UPD858 chassis . Where's the 7.8000 ?

[cb4ever104]

Hi All .

This one's bugging me . Simple question , which I'm sure one of you smart guys will be able to answer . How does the AM TX carrier frequency get generated ?

Example . The ADAMS CH1

VCO is 34.765 . AM Triple Crystal is 11.285 x 3 = 33.855 . Down mixer is 0.91 .

The problem I have is with the 7.8 LO circuit . There are 2 crystals . The 7.7975 for LSB , but USB and AM TX both share the 7.8025 .

Capture_2.JPG

In order to get my CH1 26.965 AM TX frequency I need the VCO 34.765 - 7.800 . But I don't have a 7.8 crystal . So , how does the 7.8025 oscillator get offset to 7.800 on AM TX ? I can't see anything in the diagram that actually does that . There's no AM TX adjustment in the service manual either . Only USB and LSB .

Capture.JPG

Thx in advance

Sammy

[14CS06]

Hello Paul

1 RX, AM, CH19

4 Same as Step 1

On Realistic TRC-449

https://www.fichier-pdf.fr/2016/10/01/r ... manual.pdf

@+
Claude

[cb4ever104]

Hello Paul

1 RX, AM, CH19

4 Same as Step 1

On Realistic TRC-449

https://www.fichier-pdf.fr/2016/10/01/r ... manual.pdf

@+
Claude

Hi Claude . I've already got that information . I don't see any adjustment for the AM 7.800 oscillator circuit at TP9 .

Thanks

[cb4ever104]

There's a mention of "AM/SSB offsets" coming into the Carrier Osc , but I can't see that part in the schematic . It's probably staring me right in the face ....

Capture_3.JPG

[lbcomms]

The IF is at exactly 7.800 MHz in AM receive, and no carrier oscillator is needed to process the incoming signal because the needed carrier is on the signal you are receiving, the radio doesn't have to supply it as it does on SSB. So no 7.800 MHz oscillator needed. A separate 11MHz crystal is used for AM receive to set the local oscillator so the incoming signal is centred on the 7.8MHz IF (crystal filter) frequency.

They could have used a separate 7.800 oscillator for AM transmit, but that would have cost them an extra crystal. By bypassing the crystal filter on AM transmit and high level modulating it, the USB crystals can be used for AM transmission, saving the cost of a crystal.

Hope this helps...

[cb4ever104]

Hi Sue .

Thanks for that . I think the part I need to understand more is your "high level modulating" comment . I'll see if I can find anything more about that on-line .

Thanks

[lbcomms]

There are two different ways of modulating an AM signal, low level and high level.

Low level is AM generated in the same way as SSB, at the IF level and fed to the mixer where it combines with the LO to create the signal you'll transmit, it then goes to a power amplification stage (i.e. driver and final) and then to the antenna. The HR2510 is a good example of this method, as are most "ham" transceivers.

High level just sends a carrier, unmodulated, to the driver and final. The DC supply to the output stage is then varied at an audio rate, modulating the transmission. Most CB type radios with AM work this way, such as the Cybernet AM/SSB sets, and nearly all AM-only rigs. The audio power stage - needed to drive a speaker when receiving - does the double duty on TX to modulate the supply to the RF output stage.

[cb4ever104]

Hi Sue .

OK . Thanks again for that . Before I get into that more , do the Uniden/Ranger models with the dedicated 10.695 AM/FM IF crystal use the same high level modulation ? Because if they do , then I'll need to go back to square one and see how you can generate a 26.965 carrier from a 34.765 and 7.8025 signal . Thanks for bearing with me on this one . I'm feeling a bit thick today

Paul

[lbcomms]

The later radios (uPD2824 and onwards) use 3 coils for each crystal (carrier oscillator and downmix). Some radios use a separate 10.24 (ranger etc), some use a tripler (AR144 and the like), and "stretch" the crystals further using inductance. It means the newer radios only need 2 or 3 crystals, whereas the 858 radios had 6 or 7. Crystals are relatively expensive so it saves them a few dollars, but at the price of stability.

The inductors used for frequency alignment in the newer radios vary in value a bit with temperature change, so they take a while to stabilize and drift when they get hot - unlike the 858, which is usable from cold turn-on and is stable even on a hot day down here in Oz.

