optimal ratio

Discussion in 'General Electronics Chat' started by mik3ca, Feb 11, 2008.

  1. mik3ca

    Thread Starter Active Member

    Feb 11, 2007

    I was just curious, for this amplifier, how extreme can I go with R1? (what is the maximum value of R1?)
    I'm curious. Would omitting R2 produce any negative effects in terms of input sensitivity?

    My Q1 in this case is a Pn3563 with a measured HFE of about 130.

    I find that a high R1 increases distortion, and a low R1 reduces sensitivity.
    Please tell me what you suggest for R1 and why.
  2. SgtWookie


    Jul 17, 2007
    It would be helpful to those trying to help you if instead of showing node numbers, you displayed the values of the components you've selected thus far, including the frequency range and amplitude for VS.
  3. beenthere

    Retired Moderator

    Apr 20, 2004
    I have to agree with SgtWookie that no component values makes an answer quite difficult.

    In addition to that, the schematic is that of a textbook AC amplifier. Many component values are selected with specific goals in mind, like gain or operating frequency.
  4. Audioguru


    Dec 20, 2007
    The current in R1 should be about 1/10th the transistor's collector current.
    Then the current in R2 is the current in R1 minus the transistor's base current.

    The base voltage is determined by the ratio of R1 and R2. It must be selected so that the collector of the transistor has the correct operating voltage to produce the amount of voltage swing that you need. The collector voltage should be about half the supply voltage when the bias voltage is correct. The collector voltage can be lower to reduce distortion but then the max output level is reduced.

    The emitter resistors should total about 1/5th the collector resistor.

    The input impedance of the transistor and its bias resistors is causing an attenuator if the value of RS is high.

    RE1 is not bypassed by a capacitor so it applies negative feedback which reduces gain and reduces distortion.
    The voltage gain (not inlcuding the attenuation caused by RS) is determined by the ratio of the collector resistor in parallel with its load resistor to the value of RE1 plus the transistor's internal emitter resistance.
  5. mik3ca

    Thread Starter Active Member

    Feb 11, 2007
    So if I grounded node 2 and forget about the input resistor that would be better (I think).

    If I were to use 1.5 Megaohms for R1, what you are saying is that RC should be 150K even though the HFE is 130?

    I always thought that R1 = RC * HFE.

    I guess in my books, R1 = R2 since I'm grounding the input capacitor.

    I'm a little lost here.

    based on what I learned on voltage dividers, it would make sense for me to make RL = RC.

    Why only 1/5th?
    Why not 1/10th for 10x gain?
    Is there a property of a transistor I need to be educated on?

    It would be nice if I can increase gain and reduce distortion.

    If I can add large valued inductors and maybe capacitors across them, and know exactly where to put them and why, then maybe I can increase sensitivity and further reduce distortion in the amplifier. Maybe I should put a tank circuit in series with R1.
  6. Audioguru


    Dec 20, 2007
    You do not want to ground Node 2 because it is the circuit's input.
    The input resistor is needed for testing this circuit because it is the resistance of the source signal.

    Pick RL first. Then RC must be half or less to avoid loss of output level and loss of gain. Your RC of 150k is extremely high so the load should not be less than 300k which is also extremely high.

    The current in R1 is the collector current /hFE, plus the divider's current. The value of R1 is determined by how much voltage is across it. Then Ohm's Law calculates its value.
    With a 9V supply, an hFE of 130 and a collector resistor of 150k then the value of 1.5M for R1 is twice as high as it should be. Therefore the bias voltage will be wrong if the transistor has low hFE (its minimum is only 20).

    The input capacitor is the input that you don't want grounded.
    If R1 = R2 then the base voltage is about half the supply voltage which is too high so the transistor would be saturated.

    1) I used your very high value resistors then selected 9V for the supply voltage and 4V for the collector voltage.
    2) I selected the RL to be twice the value of RC so that the transistor is not loaded down too much.
    3) I selected the total of the REs to be 1/5th of the collector resistor.
    4) I selected 1k for RE1 so that the voltage gain will be about 50 (150k//300k)/(RE1 + internal RE)= 50.
    5) I calculated the emitter voltage.
    6) I saw a Vbe of 0.63V on the datasheet for the transistor at the low current and added it to the emitter voltage to determine the base voltage.
    7) I calculated the current in R1, the base current then the value of R2.

