PIN Diode purpose

[K6JEK]

What is the PIN diode in the emitter circuit of the driver transistor doing? R6 sets the drive level but what is the function of the PIN?

This is a cute little QRP amp from W8DIZ which works very nicely.

[Rob K2CU]

R6 at 10K is not affecting the bias level of the transistor. Ait is affecting the current through the pin diode and therefore its resistance. So the control adjusts the amount of RF shunting across the emitter resistor and hence the gain of the stage.

[AB2EZ]

[This analysis is not correct... see my comment further down this thread...Stu]

Hi!

 D1, C3, and the series combination of R5 (180 Ohms) and R6 (adjustable from 0 - 10k Ohms) are forming an envelope detector. One can also view it (equivalently) as a peak detector with a decay time constant given by: C3 x (R5 + R6)

Since R5  (180 Ohms) is small compared to the maximum value of R6, R5 serves to set a lower limit on the value of R5 + R6... in order to protect the diode from excessive peak current.

If we pick an example value of 5k Ohms for the series combination of R5 and R6, then the RC time constant (i.e. the decay time constant) of the peak detector would be 47nF x 5k Ohms = 235 us. This corresponds to an envelope detector bandwidth of somewhat less than 1000 Hz. If R5+R6 is set to 1k Ohms... then the envelope detector bandwidth would be 5x larger.

Since the output of the envelope detector is not shown as being connected to anything, it would not (at least as per the schematic) have any effect on the behavior of the circuit. [There will be a small amount of average current flowing into the diode... but this will be negligible compared to the current flowing into the 50 ohm emitter resistor.

[WD5JKO]

Looks to me that R6 at Max means the pin diode is high impedance, and that means that Q1 emitter is not bypassed much at all by C3. This lowers the gain of Q1 via emitter degeneration, a form of negative feedback.

Looks like Q1 emitter is at ~ 1.9v based upon the biasing resistors R1, and R2. This means Q1 emitter current is ~ 37 ma. Start making D1 conduct ( by cranking R6 to a lower value) will boost the stage gain of Q1 because C3 is effectively in parallel with R3.

Could have done this without a diode....

Jim
JKO

[w3jn]

What Rob and Jim said....

Pots are very poor at RF - too much distributed capacitance, hence the PIN diode.  I put a bunch on a network analyzer and I had one 10K pot that had about 5 ohms of reactance at 2 MHz.

[K6JEK]

What Rob and Jim said....

Pots are very poor at RF - too much distributed capacitance, hence the PIN diode.  I put a bunch on a network analyzer and I had one 10K pot that had about 5 ohms of reactance at 2 MHz.

OK, now I get it.  I should have simply asked, why not just a pot?  Answer: At RF a pot is crummy.

Thanks everybody.

Jon

[Tom WA3KLR]

Designers don't resort to over-complicating the circuit (normally) unless this is proven by prototyping with a simpler design first.

It’s possible that the low resistance pot required was erratic or couldn’t handle the current?  Also the stage may have started to oscillate with the inductance of the pot.  

Or, with the r.f. path of Pin diode D1 and bypass cap C3 contained at the stage, the gain control pot R6 can now be remoted to the front panel.

[AB2EZ]

[This analysis is not correct... see my comment further down this thread...Stu]

All

The diode is going to be reversed biased. The reverse bias voltage will go up and down as the RC circuit charges (quickly) and decays (slowly).

According the the data sheet, the reverse biased capacitance of the diode is 1pf or less.

http://www.datasheetcatalog.com/datasheets_pdf/M/P/N/3/MPN3700.shtml

At 3.885 MHz (for example) the impedance of a 1 pf capacitor is ~40,000 Ohms.

It is hard for me to see how the diode and the circuit to the right of the diode will have any effect on the behavior of the amplifier.


Stu

[K6JEK]

All

The diode is going to be reversed biased. The reverse bias voltage will go up and down as the RC circuit charges (quickly) and decays (slowly).

