According to the excellent site of AB4OJ, the Icom 7300 in summary has a 14bit analogue to digital converter with a 1.5 volt peak to peak input (LTC2208-14). I make this to be 91.5uV per quantisation level. There is a 20db gain amplifier (LTC6401-20) ahead of the ADC so best case this would give 9.1uV per quantisation level.

Wikipedia gives for HF a S5 signal to be 3.2uV (rms relative to 50R). Equivalently this is 9.1uV peak to peak, resulting at best in a S5 signal at most only changing the Least Significant Bit of the ADC. Not much can be hoped to demodulate from this surely?

By the same logic, S9 would only move between 15 levels. Again how much can be demodulated from this strong signal?

As the 7300 works very well, clearly I am missing something but I don’t know what. Can anyone fill me in?


2 Answers 2


You are missing that a correctly dithered quantiser is LINEAR, and that there is more then enough uncorrelated noise in the ADC input bandwidth to correctly dither the quantiser.

You are not digitising just the signal you care about, you are digitising a whole bands worth of mostly uncorrelated stuff, and then reducing the bandwidth in the digital domain. The bandwidth reduction strips off noise outside the range of interest, but that noise includes quantisation noise.

Every time you halve the bandwidth you gain 3dB on the signal to (total) noise ratio, so that every time you drop the bandwidth by a factor of 4 you gain an effective extra bit.

If the ADC is running at say 125MHz and you are downconverting to an 8KHz sample rate for the baseband processing then that gains you an effective 8 bits of extra meaningful word length if you do it right, putting the noise floor at about 20 bits or so (The number you want for the ADC is the ENOB, not the output word length).

If the thing basically follows the HPSDR architecture there is probably also a 1:2 (voltage ratio) transformer feeding the preamp, so another 3dB on the voltage there (But obviously no power gain).

  • \$\begingroup\$ @ Dan Mills Nicely explained. \$\endgroup\$ Commented Oct 26, 2018 at 12:49

Thermal Floor (Boltzmann noise, et al) for 1,000 Hertz bandwith is

-174dBm + 30 dB = -144dBm

What the voltage level, across 50 ohms from a standard RF interface, for -144 dBm?

Given 0dBm across 50 ohms is 0.632 volts peakpeak, and -120 dBm is down by 1,000,000 to 0.632 microVolts PP, and -140 dBm is down to 0.0632 uV (or 63 nanoVolts), the extra 4 dB drops the level another factor of 2.5X to 25 nanoVolts PP

Summary: the input signal, for 1,000 Hertz bandwidth signal at 0dB SignalNoiseRatio, is 25 nanoVolts PeakPeak.

Thus some amplification, at RF or in an active-mixer or in the Intermediate Frequency filtering stages, occurs prior to feeding the 14-bit ADC.

Or some signal-processing methods are used. Read on.


If you use a transformer to stepup the signal, converting from 50 ohms to 5,000 ohms impedance and then BUFFER (Unity Voltage Gain) to feed the ADC, you'll have 10X higher voltage or 250 nanoVolts into the ADC.

Still a factor of 40 too low.

You can dither the ADC sample amplitudes (random noise is one ditherer). Then post-process digitally, collapsing the samples from 100,000,000 per second (for example) down to 100,000 per second and pick up a factor of SQRT (1,000) = 31X improvement, which is getting close to the factor of 40x.

[by the way, this averaging, this collapsing of frequency from 100,000,000 to 100,000, is exactly what a MIXER does, in downconverting to a much lower IF frequency.]

Note the original energy level of -144 dBm produces ZERO dB SNR; assume you want 10 dB SNR, which raises the voltages by sqrt(10) or 3.1X, requiring -134 dBm from the Antenna.

Now..... is this what iCOM 7300 uses? You have some of the fundamental limits to muse over.


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