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voltage divider transistor biasing...

Discussion in 'General Electronics Discussion' started by Mor, Jan 16, 2014.

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  1. Mor

    Mor

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    Dec 2, 2011
    i am intent on building a simple heterodyne bat detector based on this one by Bertrik Sikken (http://bertrik.sikken.nl/bat/simphet.htm).

    [​IMG]

    But i am using a 9V battery, instead of the 6V he has. Looking at the circuit, and reading his article, i realised he was biasing the transistor, and that i would have to redesign for 9V. This is new for me so, have i got this right?

    deciding on 1mA for the max Ic, and 1k emitter resistor, i get a max voltage drop across the collector of 8V, and therefore Rc=8k, and Qpoint of 4V and 0.5mA. From there, i am looking to get 0.5mA/Beta (500 or so depending on manufacturer for the BC550C) for the base current of the transistor, giving me 1uA. so far so good (i think...).

    for R2, the voltage drop must be the drop across Re + Vbe, which gives 1.7V, and the current through it must be AT LEAST 10xthe base current. (if it has to be AT LEAST 10x, can it therefore be lot more? I will come back to this.)
    with 10xIb i get R2 = 180k, and R1 = 7.3/(10xIb + Ib) = 691k

    I was looking at these values, and thought they were quite high for R1 and R2, then i remembered something about Stability factor, found the equation, and got a value of about 113 - not good. To reduce this, you can increase the value of Re, which in turn reduces the possible voltage swing, or increase the thevenin value for the divider R1 R2, but that increases the current and therefore decreases the life of the battery.

    assuming i have got everything right so far, this is my first question: can the current through R1 and R2 be not 10xIb, but 100xIb? Everyone always uses 10x in every example I've looked at. Doing the calculations again, but with 100xIb instead of 10x, I get R1= 75k and R2=18k. The stability factor then becomes a much better 15!

    Have I got this right?

    I have one other question about the circuit. The parts circled in green (controls the multiplier frequency) and blue (low pass filter) are both related to frequency. Does this mean they are essentially voltage independent? - can i leave their values as they are even though i am using 9V instead of 6V? and if not, what do I need to work out. I afraid my understanding of the TCA440 is abstract at best, and i cannot fathom the datasheet.


    BC550C transistor
    TCA440 am reciever
    marked circuit
    where the circuit came from by Bertrik Sikken
     
    Last edited: Jan 16, 2014
  2. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    If R1 is 75K, R2 is 18K, Rc 8K and Re 2K then you will have 4.5V on the collector. If this is what you want then you are correct. I chose Re for you because I couldn't see it anywhere. Base current is 1.17uA so as long as you keep the same transistor which has enough gain then it should work because you have increased the base current from the old design. The only other issue you may have is the input capacitance was calculated for the old resistors and produces a HP filter with the bottom base resistor. You may have to modify this if the too higher frequencies getting through enough. The cut of frequency with the new components and old 10nF is 884.19Hz. the old cut off was 589.46Hz
    Adam
     
  3. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    I don't know what you mean by a maximum "voltage drop across the collector" and you can't get 8V from a 6V source.
    Base current is not a critical number in the calculations. Current gain for an individual component type can vary over a 2:1 ratio or even wider, and is affected by temperature as well. As long as the current through R1 and R2 is "much" higher than Ib, you can neglect Ib, especially with a high-gain transistor like the BC550C.
    Yes. Increasing the R1,R2 current will make the voltage divider more stable with respect to variations in Ib, as you say later. But it reduces the input impedance of the circuit, which can be undesirable. Actually, with the decoupled emitter design you have there, the small-signal base-emitter impedance will dominate the circuit's input impedance anyway.
    OK.
    I don't know how "stability factor" is defined, but you're right that reducing R1 and R2 will make the operating conditions more stable, because the base voltage will be less affected by the base current (which is somewhat variable).
    The parts marked in green are a custom change that the designer has made to the circuit. According to the data sheet, pin 6 is supposed to be connected to a tuned transformer that feeds the next stage (pin 5). I don't see how varying the resistance from pin 6 to VCC will affect the tuned frequency, since there is no tuned circuit in the design you are using. But the simple answer is to try it; if you can't tune over the full range you need, or if the tuning range is wider than it should be, change the resistor and/or the potentiometer accordingly.

    The blue components don't need to be changed.

    Thanks for the links to the data sheets. It's common courtesy, but not so common :)

    BTW here is how I would do the full calculations for the amplifier stage.

    I would set Ve around VCC/10, i.e. 0.6V. You suggested 1.0V so let's split the difference. Ve = 0.8V.

    Let's keep Re as 1k. So Ice will be 0.8 mA.

    With 0.8V on the emitter and a Vce(sat) of around 0.2V, the collector will pull down to about 1V. So the collector voltage swing is from 6V to 1V. You want to bias the transistor so its collector is near the centre of that range, which is 3.5V. This leaves 2.5V across the collector resistor, at 0.8 mA, so Rc should be 3.125 kilohms. 3k0 or 3k3 will be close enough.

