what is the topology of this receiver?

what is the topology of this receiver and how does it actually work?especially p1,c14

http://www.qsl.net/va3iul/Homebrew_R...r_PLL-4046.gif

Interesting! If I understand it, and it is past my bed time so my brain is shutting down:

The PLL drives a divide by 4 quadrature generator, the outputs are 90 degrees shifted (like I and Q). IC1 and IC2 are balanced mixers taking one phase as local oscillator and the antenna signal via T1 as the other signal. C14 and P1 are to cancel residual carrier before it passes to the two ceramic IF filters then IF amplifier IC3 which also does the envelope demodulation. T3, T4 and T5 are a simple class AB audio amplifier.

Perhaps someone where it isn't so late at night can describe it better!

Brian.

My problem is with the understanding of the operation of C14 and P1. The mixer outputs should be I/Q signals shifted by 90deg.
In direct conversion, these are normally fed to a set of opamps that select either two of the sidebands. The carrier is supposed to be canceled I think by the I/Q mixer.

Now this design differs a bit in the sense that the output of the I/Q is an IF. Like you say, C14 and P1 are to cancel residual carrier, but since it is an AM receiver, why one should want to suppress the carrier? Or do you mean the carrier out of the PLL that somehow manages to pass to the output of the I/Q?

Also what frequencies do the ceramic filter filters exactly?

The filters are 455KHz standard ceramic filters as commonly used in AM receivers. I do not have my data sheets with me at the moment so I could be wrong about the quadrature divider, I need to check the phase relationship of the VCO outputs to be sure.

If I'm right, the VCO should run at (input frequency +455KHz) * 4 so it would tune 1.82MHz to 9.82MHz.

Brian.

Indeed.
However, it is the components operation that is confusing to me, as said in my previous post.

Receiver circuit is from Elektor 7/2010 UK edition, AM receiver with quadrature mixer.

Thanks for the info!
Unfortunately, I can't find the issue nowhere.
I can only find a small description:
This circuit is for a superheterodyne receiver where the image frequency is suppressed without the use of an input filter. Instead, it uses two NE(SA)612 type mixer ICs that each work 90( out of phase. With a quadrature front-end, the image frequency is rejected and the noise associated with it disappears. In theory, this increases the sensitivity of the receiver by 6 dB.

But since the mixer is quadrature the AM carrier should disappear, shouldn't it?

page 40 http://dfiles.eu/files/zubma3iic

Thanks!
I see what he is doing now with this network. An easier setup (though not standalone) would be to use a pc with I/Q demodulation software. That could cover also other modulation schemes (ssb etc)

The article says:

With a low lF, such as used in Software Defined
Radio, the phase shifting following the mixers
has to cover a relatively wide band, because
the lF frequency is low compared to the lF
bandwidth. This is much easier to achieve
using software rather than a complex phase
shifting RC network. With thisAM receiverthe
lF bandwidth is small compared to the centre
lF frequency of 455 kHz and the maximum
phase error is almost negligible even when a
simple RC network is used.

Does it mean that the values for the RC networks for shifting 455KHz +/- 1-2KHz (audio) do not need to be changed for this low shift?

SDR uses zero IF. Phase shifting at both receivers is done by 7474 D flip-flops and is constant 90° over all bandwidth.

I am talking about the IF not the LO

C43 and C14 also shift the IF outputs +-90 degrees before they are added, leasing to image cancellation. Its like the phasing system for generating SSB.
Frank

I think is a mistake in the Elektor article. The Image Noise in a superhetrodyne receiver can degrade the sensitivity with maximum 3dB, and not 6dB as is mentioned in the article.

I believe that so. After all, all what is canceled, is the opposite sideband, i.e half the signal (since tha AM carrier is suppressed)

What I cannot understand, is how the schematic allows the AM carrier to pass through (because it is an AM RX). It should only allow one of the two sidebands to pass through (since the other is canceled) and the carrier should be suppressed. At list this is how a quadrature mixer works, am I wrong?

How does he receive AM without a bfo in this schematic?

It looks like SSB Receiver..

This receiver topology uses the principle of Image Rejection Mixer.

In standrd AM receivers, the image frequency (that is 2*IF apart of the receiving frequency) is not always outside the operating band.
So, image can not be rejected by a fixed filter. For this reason, a variable tuned circuit is used before the mixer, (e.g. in the antenna circuit). Some receivers have an additional tuned stage.
Many receivers use tandem variable capacitors for tuning, because tracking is needed between such tuned circuit and the local oscillator.

Using an image rejection mixer, the image is cancelled when the two brances of signal are added, and the "good" frequency is reinforced. P1 and C14 control the amplitude and phase balance in order to maximize the image cancellation.

The improvement is in image rejection and simple tuning, not in sensitivity.

If you want to know more, look for "Image Rejection Mixer", "Image Rejecting Mixer" or "IRM".

Regards

Z

Its the image of the IF component that is cancelled not the modulation.
Frank

Right...
But If I did not use an IF and alter the schematic so that the quadrature mixer downconverts directly to audio instead of 455KHz (DCR), then this quadrature mixer should cancel the opposite sideband of the signal. Ie, the "image" signal would be just the opposite sideband, and the carrier would be also suppressed, am I right?

Because of the above, there has been a second question that rises here. In the case I mentioned (DCR) the AM carrier will be also suppressed.
But why this is not happen also with the Image rejection? Just because the image and real signals are far apart, or is there another reason?

I guess "DCR" stands for Direct Conversion Reciver (right?).
In that case, the 90 degrees filter is at baseband and must have a very wide relative (percentual) bandwidth. At FI is very simple because it covers a narrow relative bandwidth.
Such a filter (a Hilbert filter) has ideally 0 transfer at dc, so carrier is suppressed if the LO has exactly the frequency of the carrier.

Z

So why is not the carrier suppressed in this AM receiver? That is my whole point.