Is there a mistake in this layout?

Started by Canucker, November 09, 2013, 01:37:28 AM

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Canucker

On the top of this image where it says the Negative lead from the power supply should it not be the positive? I don't see where the positive lead attached. Also...the 1 Meg resistor by the input...what does that achieve?


LucifersTrip

always think outside the box

Jdansti

Translation:

You have a PNP transistor. Your positive battery lead should be connected to ground and the negative battery lead should be connected where it is shown on the schematic.
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R.G. Keene: EXPECT there to be errors, and defeat them...

Thecomedian

Quote from: LucifersTrip on November 09, 2013, 01:41:52 AM
it's pnp
no pop

Um, to expand. PNPs have Positive-Negative-Positive doped sandwich material. A PN junction is forward biased when positive voltage touches P and negative voltage, otherwise commonly ground, touches N. The effect is that the positive voltage pushes holes towards the center, while the negative voltage pushes electrons towards the center of this junction. These holes and electrons then jump across, causing current flow, and continue on for electrons to the positive voltage source, and holes to the negative voltage source. This is theory so it's just a way of thinking about it, not technically 100% reality, but it's the way it's taught.

So anyway, you need to forward bias that junction of NP for the base-emitter, to turn it on. A forward bias means that you need negative voltage, or voltage more negative than, positive voltage going to the emitter of NP junction. the PN junction of collector to base is REVERSE biased, which means you need negative voltage, or voltage more negative than, the more positive voltage going to the base.

So, to end,  the voltage needs to be more positive for base, and most positive for emitter. The collector needs to be least positive in voltage. That is why it's a -9v supply and 0v ground (which is going to be called "positive ground").

taking this and doing the reverse for NPN. It's NP for C-B and PN for B-E. We forward bias B-E so + on P(B) and - on N(E). We reverse bias C-B so + on N(C) and - on P(B).

And the 1meg resistor is to eliminate popping of switching the effect on and off, as lucif implied.
If I can solve the problem for someone else, I've learned valuable skill and information that pays me back for helping someone else.

Jdansti

Thanks for the expansion.  I learned something.   :)
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R.G. Keene: EXPECT there to be errors, and defeat them...

Canucker

Quote from: Jdansti on November 09, 2013, 01:53:07 AM
Translation:

You have a PNP transistor. Your positive battery lead should be connected to ground and the negative battery lead should be connected where it is shown on the schematic.

My brain was being kinda slow....I basically looked at this layout after looking at the original lofo momofo and oversimplified with an "oh thats not right" rather then "look at all the differences...oh thats why thats different too". I need to be more careful when surfing new projects after midnight. Thanks for catching my oversite.