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Voltage or Amplitude?

Discussion in 'General Electronics Discussion' started by Truckdrivingfool, Sep 14, 2013.

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  1. Truckdrivingfool

    Truckdrivingfool

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    Sep 30, 2012
    Working on a preamp for a piezo/microphone pickup for a guitar. I've found and read plenty of how amps work but no answer on if the volume boost comes from the higher voltage of the output signal or the raise in amplitude from the gain achieved. Both?

    I had a longer post almost typed out but closed the window on accident and lost it, so I have many more questions for the project but this is my first hurdle.
     
  2. john monks

    john monks

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    Mar 9, 2012
    What do you mean by "volume boos", "output signal", and "raise in amplitude"?
     
  3. Truckdrivingfool

    Truckdrivingfool

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    Sep 30, 2012
    The piezo output works to drive my guitar amp but by itself it's too weak/quiet. I've made and tried the amp from this schematic which made it louder(volume boost) but it's too much amp for a preamp @ 9v. I thought of just lowering the voltage so I started trying to model it in LTSpice. I had to look back at this tutorial to remember how to run the simulation in LTSpice. Which now has me thinking of just using the circuit from the tutorial as it is more adjustable (especially since it's explained right there for me) and I think will have less distortion to it.

    Below is what I'm looking at in LT

    [​IMG]

    The top schematic is the one I've made and used the bottom one from the tutorial. The input signal is the 500mv signal along the bottom. The output signal which is what will be coming out the guitar jack going to the actual guitar amp is the one across the top peaking at 7.2v and 6.3 respectively. The raise in amplitude is the difference in the waveform spread, ie. the top one the input swings from .5v to -.5 and the output swings from 0 to 7.2v.

    So I'm trying to figure out if the extra volume to my ears comes from the higher output voltage or the gain/raise in amplitude/gain. If it's the higher voltage then both amps will be roughly the same, if it's the amplitude then the way I have it set, they should be different and the bottom one will be better suited for my purpose.
     

    Attached Files:

  4. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    I think you've got mixed up terms.

    Your raise in amplitude is commonly called gain.
    gain is the ratio of output voltage/input voltage.

    Amplitude is the difference between max. and min. of the same signal. For your sine waves it is the difference between the peak and the trough of the signal. Both input and output voltage have an amplitude, albeit a different one.

    The output voltage of an amplifier is Vout= gain*vin.
    As long as the supply voltage (minus some internal voltage drop) is high enough to allow correct operation of the amplifier.
    The output volume is controlled by changing the gain of the amplifier (or equivalently reducing input voltage while gain is fixed - a potentiometer can be used for that). Reducing the operating voltage of the amplifier will not reduce the output volume as long as the amplifier is still in its linear operating range. At some point, the amplifier will stop operating linearly and distortions (clipping etc.) will appear at the output.
    Vice versa it does not help to increase the operating voltage to achiev a higher output volume. You need to increase the gain, too.

    In your circuit, you can adjust gain by changing R2. Increasing R2 will increase the gain, making R2 smaller will reduce the gain.
    Or you keep R2 as it is and put a voltage divider (potentiometer) at the input of the amplifier. This gives you adjustability while keeping the amplifier at its designed operating point.
     
  5. duke37

    duke37

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    Jan 9, 2011
    R5 will contol the gain of the seond circuit.
     
  6. Truckdrivingfool

    Truckdrivingfool

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    Sep 30, 2012
    Thanks Duke, the tutorial that circuit is from demonstrated that.I have a grasp of controlling gain with that one but not why it works but I'll leave that for another thread.

    Let me say I'm obviously an amateur layman at electronics and I'm hoping posting pics of what I'm looking at helps you guys understand what I'm asking. If it's a hindrance let me know.

    Harald thanks to you too. Referencing the above posted amps. In the top one since the amplitude of the input wave is 1v peak to peak and the amplitude of the output wave is just over 7v the gain is 7 correct? In the bottom one same input wave but the output is roughly 4.3v to 6.3 peak to peak so it would have a gain of 2 correct? Thus even though the peak output voltages are close between the two, the 2nd should create less volume because it has less gain correct?

    With your pointing out adjusting gain of the first circuit I played with that and noticed that changing either R2 or R3 effects the gain, is this a voltage divider or just that it acts as one or similarly?
    [​IMG]

    Also is it considered as clipping the way that the output voltage cuts off at zero? Wouldn't this be distortion that in an audio use I might want to get away from especially in a preamp where I want to closely maintain the original signal.
     

    Attached Files:

  7. BobK

    BobK

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    Jan 5, 2010
    Yes, that amplifier is clipping. You need to reduce the gain.

    Bob
     
  8. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Yes, your simulation in post #6 is clipping. Part of the problem is that the transistor isn't biased properly.

