information in the National Semi Voltage Regulator Handbook.

    Heatsinks are to prevent the regulator/transistor from thermal damage. It can get very hot to the touch and still be operating in it's safe range.
     
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  15. Domskis

    Thread Starter New Member

    May 21, 2015
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    Thank you all for great answers. I got it working using 3 Ohm resistor. Nothing else would bring the voltage up to required level. I measured the hard drive and it takes steady 0.5 Amp with jumps to 1 Amp.
    What is the purpose of the 0.33uF capacitor there ?
     
  16. MikeML

    AAC Fanatic!

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    This shows how much of the 5A load current is delivered by the 7805 as R5 is varied from 1Ω to 4Ω (red trace). The green trace shows the power dissipation in the 7805, assuming you were starting with clean 12V input (which you aren't).

    boostreg.gif
     
  17. MikeML

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    Stability of the regulator, and to improve dynamic response. Without it, the regulator may oscillate, especially with the current booster.
     
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  18. ian field

    Distinguished Member

    Oct 27, 2012
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    How did you calculate the value of the current sensing resistor? It can be a bit of a juggling act to get it right - the transistor must start to conduct because of increasing B/E bias before the 7805 reaches its maximum current ratin (I believe there are versions that can handle 2 or 3A). If you have a large voltage overhead from the rectifier, the 7805 will be dropping a lot of voltage and therefore dissipating a fair bit. You can reduce the 7805 dissipation by increasing the sense resistor to shift more of the current into the external transistor, but its easy to run into limitations if the transistor hasn't much gain.

    The appnotes circulated by some of the many manufacturers show an even bigger bypass regulator, the PNP external transistor is shown driving a bank of parallel NPN transistors.
     
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