I saw on Hackaday that Ted Yapo made a digital clock out of diodes and oscillators. This ought to be impossible, since diodes can’t invert or amplify, but it turns out that it isn’t, because of reverse recovery time.
Yapo is using regular power rectifier diodes as RF switches, like PIN diodes. When they’re forward-biased or even zero-biased, they pass RF AC with no problem. When they’re reverse-biased, the depletion region grows, the junction capacitance drops, and they block RF AC (up to some rolloff frequency, anyway). So you can use DC voltage biases to get them to block or pass RF. Then you can use a faster small-signal diode to rectify the RF AC into DC, which you can use for further control signals.
Essentially, you carry two signals from two different power supplies on the same wire at the same time, allowing you to do something thought impossible for decades.
You could argue that this isn’t “really” diode logic because you need an external oscillator to drive it. Yapo says this is a bogus criticism — the external oscillator is just another power supply, like how you need regulated 5VDC for TTL and regulated 3V3 or 1V8 for many modern CMOS chips. You don’t need more oscillators as your logic gets more complicated.
(Incidentally, I’ve seen an avalanche relaxation oscillator LED flasher that works by back-biasing the collector-base diode junction of a transistor until reverse breakdown, so I suspect you can get an RF oscillator out of relatively ordinary diodes too, not just exotics like Gunn diodes and tunnel diodes.)
It occurred to me that maybe you don’t even need two different kinds of diodes for this trick. You can use different regions of the response curve of a single kind of diode. You can use a capacitor (with more capacitance than your back-biased diodes) to move an AC signal from riding on one DC voltage level to another. If the DC level is close to the diode’s forward drop, it rectifies the RF signal. If it’s above the diode’s forward drop, it passes the RF signal; if it’s below it, it blocks it.
This is kind of tricky because it seems like it could be hard to get amplification with just one kind of diode this way. Unless you’re going to use transformers, the RF voltage needs to be high enough to, when rectified, switch a diode from blocking RF to passing RF. I was thinking that this wasn’t really a problem because you can use an arbitrarily small DC voltage to build up an arbitrarily large charge as you back-bias a diode, but it’s actually a bit trickier than that — the small DC voltage will stop building up the large charge once the voltage created by that charge gets high enough. But I think you can use a standard voltage multiplier circuit (Cockcroft-Walton generator) made of diodes and capacitors to solve this problem. (I say “I think” because I’m not sure if this would require a different kind of diode.)
A standard 1N4148/1N914 handles 10 mA at .75 V forward and can peak at 100 mA at 1.1 V forward, withstands up to 75 V reverse, and has .9 pF junction capacitance and 4 ns reverse recovery time. I think this means it can usefully rectify signals up to about 60 MHz. Common LEDs can also pulse up into the tens or hundreds of MHz. Schottky diodes don’t have a reverse recovery time but have more capacitance; a 1N5819 has 150pF and can handle 1 A at .4 V forward. That capacitance will pass 1 A at .4 VAC at 2.7 GHz, although presumably their lead inductance will start to be a problem well before that, and you probably don’t really want to run your logic circuits at 400 milliwatts per diode anyway, and lower currents will make the capacitance proportionally more important — 27 MHz already passes 10 mA.
(2.7 GHz? Is that really true? To charge 150 pF up to a .6 volt peak, I guess you need 90 pC, which an amp would deliver in 90 picoseconds, which gives you a waveform period of about four times that, which is indeed about a third of a nanosecond. Wow.)
There are smaller Schottky diodes with correspondingly smaller capacitances. The 1N6263 offers 0.1 pF and can carry 1 mA, which hits the corresponding transition at 3.9 GHz. Amusingly, at the other end of the scale, things look better: the MBRP40045 can carry 400 amps (!) and claims only 3500 pF, which at an AC RMS voltage equal to its 540 mV forward drop, would only start carrying 400 amps RMS at 34 GHz. But 3500 pF would resonate in series with its own leads at 2.7 GHz with only one picohenry of parasitic lead inductance.