    For the widest output swing without distortion, your transistor should be biased so that when no input signal is present, its collector voltage is about half the supply voltage. In that circuit, the output waveform only extends up to 3V so the collector voltage is far too low.

    You can change that by increasing R3, which reduces the base current so the transistor doesn't conduct so heavily, and its collector voltage will rise because of R2 which is pulling it upwards.

    That design is not as good as the bottom diagram in post #3. The latter diagram uses four resistors to control the biasing of the transistor, and it even separates the DC bias from the gain setting. It will give better and much more consistent performance, because variations in gain (due to component differences, temperature variations and aging) will not affect the DC conditions much. With the diagram in post #6, all of these variations will cause significant changes in the quiescent (idle) collector voltage.
     
  9. Truckdrivingfool

    Truckdrivingfool

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    Sep 30, 2012
    Thanks Bob and Kris

    I don't know why I had it in my head that clipping was only on the +ve side but now I see it's anytime the wave distorts due to a voltage barrier.

    With the info Kris gave me I played for a while at adjusting the collector voltage up to the 4.5 range and tinkering the gain. I now see how the biasing works and that as I brought the gain up it would effect the bias to the base. After much back and forth adjusting I got the gain up to 7.4 without clipping.(at least @ 200Hz).
    If this is a fairly common situation (I'm guessing it is) can you guys point me to some math that will save me the process of sitting here changing values back and forth to get it optimized?

    For now since it's already built and in the guitar to a certain degree I think I'll just switch a few resistors and see if I'm happy with the sound. But in the future I may try the 2nd circuit.

    That said, on the 2nd circuit I don't know if it's the design or something I'm missing about the gain adjustment but I can't get very much gain from the circuit.

    I've tried adjusting R5, C1, and C2 and unless I'm figuring it wrong I can only get a gain of about 1.4.

    I'd take some math on this too if it's not too complex.

    Lastly I'd like to say thanks to everyone again I appreciate the time you guys spend to read posts and share your knowledge.
     
  10. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Right, and that's a good way to define clipping. Any time the signal hits a voltage limit, the part of the signal that "wants" to go past the limit is chopped off - "clipped".

    With the circuit that has the emitter connected straight to the 0V rail, the "right" amount of base bias current depends on the current gain of the transistor. This is a parameter that varies over a range of at least 2:1 and as much as 5:1 from one part to the next. When you simulate the circuit, the simulator assumes a typical value for the transistor in the simulation. You can fine-tune the base bias for the simulated transistor, but when you use a real transistor, the quiescent collector voltage will probably be significantly different.

    This is why the other circuit, with the "emitter degeneration" resistor, is much better. With the emitter resistor, the circuit becomes much less dependent on the current gain of the actual transistor used. This is standard design practice - the circuit should not be dependent on any characteristic that is not well controlled.

    With the grounded emitter circuit, you can only calculate ideal values if you know the exact current gain of the transistor. With the emitter resistor design, you can estimate values by calculation. I use rules of thumb to choose the emitter and collector voltages - emitter voltage 5~15% of supply voltage and collector voltage roughly half way between VCC and emitter voltage, choose emitter and collector resistors for these values and adjust the base voltage divider for proper bias using a simulator.

    The signal gain of the emitter resistor circuit is determined by the resistor in series with the capacitor in the emitter resistor bypass circuit. Lower values give higher gain. The capacitor value may need to be increased as well. If in doubt, try a large value like 100 uF. You can get much more than a gain of 1.4 from the circuit.

    I assume you're driving this circuit from your guitar pickup? If so, when you simulate the circuit, you need to include the series resistance of the pickup. The signal source has a field for "series resistance". To simulate a direct connection to the pickup, the series resistance should be something like 10 kilohms.
     
  11. Truckdrivingfool

    Truckdrivingfool

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    Sep 30, 2012
    The gain problem was my error, after your response I looked back at the tutorial and noticed that the caps should have been 1uf not the .1uf I put in.

    In the quote is VCC the same as the supply voltage? I tried to adjust the emitter and collector voltages as you said but couldn't seem to get them to separate as they always stayed proportional no matter how I adjusted them. I was able however to adjust to get rid of this clipping. (it's still considered clipping right?) Green is the collector, Blue is the emitter.
    [​IMG]

    Also glad you mentioned the series resistance as I didn't have that in there. Yes a magnetic pickup will range from 5k which might be a bit weak on the output and need a preamp to 12k which definitely doesn't need any help on the output. However I'm looking at these circuits to use with a piezo disc for a pickup which I can't get an ohms reading from and also for use with a small condenser mic. Each signal will get a preamp, then mixed via a pot, and sent out to the amp.
     

    Attached Files:

  12. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    Yes, VCC is the positive supply voltage.

    Your emitter resistor is too high. Start by choosing a quiescent collector current. For example, 1 mA. This current flows in the collector and emitter resistors. Actually the emitter resistor also sees the base current, but that is relatively small, so you can pretend that the emitter and collector currents are equal.

