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Question about signal processing devices

  1. Jul 10, 2012 #1
    Since signal processing electronics rely on a constant DC power supply with a supply current that can only be converted into a fluctuating DC current by the switching devices of the circuitry, does an AC input signal coming from a microphone or wireless receiver need to be converted into a fluctuating DC signal first by a full-wave rectifier?
  2. jcsd
  3. Jul 10, 2012 #2


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    You use an A-to-D converter to convert analog signals to digital signals.
  4. Jul 10, 2012 #3
    What about for analog signal processing, such as when an analog signal has to be amplified? The amplifiers are supplied a constant DC current from the power supply and the input signal coming from the antenna or microphone is an AC signal.
  5. Jul 10, 2012 #4


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    Obviously. But your OP asked about "switching devices of the circuitry", not about analog processing. In practice, you choose to use analog or digital processing based on the most efficient way to get the job done. Often you use a combination of both (like doing some analog filtering of a signal before converting it to digital and doing some DSP on the signal).
  6. Jul 10, 2012 #5
    So then a rectifier stage or ADC stage can be placed in the device to convert the negative cycle into a positive cycle or logical "0" in digital signalling so that the signal can be sent into the DC supplied electronics.
  7. Jul 10, 2012 #6


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    I'm not sure I'm understanding what you are saying.

    A rectifier usually is used for power conversion, like rectifying the output of a power transformer to convert AC to a mostly DC power source.

    However, a rectifier can also be used in a simple AM radio receiver, as an "envelope detector" to recover the modulating signal.

    An ADC is different from a rectifier. It is used to "digitize" an analog signal -- to get a digital representation of the analog waveform.
  8. Jul 10, 2012 #7
    To keep the question simple, what if an AC signal coming from a radio antenna needs to be amplified? Since the amplifier is using a DC power supply, how will the negative cycle of the signal pass through the amplifier since the DC supply has a fixed polarity?
  9. Jul 10, 2012 #8


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    Ah, now I see what you are asking. You use a DC blocking capacitor between the input AC signal source and the unipolar DC biased amplifier circuit.

    See this thread: https://www.physicsforums.com/showthread.php?t=425371

    http://ecelab.com/circuit-ac-amp-1a.jpg [Broken]
    http://ecelab.com/circuit-ac-amp-1a.jpg [Broken]
    Last edited by a moderator: May 6, 2017
  10. Aug 21, 2012 #9
    So the negative cycle of the signal that is opposite to the DC bias amplifier can be temporarily stored in a capacitor and then quickly discharged to the amplifier but connected in such a way that it enters the circuit with the same polarity as the DC bias source to ensure that the data of the negative cycles is not lost.

    But interestingly wired and wireless transmitters make use of DC biased circuits in their operation so the analog signal may not even have a negative cycle to begin with and simply be a fluctuating sinusoidal DC current so the signal entering the amplifier or any other DC biased transistorized circuit will not need to have this blocking capacitor.

    But is it possible for wireless transmitters to transmit radiation if powered by DC sinusoidal electrical signals?
    Last edited: Aug 21, 2012
  11. Aug 21, 2012 #10
    There is no "losing of negative cycles". The input of the amplifier is biased such that it stays in its high-gain operation region for the entire cycle (amplifier outputs can be different, but let's keep it simple).

    You seem to think that dc and ac signals are mutually exclusive. In reality the total signal is a superposition of an ac and a dc component. For example, assume the output of your antenna is a 1V peak-to-peak sine wave. It is put through a dc blocking cap so it is pure ac. Then, if the input of the amplifier that receives the signal is biased at, say, 6 volts, then the input signal will actually swing between 6.5V and 5.5V. In all cases the input amplifier remains appropriately biased. The negative cycles are amplified exactly the same as the positive cycles. A purely linear amplifier needs no storage capacitance in principle.

    I think you might be confusing "dc biased" with "dc coupled". All linear circuits are dc biased. Biasing means setting voltage or current levels such that the device is working in the desired operating region (usually a region where it provides a high gain). A receiver that is "dc coupled" means there is no ac coupling capacitor so the dc bias of the signal at the antenna is the same as the dc bias at the input of the amplifier. This obviates the dc blocking capacitor, but is more challenging to design.

    This statement actually makes no sense. dc is a signal at zero frequency, and there is no such thing as a zero frequency sine wave. If you mean can a transmitter transmit radiation if it is power by a dc bias, the answer is yes, and in fact more antennas require some kind of biasing. Think of the dc bias as a tank of electrons.
  12. Aug 21, 2012 #11
    No, the question does the negative cycle switch on the transistor is where the confusion begins because the DC power supply has a fixed polarity and what the signal does is to simply operate the transistor so that it allows the DC supply current to pass through. But what happens if the polarity of the AC signal is reversed? The DC current cannot possibly reverse direction can it?

    But it is known that when voltages of opposite polarities are connected in series that they will oppose each other and there will simply be a deduction in total voltage. So assume that there is a 6V DC supply and a 1V peak to peak AC signal, the positive cycle of the AC signal will add to the supply voltage and produce a 6.5V output while the negative cycle will deduct from the supply voltage and produce a 5.5V output so the DC current is turned into a signal that fluctuates between 6.5V and 5.5V.

    Another consultant on this site says that DC can also have fluctuations as long as it is between a positive value and 0 and no reversal of polarity. Here is the quote:

    The link to the quote:

    Last edited: Aug 21, 2012
  13. Aug 21, 2012 #12
    I'm not trying to be rude but I'm having a very hard time understanding what you are asking. If the transistor is operating in its correct region then the transistor is "on" during the whole cycle. No, the dc current does not reverse direction. The total current through the device is the superposition of a dc current and an ac current. The dc current stays constant always if the transistor is operating correctly (i.e. if it is a BJT it stays in forward active mode). The ac current responds to the ac voltage at the base of the device. What do you mean polarity of ac signal reversed? How does the biasing of the device change during the cycle?