According the the data sheet, the reverse biased capacitance of the diode is 1pf or less.

http://www.datasheetcatalog.com/datasheets_pdf/M/P/N/3/MPN3700.shtml

At 3.885 MHz (for example) the impedance of a 1 pf capacitor is ~40,000 Ohms.

It is hard for me to see how the diode and the circuit to the right of the diode will have any effect on the behavior of the amplifier.

Stu

But they do, I assure you.  I built this little amp and adjusting R6 does indeed vary the output  from 0 to 5 watts just fine.

[AB2EZ]

[This analysis is not correct... see my comment further down this thread...Stu]

Have you checked to see if the actual amplifier is consistent with the schematic that you posted? For example, is there really a 51 Ohm resistor between the emitter of Q1 and ground?

Stu

[WD5JKO]

I don't like how the pot setting adjusts the DC operating point of Q1. Separating gain adjustment from DC operating point is illustrated by the file attached from VE7BPO. See attached file, or figure 11 at:

http://www.qrp.pops.net/Cascode_BJT.asp

Jim
JKO

[AB2EZ]

Jim

A very interesting and instructive reference.

All

I think I now understand how the circuit works... and what Rob and Jim were saying. What I was overlooking is the Q1 base bias voltage divider.

Roughly 1/3 of the Q1 collector voltage will appear on the base of Q1... and therefore about 3.5V of DC will appear from the emitter of Q1 to ground. Thus Q1 will always be biased on (since the input RF signal is only a few hundred mV).

This will result in an average current through the diode of roughly 3V /(R5 + R6).

If R5+R6 is (for example) 1k Ohms... then about 3mA of average current will flow through the diode. The rf impedance looking from the emitter into the diode (and from the output of the diode through the 47nF capacitor to ground) will be around 26 Ohms/ 3 ~ 8.7 Ohms. If R5+R6 = 10k Ohms, then the rf impedance looking from the emitter into the diode ((and from the output of the diode through the 47nF capacitor to ground) will be around 26 Ohms/ 0.3 ~ 87 Ohms.


When combined with the 51 Ohm emitter resistor... the effect of varying R6 will, as per the design, be to vary the total emitter rf impedance from around 7.4 Ohms (8.7 Ohms in parallel with 51 Ohms) to around 32 Ohms (87 Ohms in parallel with 51 Ohms)... and thus vary the voltage gain of the amplifier over more than a 10dB range.

Stu

[w3jn]

Stu, you're neglecting the resistance contributed by the PIN diode at low forward current.  Note the sharpness in the upward curve in the resistance as the current hits around 1 mA.  The adjustment range is quite a bit more than it might appear due to the variable resistance nature of the PIN diode.  R5 and R6 can be largely neglected for the purposes of the emitter bypass, all R6 does is adjust the diode forward current; R5 is there to limit it to a reasonable level.  The resistance of the PIN diode + the reactance of the capacitor are the major players here.

Here's the data sheet on the PIN diode http://www.datasheetcatalog.org/datasheet2/1/03hzwxo0wayo0q15ltp4f2i0gw3y.pdf

[AB2EZ]

JN

I think we are saying exactly the same thing at this point.

The numbers 26 Ohms/3 and 26 Ohms/0.3 are the values of diode resistance I calculated using the rough rule of thumb: R= .026V/I;  where I is the forward current in A and 0.026V = 26mV = kT/e = Boltzmann's constant x the temperature in Kelvins / the electron charge.

Stu

[w3jn]

The resistance rise of the diode at low currents appears to rise exponentially, although the data sheet stops at about 3 ohms.  This paper from On Semi indicates a series resistance of about 30 ohms at low forward current.

[W4AMV]

The PIN diode (p- Intrinsic-n) does not obey the p-n diode equation wrt to Resistance vs. forward current. However, its R value is directly affecting the gain in the posted circuit. A nice writeup on the PIN diode operation is at:

http://www.microsemi.com/micnotes/701.pdf

[w3jn]

That's the paper I intended to post, and neglected to paste the link.