    With a 9V supply, let's make the emitter voltage 0.9V. The collector voltage needs to be roughly half way between VCC and VE; call it 5V.

    Now you can calculate the collector and emitter resistors. The collector resistor carries 1 mA and has 4V across it, so it's 4k. The emitter resistor carries 1 mA and has 0.9V across it, so it's 900 ohms.

    Now you need to adjust your base resistors to get roughly 5V on the collector. You can calculate approximate values because the base voltage needs to be about 0.7V higher than the emitter voltage, i.e. 1.6V. The top resistor will have 7.4V across it and the bottom one will have 1.6V so their ratio is 7.4/1.6 which is about 4.6. So you could use 47k and 10k and adjust them if necessary. But I would use higher values than that. Especially if you're feeding the amplifier from a medium-impedance source!

    Your small-signal transistor has a current gain of at least 100 so the base current will be less than 10 µA. The current through the base resistor voltage divider should be a few times higher than this, so that the base current doesn't affect it much - say 50 µA. With 9V across them, the total resistance should be something like 180 kilohms. 150k and 33k work out about right.

    Piezo discs are capacitors so you won't measure any resistance. You can measure the AC resistance by feeding a sound into it, and loading it with two different resistances alternately and measuring the difference in voltage in the two cases, then calculating the piezo's impedance using Ohm's Law. The piezo's impedance is highly dependent on frequency - it's lower at higher frequencies.

    Generally piezos are used with relatively high impedance inputs. You can increase the impedance of that amplifier using an emitter follower stage at the start.
     
  13. Truckdrivingfool

    Truckdrivingfool

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    Sep 30, 2012
    Is it 1mA just for example to make the math easy, or is there something that made you pick it?
    Understood from your other post

    Where does the 0.7v value come from?

    What qualifies as a medium impedance source?

    Where is the gain found on the datasheet?
    Where does the 50uA and how are getting the need for 180K?
    Understood
    Since you mentioned it I started to look at this but want to get a better grip everything up to this point before I complicate it any more.
     
  14. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Both. 1 mA is a reasonable collector current for an audio preamplifier with medium-impedance output - it yields a collector resistor of 4k which is typical for an audio preamplifier stage.
    The 0.7V value comes from the base-emitter voltage of a small-signal bipolar junction transistor with moderate base bias. It's only approximate.

    Google bipolar transistor base-emitter voltage. Most of the hits describe the construction of a transistor, or define its behaviour using heavy mathematics, but there's one that's useful: http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/info/comp/active/BiPolar/bpcur.html

    As you can see from the second graph on that page, the base-emitter voltage is affected by the base current, but it's generally around 0.7V (at the relatively low currents we're talking about). (It's also affected by temperature.) So 0.7V is an approximation that we can use to calculate rough component values in a low-current circuit.
    Well, everything is relative I suppose. In the context of audio signals, an impedance around 10k would be "medium".
    Gain is called hFE (with the "FE" as a subscript). Download a few data sheets for small-signal bipolar junction transistors such as 2N3904, BC547B, 2SC1815 and see how the gain is specified. The data sheet usually defines limits for current gain in the tabular section, and has a graph of typical values in the typical characteristics section. Both of these are useful.

    Current gain is not a well-controlled parameter. It varies with collector current. It also varies with temperature. This is why the simple biasing method using a single resistor to set the base current does not give a well-defined collector voltage. That's why the emitter degeneration resistor is such a good idea.
    I calculated that the base current would be about 10 µA. This means that about 10 µA is being sucked out of the voltage divider. For the voltage divider to keep a steady voltage, the current through it should be significantly more than the load current. I chose five times higher, i.e. 50 µA. This will make the voltage divider's output voltage relatively immune to the effect of the transistor's base current.

    With 9V across the voltage divider and 50 µA flowing in it, Ohm's Law (R = V / I) tells us that the total resistance should be about 180k. That figure is just approximate.
     
    Last edited: Sep 21, 2013
  15. Truckdrivingfool

    Truckdrivingfool

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    Sep 30, 2012
    Thanks again, I got my resistor assortment and went to tinkering. Using the circuit (since I already had it built) with bit of trial and error switching resistors to adjust the gain I got the outputs (volume wise) of the piezo and mic roughly the same. I have to say it works great and gives a nice clean sound without (as much) feedback. Both were problems before.

    Going forward I'll probably keep the piezo but get rid of the mic pickup in favor of a magnetic one. I'll probably also redo the preamp in favor of the "emitter degeneration" resistor design as (atleast using it in the simulator) I think its easier to adjust the gain without having it go wonky. I think I now understand what is meant by circuit stability after playing with the two in the simulator.:D

    If you're interested there's a quick video of the project here.
     
    Last edited: Sep 24, 2013
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