    I simulated it.
    I couldn't find a transistor with an hFE of only 130 so I used a 2N3904 with a typical hFE of 200 but it doesn't matter much since this circuit has plenty of negative feedback.
    The collector voltage turns out to be exactly 4.0V.
    The output clips the bottom of the waveform when the input is 51mv peak and the top of the waveform is very distorted because the signal is too high.
    The output is trying to be 5.1V p-p but the distortion reduces it to 5.0V p-p.

    Then the output voltage swing is reduced to half and the voltage gain is also reduced to half.
    I try to make the load resistance 10 times higher than the collector resistor so the loss is low.

    Um, everything.

    It is easy, just use more transistors which increases the gain then add negative feedback to decrease the distortion. Opamps have the circuit already designed and built.

    You sketched an audio amplifier. you don't use a tuned circuit (tank) in an audio amplifier.

    Your super-regen radio does not have an RF amplifier, an IF amplifier nor an FM detector. Therfore is has poor sensitivity and high distortion.
  7. mik3ca

    Thread Starter Active Member

    Feb 11, 2007
    The reason why I want to is because I'm going to turn it into a common-base amplifier.

    My circuit tests seem to agree with that. I think I will stick with 30K for RL and 15K for RC.

    I actually didn't use this value in my circuit. I'm only using it because I was trying to follow your math.

    I'm trying to figure your example out:

    9 / 150000 = 60uA (collector current)
    60uA / hFE = 0.46uA
    9 / 1500000 = 6uA (base current. = divider current? I didn't determine R2)

    60 + 6 = 6uA + 0.46 = 6.46uA
    9 / 6.46uA = about 1.4M

    I'm going to need more equations. I know that current = voltage / resistance, and resistance = voltage / current, but I don't understand why 1.5M is double the correct value.

    What voltage should appear at the base compared to the collector, and what is the best way to calculate the values of the resistors?

    How were you able to just select 4V for the collector voltage? Is it because it is close enough to 4.5 which is 1/2?

    I will have to do the same.

    I think that won't give enough gain.

    Seems that higher values can make a difference.
    In my version, my RC is 15K and RL is 30K, and based on the parallel of those resistors alone, the result is 10. With 150K ad RC and 300K as RL, the result is 100. So there must be a way to define the absolute maximum resistances I can use in the above circuit given my transistor hFE is 130 and the transistor is PN3563.

    I forgot how you did that.

    Just how nuts can I go with the transistors. I wonder if someone made a ridiculously sensitive receiver with 100 or more transistors (not bound into an IC either).
  8. Audioguru


    Dec 20, 2007
    The collector resistor does not have 9V across it. It has 5V across it. So its current is 33.3uA.

    R1 does not have 9V across it. It has 9V minus the base voltage across it.

    The collector current is 33.3uA. The hFE is 130 so the base current is 0.26uA. the current in R1 should be about 10 times the base current so its current should be 2.6uA.
    R1 has 7.37V across it so its value should be 7.37/2.6u= 2.8M. Sorry, the currents are so low that I lost the decimal point, but it doesn't matter because the base voltage is correct.

    I selected the total of the emitter resistors to be 1/5th the value of the collector resistor. Therefore they have almost the same current as the collector resistor so they have 1/5th the voltage drop, which is 1.0V. The datasheet shows that the base-emitter voltage is 0.63V at this current so the base voltage must be 1.63V.

    It is close to half. I know that if the collector voltage is high then the top of the waveform will be very distorted so I select a collector voltage less than half the supply voltage.

    The AC voltage gain of a common-emitter transistor is the total collector resistance (including its load) divided by (the unbypassed emitter resistance plus the transistor's internal emitter resistance). I selected 1k for RE1. The internal emitter resistance calculates to be about 780 ohms. The 150k collector resistor in parallel with the 300k load gives a total collector resistor of 100k. The voltage gain is 100k/1780= 56 minus 10% because the currents are too low.