    The BC550C has a current gain between 420 and 800 at low collector current. Let's assume 600. The actual current drawn by the base from the base bias transistors will be around 1.3 µA which is scarcely worth including in the calculations.

    R1 and R2 need to be chosen to set the base voltage at 0.8V plus the Vbe voltage. For a given component type number, Vbe (for a given collector current) varies from manufacturer to manufacturer, batch to batch, and component to component, and varies with temperature, so it's not an exact value.

    Data sheets for the BC550C give typical Vbe values for Ic = 1 mA, at room temperature, of 0.65V Fairchild) and 0.625V (ON Semi). NXP don't have a graph for this parameter. So let's assume Vbe will be 0.64V.

    Therefore Vb needs to be 1.44V. With VCC = 6V, the ratio R1/R2 needs to be about 4.56 / 1.44 or 3.1667.

    Regarding the ratio between R1,2 current and base current, let's go half way between 1 and 2 orders of magnitude - that's 1.5 orders of magnitude, or about 32 times. 32 x 1.333 is about 42 µA which gives a total resistance (R1+R2) of about 142 kilohms.

    So we need two resistors with a ratio of about 3.1667 that add up to about 142k.
    68k, 22k
    120k, 39k
    150k, 47k

    Any of those options will be fine. I would go for the first one, since the stage's input impedance is already around 10k due to the small-signal base resistance, so adding 17k in parallel with it won't make much difference.

    You can do a worst case analysis to see how far the quiescent collector voltage will vary with worst case values for resistances and gain (but useful Vbe limits aren't specified in data sheets). As a rule of thumb though, with more than 10% of the supply voltage across the emitter resistor, it will be fairly resistant to real-world variations.
     
  4. Mor

    Mor

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    Dec 2, 2011
    sorry - perhaps i didn't make myself clear. I am redesigning this circuit from a 6V source to a 9V source, and by voltage drop across the collector, I meant Rc when Vce = 0, given that the drop across Re is 1V.

    But i will try your calculations (exchanging 6V for 9V) and see how I get on. I think I'm on the right track.

    no, i don't get it either - but i simply don't have the equipment to test the circuit, other than to use the corresponding 40kHz transmitter i have, but of course, that won't help in testing much above or below that level.

    Thanks for you help.
     
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Sorry, my mistake. You were clear; I didn't read properly.
    Cool :)
     
  6. Arouse1973

    Arouse1973 Adam

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    You will need a 3.7K I think for 9V.
     
  7. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    The new suggested design should work ok but after a closer look there may be some issues you need to be aware of.

    1) the lower frequency cut off has been shifted up to over 25K. Batt sounds range from 20Khz - 80Khz. The first modified design is around 12Khz.. The amplifier should be ok with this as long as the input is high enough in amplitude.

    2) There is a slight gain drop with re in effective mode of 284 versus 296 for the old design. But this should be ok.
    Note:
    The small signal input impedance is determined by hfe*RE so at lower frequencies the gain is not dependant on re if the CE is not in effective mode.


    Thanks
    Adam
     
  8. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Re the change in cutoff frequency: the "cutoff frequency" is not the frequency below which signals will be "cut off"; it's only the -3 dB frequency. The stage has a lot of gain so having a -3 dB frequency that's not outside the designed frequency range is not necessarily a problem at all.
    Not with the decoupled emitter. The circuit's AC input impedance is dominated by the incremental resistance of the base-emitter junction which I estimated at less than 10k. I did this calculation from the slope of the graph of Vbe vs. Ic assuming a gain of 600.
    I can't make sense of that.
     
  9. Arouse1973

    Arouse1973 Adam

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    Kris that is what we call in the UK the cut of frequency -3dB.
     
    Last edited by a moderator: Jan 17, 2014
  10. Arouse1973

    Arouse1973 Adam

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    Effective mode is when the capacitor is so low that re is dominant
     
  11. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    What about the input bandwidth? Any ideas on that?
     
  12. Arouse1973

    Arouse1973 Adam

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    Kris can you please supply me the formula for the input imedance I am very interested in this.
    Thanks Adam
     
  13. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    I calculated an estimate using the slope of the line on the Ic vs. Vbe graph in the Fairchild BC550C data sheet: http://www.fairchildsemi.com/ds/BC/BC546.pdf - Figure 2 on page 2.

    The Y axis is Ic, not Ib, so you have to assume some figure for hfe.

    I looked at the section of the graph between Vbe values of 0.6 and 0.7V. Ic varies from 200 µA to about 4.5 mA over that Vbe range, and the circuit's operating point is around the middle of that range.

    Assuming a current gain of 600, the base current will range from 0.333 µA to 15 µA over that range. That current range, over a voltage range of 0.1V, works out to about 6.8 kilohms.

    This calculation only gives a ballpark figure. The voltage vs. current graph for Vbe isn't linear (the line on the graph is pretty straight, but that's only because the Y axis is a log scale and the X axis scale is linear. So any figure you can give for input impedance is just approximate.

    I think you would get a more useful figure by measurement, using a smaller signal, instead of calculation from a typical characteristics graph over a wide voltage and current range.
     
  14. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    Thanks for that Kris.
     
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