Using higher radio frequencies allows you to use smaller capacitors to pass them or smaller inductors to block them. At 10 MHz, for example, 22 pF is 700 Ω, and 100 μH is 6000 Ω; at 20 MHz, 10 pF is 800 Ω and 47 μH is 6000 Ω.
The effective switching speed of logic circuits built this way might be able to reach a tenth of the carrier frequency of the RF power supply. The rectifier output needs to be smoothed over intervals in that ballpark, and the “DC” signals need to be separated from the “RF” by at least that much.
(Speaking of inductors, square-hysteresis-loop magnetic logic had somewhat similar characteristics, using a DC signal to push a magnetic core into saturation and thus pass or block an AC signal, also restoring the signal edges with the sharp hysteresis. It was much slower, though, because magnetic domains take a while to move around, like nearly a millisecond. This approach seems like it might have the potential to reach into the MHz.)
The AC current produced by a small AC voltage across a diode riding on some DC bias is an exponential function of that DC bias, which will then produce a nicely exponential AC voltage if fed into, say, a resistor or capacitor. This suggests a sort of alternative approach: rather than trying to suppress the RF entirely by back-biasing the diode, just move down closer to the threshold voltage to attenuate it more, but not so close that you’re effectively rectifying. I am not sure if this will work. Let’s try to work it out.
Let’s see if we can fit a curve to the 1N4148 figures I gave earlier. The current should grow by a factor of e every 25.3 mV, according to my understanding of the Ebers-Moll equation; in fact, though, it grows only by a factor of 10 over the 0.35 volts from .75 to 1.1 volts, when actually it should be growing by a factor of a million. Horowitz & Hill has some diode curves, including LEDs, in figure 2.8; it gives the 1N914 (supposedly the same as the 1N4148) as ramping up to about 30 mA at something like 0.7 volts.
I measured some points on the curve in the Gimp. 0V is 97 pix and 1 V is 240 pix and 2 V is 384 pix, for 143 or 144 pixels per volt; 5 mA is 374 pix, 10 mA is 301 pix, and 15 mA is 230 pix, for about 144 pixels per 10 mA. So using the formula ((x - 97) / 143.0, (301+144-y) / 14.4) we can transform points measured on the curve with Gimp’s crosshairs; we get 0.48 V, 0.1 mA; 0.59 V, 0.6 mA; 0.66 V, 1.4 mA; 0.71 V, 2.5 mA; 0.75 V, 5 mA; 0.79 V, 7.8 mA; 0.80 V, 10.3 mA; 0.85 V, 20 mA; 0.86 V, 30 mA.
This suggests that, at some temperature, a voltage wiggle between 0.59 V and 0.66 V (70 mV) will result in a current wiggle of 0.8 mA (11.4 mʊ); a voltage wiggle between 0.66 V and 0.71 V (60 mV) will result in a current wiggle of 1.1 mA (18.3 mʊ); a voltage wiggle between .71 and .75 V (40 mV) will result in a current wiggle of 2.5 mA (63 mʊ); a voltage wiggle between .75 and .79 V 2.8 mA (70 mʊ); between .79 and .85 V 12.2 mA (200 mʊ); and presumably continuing from there.
(This amounts to a factor of 5 increase in current over 270 mV, from 590 mV to 860 mV, which would be an increase by a factor of e every 168 mV, not evey 25.3 mV.)
Let’s leave aside the delicate question of biasing the circuit to 0.48 V or 0.59 V or whatever rather than 0.45 V or 0.65 V, which are points that shift with temperature. Let’s suppose it’s feasible.
The thing I’m not sure about here is that I feel like we don’t really get a place with more or less rectification along this curve. As long as dI/dV keeps growing exponentially, an AC waveform of some fixed voltage will always have less conductance on its lower half, in fact by always exactly the same ratio, so it will always be the same percent rectified, leaving aside frequency-dependent effects like reverse recovery time. But I guess that if we put a resistor in series with the diode, the resistor-diode series unit will gradually become more symmetric as more of the voltage is across the rectifier. And of course diodes have their own internal resistance, which has the same effect, even if it isn’t visible from the curve there yet.