    The voltage gain is RC/RE and the current doesn't matter much.

    A cheap radio has no RF transistor, a mixer/oscillator transistor, 2 transistors in the IF amplifier and 3 transistors in the audio amplifier. It has a total of 6 transistors. I had one.
    My friend showed me his new radio that had molded into its cover, "14 Transistors". It worked the same as mine, no better.
    We looked inside. It had 6 transistors used for the radio, one transistor used as the AM detector diode and 7 transistors connected together in a circle but not connected to the radio!

    An old FM IF amplifier IC has about 50 transistors and a voltage gain at 10.7MHz of about 22,000. The RF amplifier has a max voltage gain of about 10 so an FM radio is extremely sensitive.

    The most sensitive FM radios have an input sensitivity of about 1.3uV. Ambient and electronic noise is about the same level and the curvature of the earth stops the VHF signals from going past a certain distance.
  9. mik3ca

    Thread Starter Active Member

    Feb 11, 2007
    how did you calculate 5V?

    How can I calculate base voltage before making a decision of what R2 is?

    This part DOES make sense to me. (Collector current / HFE = Base current)

    How can you get 7.37 volts across R1 when one end of it is tied to your power source of 9V?

    That is what I need help on, math that explains how you determine each voltage.
  10. Audioguru


    Dec 20, 2007
    I selected a collector voltage of 4V. Since the supply is 9V then RC has 5V across it.

    I selected that the total of the two emitter resistors are 1/5th the value of the collector resistor. So they have a total of about 30k.
    I showed you that the 33.3uA collector current flows through the 31k emitter resistors. Then the emitter voltage is 1.0V.
    The datasheet shows a base-emitter voltage of 0.63V at this current so the base voltage must be set to 1.63V by the divider of R1 and R2.

    You must set the base voltage to 1.63V which will be across R2, so R1 has 9V - 1.63V= 7.37V across it. The small base current is subtracted from the current in R2.
  11. mik3ca

    Thread Starter Active Member

    Feb 11, 2007
    So you are telling me that the emitter resistors consume no voltage? It doesn't make sense.

    I'm still looking for simple math on this.
  12. Audioguru


    Dec 20, 2007
    The collector current is 33.3uA because the 150k collector resistor has 5V across it.
    The emitter current is slightly more than 33.3uA.
    The emitter resistors total 31k.
    Ohm's Law calculates the emitter voltage at 33.3uA x 31k= 1.0V.

    So the collector is at +4.0V, the emitter is at +1.0V and the base must be at +1.63V.
    Look at my schematic again.
  13. mik3ca

    Thread Starter Active Member

    Feb 11, 2007
    I seem to be getting more of the math. I'll take that picture of your transistor circuit home with me, and I will see if I can understand it further.

    The math is complex just by looking at it, but I think I can understand it.
  14. Audioguru


    Dec 20, 2007
    The math is simple arithmatic. Multiply or divide. Sometimes you must add. No algebra. No calculus.

    You cannot order a transistor with a certain hFE. It might be 100 or 300. The RF transistor you are using has a range of hFE from only 20 to 200!
    The base bias divider's current is about 10 times the base current so the base voltage is not affected much by the different hFE of each transistor. When the base current is double or half then the voltage at the base is not affected much.

    The emitter resistors create negative feedback. The 30k resistor has a bypass capacitor so only the 1k resistor determines AC gain in the ratio with the collector total resistance. The 30k resistor reduces the DC effect of different hFE and different Vbe of each transistor.

    If there is no emitter resistor then a transistor with a high hFE will be saturated and a transistor with a low hFE will be cutoff. A high supply voltage will saturate the transistor and a low supply voltage will cut it off.
    A transistor with a low Vbe will be saturated and a transistor with a high Vbe will be cutoff.
    The 30k emitter resistance reduces these effects so that any transistor with the same part number will work well.

    The unbypassed emitter resistor creates AC negative feedback. It reduces gain and reduces distortion.