Please advise me on the diode polarity in this transceiver circuit

Originally, before I removed it, it was placed opposite to the ground but when I read the schematic diagram it reads pointing to ground.

I broke one of its leg so I replaced it as it was placed originally (opposite to ground). According to its technical description, D8 was placed in parallel to C4 which is a part of T matching network.

So please advise me which polarity of this diode (D8) is correct or should be.



E). TRANSMIT/RECEIVE SWITCH
When the radio is in the transmit mode, pin diode switches D13 and D1
are both turned on (representing less than 0.7 ohms). D13 allows the transmit
signal to pass to the antenna and D1 shorts one leg of a T matching network
(L3, L15 and C4) to ground in the receive path. This results in a parallel tuned
circuit high impedance being presented to the transmit signal so that the
receive path does not load the transmit signal. In the receive mode, both D13
and D1 are off, resulting in the antenna signal being coupled into the receive
LNA through the 50 ohm T matching network and the unwanted load of the
transmit final amplifier is reduced to less than 1 pF by D1.

Looks the wrong way round in the schematic you show. They appear to be PIN diodes used to isolate the receiver when in transmit mode, if the line approaching from the bottom right is the PA supply, D8 will be forward biased across the supply voltage and short it out.

Brian.

betwixt said:

Looks the wrong way round in the schematic you show. They appear to be PIN diodes used to isolate the receiver when in transmit mode, if the line approaching from the bottom right is the PA supply, D8 will be forward biased across the supply voltage and short it out.

Brian.

Click to expand...

The transmit signal comes from upper left from Q3, Q4 to Q1 up there. So, the polarity of the D8 in the picture is correct?

I'm still not convinced although I haven't got the specification for the KDS114E here to check. It looks like they are PIN diodes, or ordinary diodes used in a "PIN like" mode. Essentially, when DC current flows through them they also carry any RF at the same time. When the polarity is reversed, they do not conduct and they block the RF as well. If my understanding of the schematic is right, the DC supply to the PA stage (Q1) comes from the lower right corner of the schematic and it is also used to control the diode switches. When voltage is sent to the PA, Q1 does it amplification and the signal leaves via D7/D13 on it's way to the antenna. When the voltage is removed, the PA doesn't produce any RF, D7/D13 stop conducting and isolate the PA from loading down any incoming signal.

Aside from the signal routing, with the dode as shown, there is a conduction path from the PA supply through L6, D7/D13, L3 and D8 to ground.

Brian.

ํำYes, your understanding is correct. I have looked at the spec of KDS114E. It is PIN diode and it works like what you said.
So I guess the D8 was placed incorrectly from the factory. I should change its polarity by pointing to ground then.

The schematic shows it the wrong way around but on the board it should be connected with it's cathode (striped end) toward L3. The arrow in a diode symbol always points to the cathode end.

Brian.

betwixt said:

The schematic shows it the wrong way around but on the board it should be connected with it's cathode (striped end) toward L3. The arrow in a diode symbol always points to the cathode end.

Brian.

Click to expand...

Alright, so it was originally placed correctly from factory but the diagram is incorrect. There is another file showing the placement/arrangement of components on the board and the D8 points toward L3, which is contrary to the schematic file.

So I don't have to remove D8.

Thanks.

Looks like the schematic is wrong but the board is correct. It happens sometimes, someone disdn't get it checked properly before it was printed.

If the diode was connected cathode to ground I would expect several components to be damaged as it would virtually short out the power feed to Q1 which by it's nature is a high current supply.


Brian.

Guys, after I read some info about PIN diode switching, I'm still thinking that the D8 could be placed as cathode pointing toward ground. So it will short the signal to ground when in TX mode as in other PIN diode RX/TX switch.

More advice please.

I posted a new photo, so you can see it clearer.

To DC, a PIN diode behaves like an ordinary silicon diode. If you connect it so it conducts in TX mode, it will short the PA stage supply to ground.
Brian.

This site explains how PIN diode works as a TX/RX switch. The PIN diode D2 cathode points to ground. That's why I am still curious.

https://urgentcomm.com/networks_and_systems/mag/pin-diode-antenna-switch-201107/

The missing factor is the capacitors in the link, they are not present in the schematic you showed. The link is correct, when DC is applied, the diodes both conduct and effectively become short circuits to RF. One is in series with the transmitter output so it conducts the transmission to the antenna, the other shorts out the input at the receiver to prevent it overloading. The tuned line between them stops the receiver PIN also shorting out the transmission.

In your schematic, there are no series capacitors and the full PA stage voltage is put across the two diodes forcing them to clamp the PA supply to almost ground.

Brian.

Inwhich picture did you refer to series capacitors?

The text from pdf file explains the D1 that it shorts the signal to ground but not the D8. So should I reverse the D8 by pointing cathode to ground?

Here is the full schematic file.
**broken link removed**

Here is the tecnical description.
**broken link removed**

Figure 1 in the link you gave in message #12.

Think of the diodes as simple switches that turn on (short circuit) when current is passed through them and off (open circuit) when the current is removed. The current is provided by the bias network. Take note of where the antenna is connected.

Brian.

betwixt said:

The tuned line between them stops the receiver PIN also shorting out the transmission.

Brian.

Click to expand...

Could you explain which tuned line you mentioned?

ok, I think I got it. There is no capacitor between the PA(Q1) and D1. When in TX mode, the TX signal go through D13, D7, D1 then C8 to ground. If D8 cathode was pointing to the right(GND), the TX signal would be shorted to ground which is not supposed to be. Correct?

So why did they put D8 in the circuit?

Another question is, what is the Q26 supposed to function in the circuit?

It would be sensible to protect the receiver input stage (Q6) from being blasted by the transmitter output so a forward biased PIN diode would be a normal wayto short the signal to ground at that point. As you have now realized, with the circuit drawn as it is, the maximum supply voltage to the PA stage (Q1) is clamped at 2*Vf of the diodes which according to the data sheet is 1.7V total.

The tuned line is the circuit L3, L14, L15, C4 and C7. It is necessary so the PIN diode clamping the receiver input signal doesn't also clamp the transmitted signal.
Q26 is a "digital transistor" with internal bias resistors so it certainly isn't part of the signal circuitry, I would guess it's purpose is to allow current through D1 so that C8 is added to the tuned line circuit only when in transmit mode.

Brian.