Diode and very small amplitude high frequencies signals

Right, but that part of my measurements cries out for further bench exploration. It represents only one part, and is unconfirmed. Also, what happens if the voltage is reversed? Are we to believe the diode is a 10M resistor, shunted by a diode? I'm not comfortable with that.

In article , hill_a@t_rowland-dotties- harvard-dot.s-edu says... of my measurements cries out for further bench

I'm confused. Is there some reason to expect the semiconductor material to be a perfect insulator with no resistivity at all? Nothing's perfect, and those diodes probably aren't made in the most exacting processes.

I would be blown away if you *couldn't* measure some ohmic current flow in a diode at any particular voltage level.

-- jm

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Yes. Ive been mulling that over. If there is no static bias current flowing through the diode and it magically switches to 'open circuit' when the incoming voltage reverses sign, then it would still make a perfect 'average respsonding' rectifier even at these mV or even uV levels. Just how does the reverse current start to act in the reverse direction?. On the face of it, a IN4148 seems easy enough to check out. regards john

Input transformers are all very well, but some good voltage step-up can be obtained by carefully chosen values of hi-Q capacitor and inductor in series between the aerial and the diode. Of course this makes the impedance even higher, just as the transformer would, but how strong's your signal? It might be the cheapest alternative.

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Test on a 1N4148. ForwardV DiodeR

-5mV 21megs.

-10mV 30megs.

-30mV 270megs.

All the figures are suspect as the noise filter had a long settling time and I was too lazy to wait but they do show a very sharp reverse cutoff, at about 10-15mV.

And a bonus, a rectification test feeding the diode from a 60Hz source and

Agreed. It's the rather low 10M value that raises my eyebrows. Hence my suggestion that the measurements be revisited. Picked up by John Jardine, who obtained similar values, copied below:

Test on a 1N4148. ForwardV DiodeR +50mV 8megs. +30mV 9megs. +20mV 10megs. +10mv 12megs. +5mV 21megs. ReverseV -5mV 21megs. -10mV 30megs. -30mV 270megs.

John also suggests the measurements may need further refinement.

Oops! I can think of several circuits I've designed over the years using diodes for discharge protection that might not work exactly as I intended, given this observation. And I recall several circuits where I intentionally back biased the diode a few hundred millivolts to insure an open circuit.

And others where I used a transistor collector or JFET gate instead.

Pease Porridge in the Feb 3rd issue of Electronic Design mentions this problem, and Bob suggests using a transistor. "Using 2n3904s as diodes is very important because most ordinary diodes are much too leaky around +/-60mV to work well. Ordinary gold-doped 1n914s and 1n4148s are quite unsuitable..."

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Yes if you use a serial resonant with a capacitor, inductor and a resistor as you're suggesting you get voltage amplification factor exactly equal to the Q of the circuit plus you get frequency and bandwidth selectivity

With a transformer you get all 3 of the above without having to add an inductor (as you use the inductors in the windings of the transformer) plus you get impedance level shifting of the capacitance and resistance in the secondary to the primary multiplied by the square of the turns ratio multiplied by the capacitance and resistance in the secondary of the transformer. Is it worth it. I can't tell but I see more parallel resonant circuits then serial ones .

I've actually seen implementations of the above using positive and negative feedback circuits with an opamp to get some really interesting results.

I read in sci.electronics.design that john jardine wrote (in ) about 'Diode and very small amplitude high frequencies signals', on Sat, 5 Feb 2005:

How did you measure the resistance? Is it an incremental resistance (slope of the V/I curve at the data point) or the slope of a line joining the origin to the data point on the curve.

I haven't followed this thread very thoroughly, so this might not be directly relevant. But it should be of interest to anyone trying to detect small signals with a diode.

There are several reasons why diodes do poorly with small AC signals.

The first is, of course, the forward drop. However, this can in theory be reduced to an arbitrarily low value by reducing the current to a low enough value (by, for example, making the load impedance high enough).

The second is that the ratio of reverse to forward current increases as the signal gets smaller and smaller, reaching one at the limit. This can be observed by looking at the I-V curve of a diode. At the origin, the curve is a straight line -- the diode behaves just like a resistor.

The third reason is the diode capacitance. This shunts the diode, effectively lowering the reverse impedance. It also lowers the forward impedance, but when the forward Z is lower than the reverse Z, the net effect is to further degrade the forward/reverse impedance ratio.

You can make all the DC measurements you want, but they only tell half the story. When you apply AC, you charge the load capacitor during half the cycle according to the diode's forward impedance, and charge is removed from it during the other half according to the diode's reverse impedance. As the forward/reverse impedance ratio degrades due to the two effects mentioned above, the net charge you get in the load capacitance decreases, hence the voltage it's charged to decreases. This ends up looking like a larger diode forward drop.

I spent a lot of time thinking about this some years ago when designing a QRP wattmeter, and some of the conclusions I came to appear in the resulting article, "A Simple and Accurate QRP Directional Wattmeter", published in QST, February 1990. See the analysis on p. 20, "Ac v Dc: Why the Difference?"

Roy Lewallen, W7EL

I didn't measure the resistance. The values just come from the static V and I plot points on the graph. Having had my remaining bench DVM, (good ol'e UK, Datron s**te) pack in on me and 2 battery DVMs keel over with flat batteries and the CMOS buffers floating off to la la land and 2 crocodile clips secretly fail and finally my electric pencil sharpener going t*ts up, I was not of a mind to press on :-) regards john

Right. Take a look at a diode curve across a 100 uV region. Most diodes will look pretty flat, even without considering the effects of capacitance and other parasitics.

In other words, this data is just a plot of a diode's DC I vs V characteristic, right?

What is of more interest is the slope at a given DC operating point. If we pick 0V, for example, the above data (within the limits of its precision) gives a flat line around that point (+5mV 21 Mohms, -5mV 21 Mohms). With a 100 uV signal, you might as well throw a 21 M ohm resistor in there instead.

No, it means its a better diode at low currents. See my curves again,

Excellent description - thanks.

Only one small problem - as Win pointed out, Bob Pease feels a diode-connected 2N3904 has lower leakage at low voltage than a 1N4148:

"What's All This Comparator Stuff, Anyhow?"

Does this mean a 2N3904 has a shallower slope than a 1N4148 through zero, or perhaps one or the other has an offset, such as the Agilent Zero Bias Schottky Detector Diodes shown in AN969?

Regards,

Mike Monett

Exactly. At very small currents, the diode is just a resistor (shunted by a capacitance). At slightly small currents, it's a very poor diode, with reverse current almost equal to the forward current, so on each negative half cycle you suck out nearly all the charge you delivered during the positive half cycle.

Roy Lewallen, W7EL

Your article sounds interesting. Is there a link available to see it?. The simplest approach I've seen, was is in the 'Levell TM6A broadband voltmeter'(UK). Designer chopped the low level diode output at 20Hz, allowing a 1mVac FSD. regards john

Sure there is. All diodes have reverse current.

It's the very

The gold doping is done to dramatically reduce charge storage time. Without it, the voltage across a diode continues to be in the forward direction for some time after you reverse the current through it. While a non-gold-doped diode might look good in DC tests, it makes a lousy rectifier of RF. In the extreme case, it acts like a PIN diode (which is simply a diode designed intentionally to have a long charge storage, or reverse recovery, time).

Alas, life is full of tradeoffs.

Roy Lewallen, W7EL