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8550\8050 v's 2N3904/2N3906

Discussion in 'General Electronics Discussion' started by BanksySan, Jul 22, 2012.

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  1. BanksySan

    BanksySan

    17
    0
    Jul 22, 2012
    Hi all,

    I'm building a drawdio, using this circuit layout: http://www.2ne1.com/files/3913/0968/4001/tracks.jpg

    However, I have 2N3904/3906 transistors instead of the 8550 & 8050 transistors suggested, do you think this is OK?

    Also, I was originally planning on using a 9V battery so the sound was louder, but this is making the board get very, very hot indeed. Is this a sign I've shorted the board somewhere?

    Thanks

    Dave
     
    Last edited: Jul 22, 2012
  2. john monks

    john monks

    693
    1
    Mar 9, 2012
    Not enough information
    Generally a 2N3904 and a 2N3906 make fine audio transistors.
    Without a circuit your question cannot be answered.
     
  3. BanksySan

    BanksySan

    17
    0
    Jul 22, 2012
    Hi, the link didn't post. Here's the circuit:

    http://www.2ne1.com/files/3913/0968/4001/tracks.jpg
     
  4. john monks

    john monks

    693
    1
    Mar 9, 2012
    The bases of the transistors are tied together and on the NPN transistor the emitter is going to the negative side of your 9 volt battery and on the PNP transistor the emitter is going to the positive side of the 9 volt battery. No good.
     
  5. BanksySan

    BanksySan

    17
    0
    Jul 22, 2012
    Thanks for the reply, are you saying that the circuit is wrong, or that it's OK with a 1.5V battery but not a 9V battery?
     
  6. john monks

    john monks

    693
    1
    Mar 9, 2012
    Your board shows Q1 and an NPN but your writing show Q1 as 8050 which is a PNP.
    So now I am confused. Can you simply draw a schematic of what you intend to buid so that we can take it from there?
     
  7. BanksySan

    BanksySan

    17
    0
    Jul 22, 2012
    Hi,

    That is the schematic I've been copying the track layout.

    I've used Q1 = 3904, so replacing the 8050 and Q2 = 3906. replacing the 8550.
     
  8. john monks

    john monks

    693
    1
    Mar 9, 2012
    I still the circuit the same way. You say you have a 9 volt battery and the plus side is connected to the emitter of a 2N3906 which is PNP transistor and the minus side is connected to the emitter of a 2N3904 which is a NPN transistor. The bases are tied together on your drawing.this leads to umteen amps going through the bases of both transistors. Am I seeing this wrong?
     
  9. BanksySan

    BanksySan

    17
    0
    Jul 22, 2012
    You're seeing it right, that's the way I wired it.

    The original design uses an AA battery though: http://www.2ne1.com/files/6713/0968/3989/built.jpg

    I'm following the circuit design a bit blind, I've only just started doing this stuff. My knowledge and soldering skills are definitely in the "improver" stage.
     
  10. CocaCola

    CocaCola

    3,635
    5
    Apr 7, 2012
    You need to limit the current to the bases of the transistors, you will kill them if you flood them with the umteen Amps that John is talking about...
     
  11. BanksySan

    BanksySan

    17
    0
    Jul 22, 2012
    OK... Any advise how I work out what my resistor value should be?

    Also, if I just use a 1,5 V, does that solve the problem?
     
  12. john monks

    john monks

    693
    1
    Mar 9, 2012
    Let's start all over. I don't mind spending the time no matter how long it takes.
    So first how much voltage gain do you need?
    What kind of a load do you intend to drive? A headphone?
    Do you intend to just use two transistors and some resistors?
    I can guide you through it every step of the way.
     
  13. davenn

    davenn Moderator

    13,669
    1,891
    Sep 5, 2009

    and in addition to what John said above, whilst you are at it.... what is the IC that you show in the construction ? :)

    Dave
     
  14. BanksySan

    BanksySan

    17
    0
    Jul 22, 2012
    Thanks very kind of you, John. Thanks. I started again last night.

    The IC is an NE555.

    the instructions suggest an AA battery, I tried a 9V as I have a 9V holder and I thought it would give a louder tone and the datasheets for the IC and the timer suggested they were tolerant to 9V.

    I'm driving a Peizo speaker (http://www.coolcomponents.co.uk/catalog/66mm-64ohm-speaker-p-733.html).

    I've replaces the suggested transistors with 3904 and 3906's as I had those about.

    As for my knowledge level... I've remembered Ohm's Law and Kirchoff's Laws and an slowly remembering how capacitors work. The maths is returning, but it's a painful process.

    This is the first soldered circuit I've made.

    Thanks

    Dave
     
  15. john monks

    john monks

    693
    1
    Mar 9, 2012
    That's exactly why I want to help you through this.
    Give me an idea of what you expect the circuit to give you and I will help you make it happen.
     
  16. BanksySan

    BanksySan

    17
    0
    Jul 22, 2012
    The end product should make an audible tone which is controlled by the pencil graphite. It's easier to see in this video then for me to explain.

    I assume that it's the resistance through your body and the graphite that changes the frequency of the 555.
     
  17. john monks

    john monks

    693
    1
    Mar 9, 2012
    This particular circuit will not operate properly with a 9 volt battery. Therefore I recommend replacing the 2N3904 and 2N3906 because you no doubt damaged them with the 9 volt battery. Use a 1.5 volt battery with a 2.7 ohm resistor in series with it to limit the current. Then we will take it from there.
     
  18. BanksySan

    BanksySan

    17
    0
    Jul 22, 2012
    I've replaced them with new transistors and even bought a new 555 (just on case).

    I looked at the datasheets for the transistors though and both seemed to be able to handle those voltages.

    Can I ask how you got the figure 2.7 for the resistor? (I#m not challenging it, jst want to know how it was reached).

    Cheers

    Dave
     
  19. john monks

    john monks

    693
    1
    Mar 9, 2012
    2.7 ohm is a good nominal value. At worse with a short circuit you will draw about 556 milliamps. If the transistors base emitter voltage is at .7 volts the current draw will be about 33 milliamps. That is (1.5-.7X2)/2.7. The actual base emitter voltage for boathouse the transistors should be some where around 0.75 volts to the actual current due to the transistors should be 0 amps. And the resistor is small enough to look like a short compared with the rest of the circuit. You may not need the resistor but I only suggest using it to prevent damaging the transistors.
    Now after building up the circuit it will be interesting to see what the 555 one-shot is doing. Let's see what happens.
     
    Last edited: Jul 25, 2012
  20. BanksySan

    BanksySan

    17
    0
    Jul 22, 2012
    Hi John,

    Thanks, I can see that the small resistor would provide protection from a short. I've only got 3.4 Ohms, but I think that'll be OK.

    The speaker is 64 Ohms, so I had assumed that the circuit was flicking between infinite resistance and 64 Ohms.

    The bases are connected to pin 3, so they will be at the full voltage supplied (minus anything lost to the safety resistor), Could you explain this line to me: The actual base emitter voltage for boathouse the transistors should be some where around 0.75 volts to the actual current due to the transistors should be 0 amps.?

    Again, thanks for your help, I'm really appreciating it.

    Dave
     
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