With the 3+3 inductor radios, the carrier oscillator in AM transmit runs at exactly the first IF frequency of the radio, generally 10.695 MHz.

[cb4ever104]

Hi Sue . OK . Many thanks again for your explanation , and for bearing with me .

Paul

[cb4ever104]

Hi .

Just want to pick this up again . The problem I had with this circuit was that I thought AM (TX) used the 11.2850 Xtal . It doesn't . It uses the 11.2858 . So , 3 x 11.2858 = 33.8574 + 0.91 (CH 1) = 34.7674 - the 7.805 = 26.6949 . Which is close enough Dilema solved .

S.

[lbcomms]

Detailed explanation

An AM signal contains a carrier along with both USB and LSB, so it's is twice as wide as a single sideband transmission.
Voice baseband contains frequencies up to 3500Hz, so the bandwidth needed is 7KHz ((3500/1000)x2) for AM.

It's an inefficient mode though, as the carrier can be replicated at the receiver end and both sidebands contain the same information.
Enter SSB - the carrier and one of the sidebands is removed, and all of the transmitter power is packed into the remaining sideband.
The required bandwidth is reduced to 3.5KHz to send the same information, but it comes at a $$$ cost for the additional circuitry needed.

A radio capable of both AM and SSB would need two IF filters - a 3.5KHz wide one for SSB and a 7KHz wide one for AM.

The crystal filter in a radio is one of most expensive components in a radio, so one way to save cost is to use a single filter for both modes.

By using a single filter 5KHz wide it's usable, but with a performance hit. The single filter will be too wide for SSB (more noise as it lets through audio in the 3500 to 5000 Hz range that doesn't contain any voice) and too narrow for AM (the filter cuts off the audio in the 2500 to 3500Hz range, reducing clarity).

Because SSB carrier reinsertion has to be done at the edge of the filter bandpass, and at opposite ends for each sideband, you can use the specified carrier oscillator frequencies to calculate the bandwidth:

"Compromise" single conversion radios:
President Adams / TRC449: (7.8025 - 7.7975) x 1000 = 5KHz bandwidth
AR144 / TRC-451: (10.6975 - 10.6925) x 1000 = 5KHz bandwidth

Dual conversion radios:
Courier Gladiator 858 PLL: (7.8015 - 7.7985) x 1000 = 3KHz bandwidth
Cybernet PLL02A SSB radios: (10.695 - 10.692) x 1000 = 3KHz bandwidth

The last two are dual conversion on AM only (why is another subject altogether), but it means there is the correct bandwidth on all modes. The dual conversion on AM bypasses the 3KHz filter, converts it down to 455 KHz (i.e. the Cybernet does 10.695 - 10.240 = 0.455 MHz, in other words 455 KHz) and sends it through a 7KHz wide ceramic filter before decoding it back to an audio signal.

Back to the single conversion 858 radios (frequencies are shown for FCC channel 1, 26.965 MHz with the clarifier centred)...

IF is 7.800MHz, so an incoming signal on 26.965 needs a high side local oscillator at 34.765 MHz.
The PLL downmix output needed to lock the loop is 0.91 MHz, so the other input needed is 33.855 MHz.
This is too high for a crystal, so an 11.285 MHz crystal is tripled (over-driven and its third harmonic filtered out).

For LSB it has to drop 2.5 KHz, so 33.855 - 0.0025 = 33.8525, divided by 3 gives the crystal frequency needed: 11.284167 MHz.

For USB it has to raise 2.5 KHz, so 33.855 - 0.0025 = 33.8575, divided by 3 gives the crystal frequency needed: 11.285834 MHz.

The same USB math is used for AM transmit, which saves the need for a 7.800 MHz crystal - a couple of diodes and even another transistor is way cheaper than another crystal.

[cb4ever104]

Thanks Sue. Yeah , I figured out the TX/RX switching for that AM TX crystal circuit . Think it was TR40 , or 41 . As I stated before. The problem I had was just a simple maths one. I kept coming up 2.5khz short , because I was looking at the wrong trippler crystal in the diagram. Anyway. It's resolved now. Thanks again for your explanation.

Paul