Low-voltage rectification

I have a 19kHz square wave line-level audio signal (~2Vpp) that I hope to rectify in order to drive an infrared LED at 38kHz.

So far, I've been using a pair of LEDs, wired in parallel in opposite directions, to achieve the same effect, but the resulting system is incredibly sensitive to misalignment and has a very short range (but does work!).

My theory is that if I can rectify the signal in some other way and then drive a single LED, this should eliminate the alignment issues.

Ordinarily, I'd use a standard bridge-rectifier arrangement of diodes, but I'm concerned that the voltage drop may be too high. I've looked for diodes with a low Vf and haven't found much.

What would you use?

• How do you intend to convert the 19KHz signal to 38KHz? – Bruce Abbott May 28 '16 at 10:46
• Please clarify: if the square-wave is on-off type then rectifying it won't change it. If it is alternating polarity then rectifying it will give a constant 'on' - 0 Hz and not 38 kHz. 19 kHz is on the upper threshold of human hearing. Why don't you explain the application a bit more. What is the data to be transmitted. What is the receiver. – Transistor May 28 '16 at 10:46
• I would try 4 Schottky diodes, these have a lower voltage drop than the Silicon diodes generally used in bridge rectifiers. So build your own bridge rectifier using Schottky diodes ! If you do not know how to connect the diodes, google for "bridge rectifier schematic"and you'll know. – Bimpelrekkie May 28 '16 at 10:54
• @BruceAbbott It's a well know fact that if you rectify a 50 Hz sinewave from a mains transformer using a bridge rectifier you get a 100 Hz signal which looks like a sinewave with the negative part "mirrored up". The same applies for a 19 kHz sinewave. – Bimpelrekkie May 28 '16 at 10:57
• But you said that you have a 19kHz square wave. Is it not actually square? – Bruce Abbott May 28 '16 at 10:59

A bridge rectifier may not work properly because:-

1. Full wave rectifying a square wave results in an almost constant DC output, not the doubled frequency that you want.

A square wave can be shaped into a more useful waveform by low pass filtering. However if your audio source is bandwidth limited then your 'square' wave may already be a sine wave.

1. Even with a perfect lossless rectifier you may struggle to get sufficient amplitude. A 2Vpp signal peaks at 1V, but most infrared LEDs require at least 1.1V to produce useful output. You need some way to increase the amplitude, or add a bias voltage to the signal.

If an external power supply is not available and you must power the LED directly from the audio signal then amplifying the voltage will be difficult. However developing a bias voltage is relatively easy.

The following circuit has positive and negative half-wave rectifiers (D1 and D2) with low value filter capacitors creating a small DC bias voltage with high ripple. High ripple is usually bad, but in this case we want the ripple because it is the signal!

On positive half cycles C1 charges up to about 0.6V, then discharges (through the LED) to about 0.4V during negative half-cycles. C2 performs the same function but on opposite half-cycles. This bias voltage is enough to make up for rectifier voltage drop and provide a small boost to get into the LED's operating voltage range. R1 and R2 reduce loading on the input, and limit diode current if a larger signal voltage is applied.

simulate this circuit – Schematic created using CircuitLab

I have a 19kHz square wave line-level audio signal (~2Vpp) that I hope to rectify in order to drive an infrared LED at 38kHz.

An LED is a diode, and as such, a rectifier!

But I assume you want to make use of both negative and positive half-waves of your signal.

So, use a bridge rectifier?

I've looked for diodes with a low Vf and haven't found much.

Um, Schottky diodes, mumble mumble, Wikipedia type of semiconductor diodes, mumble mumble:

Schottky diodes

Schottky diodes are constructed from a metal to semiconductor contact. They have a lower forward voltage drop than p–n junction diodes. Their forward voltage drop at forward currents of about 1 mA is in the range 0.15 V to 0.45 V, which makes them useful in voltage clamping applications and prevention of transistor saturation. They can also be used as low loss rectifiers, although their reverse leakage current is in general higher than that of other diodes.

Notice that building a perfect rectifier will not give you an 38 kHz signal from your ideal 19kHz rectangular wave – just DC. However, your wave not being perfectly rectangular with infinitely sharp edges, and your rectifier still having "dead" regions, yes, you can get something that blinks with 38 kHz this way.

Another option, if you're after the signal, not its power, is to simply AC couple it:

That way, your diode will "pulse" with 19 kHz, as the voltage at the connection between the voltage divider, the cap and the LED fluctuates with the input voltage.

Also, note that spectrally, any rectangular signal has not only its frequency, but also all of that frequency's harmonics. That's simply because the Fourier Transform of a unity rectangular signal is the (scaled/stretched according to amplitude/frequency of the rectangle) is a sinc function, $a \frac{\sin \pi f x}{\pi f x}$. That one has sidelobes at every multiple of $f$:

You can use the RC high pass in the schematic above to select the first sidelobe you want to visualize with the LED by selecting appropriate values for the two R and C (100nF/47KOhm would be a start); that will not be sharp enough to fully suppress 10 19kHz main lobe, but just add another stage of RC (or LC, if you need to keep more energy) filter afterwards, and you should be fine.

• All very nice, but rectifying a 19kHz square wave will give you DC (if zero centred) or just the original signal (if positive biased, which seems unlikely given the audio output. In fact, depending on the source we may find it bandwidth limited so not really square anyway (if the source is, for example, a PC sound card). – stefandz May 28 '16 at 11:16
• That's why you will not zero-center it, but bias it, e.g. to half V_supply. Biasing is exactly why I'm using AC coupling here. and a square wave definitely, as explained, has not only 19kHz frequency components – if it's sufficiently "truely" a square wave. If it is not, an active input stage (a.k.a random transistor) might be used to push its shape to look more rectangular. – Marcus Müller May 28 '16 at 11:25
• But I fully agree with the remark on rectification. Let me come up with an edit. Like I put this, it's not clear. – Marcus Müller May 28 '16 at 11:27
• Still doesn't get you the requested 38kHz – stefandz May 28 '16 at 11:28
• @stefandz it does, see my edit – Marcus Müller May 28 '16 at 11:30

If you have access to another power source besides the 2V signal itself then I see two options.

1) you can run the 2V signal into a non-inverting amplifier with a gain of 10 (an op-amp, a 10K resistor, and a 90K resistor) to make a copy of your ±1V signal that is ±10V. Then use a normal bridge rectifier. Due to the higher signal level the diode drops wont be as much of an issue.

2) You can make some "ideal diodes" from some op-amps and normal diodes. The op-amp can keep the drop to nearly 0, but only with the help of some other power source.

• There's no other power source available, unfortunately! – user111414 Jun 3 '16 at 11:01

with a square wave, you won't get 38 kHz from a 19 kHz signal. If it is a sinusoid (or if you can filter it to approximate one), then you would in theory get 38 kHz, but in practice, the conversion wouldn't be very efficient, an the LED signals not very clean.

A bridge rectifier with very little drop is this one used here - Storing the the charge from a MOSFET Bridge Rectifier

• I'm a little off square - see the yellow signal trace at i.imgur.com/xi6CckC.png – user111414 Jun 3 '16 at 11:02

If you want to have very low drop, and you have a power source available, you can do this (bipolar supplies but it could be modified for single supply):

simulate this circuit – Schematic created using CircuitLab

Here is a sim of the LED current with a sine wave input:

• There's no other power source available, unfortunately! – user111414 Jun 3 '16 at 11:02