    I believe you may be confusing signal with power supply. Take a BJT configured as a common-emitter amplifier. Assume that the emitter is biased at 5.3V. The collector is biased at 10 V. This is the dc supply voltage. The base is biased at 6V. This is not the supply voltage, it is the base bias. Then the ac signal of 1V means the base varies between 6.5V and 5.5V. The emitter varies between 5.8V and 4.8V. The collector current varies around whatever its dc value is based on the transconductance of the transistor. The transistor is always in forward active region. It never turns off. The input signal never exceeds the dc supply. Does this make sense?

    That is not a standard definition of dc. Most professionals would say a rectified ac signal is mostly dc with a small, residual ac signal superimposed on it. In this case, the signal is treated as dc, and the ac signal is now a second-order effect and can probably be ignore (but sometimes is important).

    Here is a link that explains transistor biasing, it may be helpful to your understanding.

  14. Aug 22, 2012 #13
    But the AC signal is supposed to turn the transistor on and off so that it in turn switches the DC signal on and off to produce a matching but amplified output signal just like if the AC signal is made to operate an electromechanical switch to turn a DC power supply on and off to make the DC current into an AC signal.

    But that maybe incorrect and maybe this is how this works. The transistor circuit is divided into many stages and has resistors, capacitors and inductors connected to it for power and signal regulation. When the AC signal is sent into the amplifier, it modulates the DC current to produce an output signal with a voltage and current that is higher than the input power of the signal. This is why there is a term in amplifier design called power gain which is the output power that the signal will have once it has been amplified.

    The amplifier basically draws power from the DC supply and uses it to increase the power of the signal.
  15. Aug 22, 2012 #14
    This is not how a transistor amplifier works. I think you need to read up on basic single-transistor stages. You are describing a class D amplifier which is a very advanced design (and requires a resonator). Stick with class A amplifiers (i.e. amplifiers that conduct over the whole cycle) until you understand them.

    Yes, the amplifier draws power from the DC supply but only a tiny fraction of that is put into the output signal. I think you are lacking a basic understanding of how transistor amplifiers work. That's ok, but it might be helpful for you to read up on some basics.

    While the dc current is "slightly" modulated to produce the output ac, the transistor always remains biased in its high-gain region of operation. The signal currents are usually orders-of-magnitude smaller than the dc biasing currents. You need to start thinking in terms of superposition and even in terms of different equivilant circuits for dc and ac operation.
  16. Aug 22, 2012 #15
    Yes it is true that the output power of the signal is only a small fraction of the total power of the DC supply. The thread originator has built a BJT audio amplifier before with a power gain of 10, and used it to amplify the signal of an MP3 player before connecting it to a 10W speaker. The input power of the player's signal is 1W while the output is 10W but the total amount of power delivered by the DC power supply is 70W so only a small amount of the supply power was used by the output signal.

    But the problem is that the understanding of the thread originator is based on mathematical network analysis and the mathematical equations were used to determine the right parts to assemble into an amplifier but the thread originator does not understand the semiconductor physics and electrical phenomena that make this combination of parts work.

    The link: http://www.electronics-tutorials.ws/blog/biasing-transistor-tutorial.html, is mostly about the network analysis for the circuits which the thread originator already understands.
    Last edited: Aug 22, 2012
  17. Mar 7, 2013 #16
    Since the signals being sent to the amplifier may be continuously variable saw-tooth or sinusoidal signals, the instant that the voltage and current drops to 0 is so short that this off time is negligible. So it is correct to say that the Class A amplifier conducts current almost the entire time it is in operation. The other classes of amplifiers such as the classes B, C, and D use various electronic components to transition into an off state for a given period of time. But because the other classes of amplifier are off for a given period of time, part of the signal is chopped off and this causes the signal to become distorted, but these classes of amplifier have the advantage of higher energy efficiencies because they and the loads that they power do not use energy during the times that they are off so these amplifiers are best suited for applications where high energy efficiency is more important than signal quality such as battery powered radios where lower energy use means that the device can operate for longer periods of time before the battery fully discharges.

    And the loading of the negative cycle of an AC signal to a DC biased amplifier is not a problem because the negative AC running in an opposite direction to the positive DC will simply cause a series opposing effect where the power of the AC signal will deduct from the power of the DC bias. Even if the AC signal is causing the DC voltage and current to fluctuate from its peak power level to a lower power level where the degree of fluctuation is equivalent to the input power of the AC signal, what is important is that the drive voltage and current at the output is higher than the input voltage and current of the signal and is powerful enough to drive the load that it is intended to operate.
  18. Mar 7, 2013 #17


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  19. Mar 8, 2013 #18
    Many of the posts including this last one have answered the questions. Thank you for all of the consultants who have replied to this thread.
  20. Mar 8, 2013 #19
    I would like to reference the diagram that Berkman Posted as a simple way to describe what I think you are having trouble with: The Input Capacitor Ci on the AC Signal input - decouples the AC signal from Ground or the voltage reference that is also used for the Vcc, and the other side feeds the amplifier. In fact - I would prefer to not use the term "AC" - just call it the SIGNAL or input signal - because it really is unknown.

    So now the Signal on the other side of the input capacitor is now biased between 0 and the + Vcc by the 2 100K Resistors.

    The Digital Case ( when you say switching) would still - decouple the Signal from the processors voltage reference - but you OP is not clear if you are talking about Digital Signal Processing ( what I tend to assume when I hear Signal Processing -not a great assumption) but you mention "Switching" - in the analog world - signals are not really switched ( State A or State B) -- it is a continuity between Limit A and Limit B.
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