From: apogosso@tpg.com.au   
      
   We have seen that:   
   1. Sensitivity determines the minimum carrier level which allows to handle   
   required modulation index. A pn diode has the best *potential* sensitivity.   
   (Only liquid nitrogen can improve it further. Sigh...) A tube is four times   
   worse.   
   2. For an unbiased detector modulation handling always reduces with audio   
   frequency, for the biased one -- it stays close to 100% till a certain   
   frequency, then starts to fall.   
   3. An unbiased detector HF high-level performance does not depend on the   
   signal level, whilst for the biased detector works worse and worse (slew   
   rate wise) with the carrier increasing. This it requires an efficient   
   (delayed and amplified AGC).   
      
   === How to properly bias the diode in an unbiased detector ====   
      
   Everyone knows that differential resistance of a pn diode is approximately   
   Rd = 25 Ohms / I,   
   where I - current trough the diode in mA.   
      
   What is the differential resistance of a diode with no bias, at low voltage   
   (<10mV)? Is it infinitely high? The answer is "No". A diode acts as a   
   resistor at low voltages. In fact, the above formula is not correct. More   
   correct version of it should be:   
      
   Rd = 25 Ohm / (Io + I), where   
   Io - so called saturation current or reverse current of the diode in mA;   
   I - current flowing through the diode in mA.   
      
   This equation is derived from I = Io * (exp (e*V / kT) - 1). Note that in   
   part 1 I deliberately omitted " -1" for simplicity.   
      
   For a silicon diode Io, the reverse current, is of the order of picoamps,   
   therefore, with no bias (I = 0) its resistance Rd is gigaohms, virtually   
   infinite.   
      
   For a germanium diode reverse current Io is in the range of 0.5...5uA. for   
   example, if Io = 1uA, then Rd = 25KOhm.   
      
   Now consider you use a germanium diode with Io=1uA in a conventional   
   booooring detector fed from a Hi-Z IFT with no cathode follower. For a low   
   signal the rectifier is virtually shunted by Rd which is as low as 25K in   
   this case. Needless to say that the low Rd=25K will be greatly shunting the   
   LC tank which normally has 100...300K of resonance impedance. The detector   
   will lose its sensitivity.   
      
   Reducing Io improves performance. You can try the following experiment. In a   
   booring radio replace a vacuum diode with a germanium point contact one.   
   Most likely sensitivity of the radio will drop. Now cool the diode by   
   spraying it with CO2 or some quickly evaporating liquid. The radio will come   
   to life. Warm the diode by a soldering iron -- the radio will become   
   completely dead.   
      
   Note that Io changes 2 times per 10 degrees, Rd changes reciprocally in the   
   same proportion.   
      
   Now it is easy to explaim why Partick ended up with 3 germanium diodes in   
   series. He connected them in series to increse Rd. Probably each of his   
   diodes had about 0.5uA of reverse current (Rd = 50K). Three diodes would   
   have 150K which is sort of OK. Connecting more diodes in series though   
   increases Rd, reduces the sensitivity as the RF voltage is divided between   
   the diodes. If one diode needs 25mV of RF to rectify, three diodes need   
   75mV, etc. So there somwhere lies the optimum number of the diodes in the   
   chain. Too few -- you shunt the LC tank and lose gain, too many -- you raise   
   the sensitivity threshold.   
      
   If Patrick happen to experiment on a cold winter day, perhaps we would have   
   heard about "two diodes". If Partick had chosen to fix his radio during a   
   scourching 45C heatwave, then today we would have heard his stories about   
   four Germanium diodes...   
      
   Mystery solved...   
      
   But what is the best Rd of a diode to use as a detector? To answer the   
   question consider the following.   
   Everyone knows that an equivalent load the detector presents to the IFT at   
   high signal is R/2. It is logical to presume that at low signal (<25mV) the   
   load on IFT shall also be R/2 -- so that IFT does not "see" a differenc in   
   load over the whole range and thus works with the least distortion and   
   losses.   
      
   In a common boooooring detector R is about 500K. Thus Rd shall be 250K,   
   which translates into 100nA of the diode reverse current. One germanium   
   diode has higher reverse current and therefore unsuitable. What about a   
   silicon diode with a negligible Io? The answer is obvious -- bias it so that   
   at no signal the current through the diode is 100nA. Or on the load it will   
   translate into 50mV of DC voltage drop.   
      
   That is the rule: bias a silicon diode to have 50mV across the load. (Use a   
   hi-Z multimeter for this purpose). The bias voltage must be temperature   
   compensated, so it must be taken from an auxiliary forward biased diode.   
      
   If you want to use a germanium diode, you need to reverse (!!!) bias it, so   
   that Io + I = 100nA. For example, if your Ge diode has 1uA of reverse   
   current, then you need reverse bias it so that --900nA is flowing, or --45mV   
   is on the load. Needless to say, it is impractical to do so given that Io   
   depends on temperature. So, do not use Ge diodes in Hi-Z detectors.   
      
   (With Io=1uA a Ge diode will be perfect in a detector with R=50K, not 500K.)   
      
   What to do if you are not looking forward to complications with forward   
   biasing Si diodes? Then... use Schottky diodes!!!! They have Io from 10nA to   
   500nA depending on the type. Buy a dozen of 1N5711 Schootky diodes and   
   select the ones with suitable Io. To measure Io use a multimeter in a mV   
   range in series with the diode under test and any battery, e.g., 9V battery.   
   Multimeters have either 1M or 10M input resistance. (To find this out   
   measure say a 9V battery with 1M in series with the DVM. It will either show   
   4.5V or 8.5V). Aim at lower than 100nA taking into account heating of the   
   chassis in operation.   
      
   What about properly biasing a tube detector. The above concept still   
   applies, but because of higher temperature, for a tube diode Rd is 4 times   
   greater. Rd = 100R / I.   
      
   So for a boooring detector with R=500K emission current at no signal should   
   be around 400nA, or 200mV across the load.   
   If your detector diode is decent, like 6AL5, you will probably have more.   
   For example, in my Lafayette HE-80 receiver I have about 390mV. The detector   
   is overbiased and is excessively loading the IFT. Therefore I need to   
   additionally reverse bias the circuit by connectin the load resistor not to   
   GND, but to some negative voltage. Even better to reduce the heater voltage   
   to have, without signal, about 200mVdc on the load .   
      
   If your vacuum diode is feeble as 6AV6 or 6Q7, etc. your emission current   
   might be not enough to deliver 200mV at 500K load. In this case you might   
   like to add slight positive bias to the circuit.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)