Again I am feeling skeptical about amplification, though. At 800 mV DC and 10.3 mA, the control signal costs 8.2 mW. If we are switching a 120 mV p-p RF waveform, which turns out to be roughly 17 mA p-p, are we really getting amplification? It’s dissipating only about 0.25 μW in that diode... but maybe we can rectify that to about 6 or 7 mA DC and run it through two or three volts of other diodes in series (three or four diodes) as a control signal? Or pump it up to 1600 mV in a 13-stage Cockcroft-Walton generator?
The thing that’s confusing me here is that I confuse the energy dissipated within the device with the energy controlled by the device. In theory we could be driving those 17 mA ac (rms) through any number of other diodes in series; but I don’t have a clear idea as to how.
By contrast, it’s totally clear to me that the mechanism of reverse bias and reverse recovery time exploited by Yapo provides amplification: the energy to maintain a diode in reverse bias is just the energy to replenish the leakage current — for the 1N4148 that’s 10 nA at 20V reverse bias, which would be 200 nW, and less if you’re only reverse-biasing it to 2V or so — and the energy to maintain it at DC ground is zero. But if you put a 125MHz sinewave at 1V RMS across it, it will pass it almost without resistance when it’s not back-biased, but block it when it’s reverse-biased. This gives you near-infinite power gain, like a MOSFET.
(Yapo is actually using the 1N4148 for his high-speed rectifier and a 1N4007, a hefty power rectifier diode, as his switch, because he’s using only 4.5 MHz as his RF frequency.)
I think I have a clearer idea of how much power you can control with such a switch now. Consider a switch with some resistance in general. When its output is shorted, it’s controlling no power, because there is no voltage across the load; similarly when it’s open-circuited, because there is no current through the load. As we add conductance starting from an open circuit, the current starts to creep up, but since the current causes some voltage to be dropped across the switch, the voltage across the load starts to fall. Load power is at its maximum when precisely half the voltage is across the load, and as we move further toward a short circuit, although current continues to increase linearly with the conductance, power starts to drop. The overall shape is a parabola; this is easily visualized as being the product of two linear ramps, one from zero current at open circuit up to the current from the switch’s conductance at short circuit, the other from zero voltage at short circuit up to all the voltage at open circuit. Those are the roots of the quadratic, from which it immediately follows that its extremum is halfway in between.
Things get a little more complex when the switch impedance is partly imaginary, but I think the difference is relatively small, not orders of magnitude — except of course that the switch doesn’t dissipate power from the imaginary part of its impedance.
So the power that a switch can control to a load is at or near its maximum when the switch itself is dropping the same amount of voltage. So, for our example earlier, if the part of our waveform across the switching diode is 120 mV p-p, it could have another part across the load of another 120 mV p-p for optimal power, which means that at 17 mA p-p it can be delivering uh 0.354² 120 mV 17 mA = 0.255 mW of power to that load, just as it’s dissipating in the switching diode, about 3% of the DC bias current. So, basically, no, that part of the curve will not work for amplification.
Yapo’s “diode-diode logic” design, by contrast, reverse-biases the switching diodes to empty out a generous depletion region around the junction in order to block the RF energy. This is essentially static — a 1N4007 can, according to Vishay’s datasheet, carry 1000 mA forward and presumably an even larger amount as RF, and leaks 5 μA when back-biased to 1000 V. At the -15 V of Yapo’s design, according to Yapo’s “DDL01 DDL Hex NOR Gate” datasheet, it should presumably be more like 75 nA of bias current and 1.13 μW, while the 12 Vpp RF ends up as 4.2V RMS at something like 8 mA, which works out to 34 mW, about 30 000 times larger. Nevertheless, the circuit barely works — he had to resort to two stages of amplification per gate and over 20 components, including inductors, and supports fanout of up to about 5.