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Fw: Re: [softrock40] mixers

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  • n7ve
    This is why this is an integrating detector. Each RF 1/4 cycle sample contributes a bit of charge to the detection cap. Over time, the charge builds up to
    Message 1 of 3 , Mar 28, 2006
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      This is why this is an "integrating" detector. Each RF 1/4 cycle
      sample contributes a bit of charge to the detection cap. Over time,
      the charge builds up to the average of the 1/4 RF input.

      It does not matter if there are many shorter pulses (28 MHz) for fewer
      longer pulses (7 MHz); the amount of total energy integrated onto the
      detection caps per unit time will be roughly the same. Thus thinking
      of this as an "IQSD" (Integrating Quadrature Sampling Detector) will
      help you understand the detector a bit more clearly.

      The detector frequency response is a function of the RF input
      impedance (50 ohms as modified by various input transformers and
      additional series R in the circuit) and the detection caps. This
      forms a simple R/C low pass filter. The actual sampling rate (7 MHz
      or 28 MHz) will not change the time constant of this R/C low pass filter.

      Also, understand the averaging ("integrating") character of this
      detector and it becomes clear why averaging over 1/4 cycle of an RF
      sinewave produces a higher peak (0.9x) than over a longer period such
      as a 1/2 RF cycle (0.707x) and makes it obvious why averaging over 1
      full RF cycle will produce zero output. It also explains why the even
      order harmonics are suppressed (zero average) and the odd order
      harmonics produce output.

      Also, if you are not concerned about frequency roll off of the
      detector caps, a normal op-amp connected differentially across the
      detection caps of a detector provides a high impedance for the "+"
      input, and a short circuit across the "-" input. I think of these two
      modes as duals of each other. One treats the input signal as an open
      circuit voltage source, the other as a short circuit current source.
      It is a very unbalanced situation, but using the differential signal
      from two sources does effectively reduce the noise figure of the
      op-amp by a factor of two, which is quite useful if you are going for
      best sensitivity (half the output noise for the same net voltage gain).

      Instrumentation amps (INAs) have great input impedance, but they have
      to be run at a higher gain to realize a low input noise. INAs like to
      be run at 500x to 1000x, and all the gain that we really need for a
      post detector pre-amp is about 50x. Extra gain just makes the pre-amp
      more susceptible to large signal overload, which nobody wants.

      - Dan, N7VE


      --- In softrock40@yahoogroups.com, FRANCIS CARCIA <carcia@...> wrote:
      >
      > When everything is sitting at 2.5 volts no signal in the voltage on
      the transformer secondary effects the direction for each sample. It
      will result in a source or sink and the z of that signal should be
      constant. The time constant is fixed but the sample time goes down
      when the frequency goes up. This means it will take more samples to
      charge the time constant to a rail. This lag prevents the rc to peak
      charge. right???
      >
      > windy10605@... wrote: If the charge time constant is always
      much less than the discharge time constant, the capacitors will charge
      up quickly to the 2.5V bias level. The time constant after the switch
      is fixed but the ....."time constant" into the capacitor, when
      including switch on/off characteristics, is not. In your example you
      will have 10 times 1/10 the sample period which should charge up to
      Vcc/2 for proper bias. The audio signal samples being read are 1/x00th
      of the RF frequency. Using a very high impedance Op Amp would be the
      best alternative and using a differential input would insure the bias
      levels presented are equal. Hey, that's what SDR-1000 does. Dang,
      Gerald Youngblood's FlexRadio group is sharp !
      >
      > 73 Kees K5BCQ
      >
      >
      >
      > carcia@...>
      > To: softrock40@yahoogroups.com
      > Date: Tue, 28 Mar 2006 08:58:53 -0800 (PST)
      > Subject: Re: [softrock40] mixers
      > Message-ID: <20060328165854.3411.qmail@...>
      >
      > nother simple mind at work....The time constant is fixed after the
      switch, pick a value.
      > the voltage on the switch is 5 volts biased at 2.5 volts. A 5 volt
      peak to peak voltage signal would just fit through the switch if you
      ignor losses. Say at the lowest operating frequency a 5 Vp-p signal
      charges each rc to 5 volts peak. Now you increase the frequency by ten
      times. a 5 volt peak signal will not charge the same time constant to
      the peak value with 1/10 the sample period. Could this be part of the
      problem?
      >
      > windy10605@... wrote:
      > Just thinking out loud again......................
      >
      > Ref my earlier notes ....seems like you need to allow an RC time
      > constant between the capacitor and the Op amp input impedance and the
      > capacitor and the charge impedance to allow the capacitor to charge
      up to
      > --and stay-- at the DC bias level .....or you drive the Op Amp output to
      > the rail since the reference DC bias level is fixed at Vcc/2. This works
      > fine at the lower frequencies but at higher frequencies this is more
      > difficult to do since the switch quadrature output time % drops a little
      > because the switch rise/fall times are "fixed". I think higher input
      > impedance Op amp configuration is worth a try on the SR-4, SR6 when used
      > above 18Mhz ? As a test, maybe a SR-7 Op Amp configuration using a 1K
      > input resistor and 50K feedback resistor (so you still have --some--
      > gain).
      >
      > 73 Kees K5BCQ
      >
      >
      >
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