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Fw: Fw: Re: [softrock40] mixers

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  • n7ve
    The time constant on the DC bias is almost a don t care . As soon as the signal stops, the residual charge freezes and becomes a gradually decreasing DC
    Message 1 of 2 , Mar 28, 2006
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      The time constant on the DC bias is almost a "don't care". As soon as
      the signal stops, the residual charge freezes and becomes a gradually
      decreasing DC charge. If the signal was at a 20 KHz offset, it
      disappears immediately when the signal stops as a signal is necessary
      to create the detected 20 KHz AC signal.

      You would need a really large signal to worry about the bleed off time
      constant of any residual DC charge.

      - Dan, N7VE

      --- In softrock40@yahoogroups.com, windy10605@... wrote:
      >
      > My comments were in reference to what is happening to the DC bias
      voltage
      > charge/discharge relative to the capacitor, not the AC signal voltage on
      > top of it .....and how the cycling of the bias voltage, based on RC time
      > constants, could cause problems if it is not allowed to go to --and
      > stay-- at Vcc/2.
      >
      > 73 Kees K5BCQ
      >
      >
      > ----- Forwarded Message -----
      > From: "n7ve" <dan.Tayloe@...>
      > To: softrock40@yahoogroups.com
      > Date: Tue, 28 Mar 2006 19:52:36 -0000
      > Subject: Fw: Re: [softrock40] mixers
      > Message-ID: <e0c464+8hnu@egroups.com>
      >
      > 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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