Fw: Re: [softrock40] mixers
- 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 firstname.lastname@example.org, 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
> 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
> To: email@example.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
> 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
> --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--
> 73 Kees K5BCQ
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