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I pulled an old computer power supply apart and salvaged a bunch of parts. One of them is a GBU806 bridge rectifier.

Wanting to learn about these things and play with my new toys, I hooked up a signal generator to the input, and my scope to the output and was able to see the half wave.

As I played with the frequency, I noticed that it got to a certain speed where the minimum voltage wouldn't go down to 0 on the scope, but the maximum would essentially stay at the peak (5 V input normal went to 2.25 V peak, eventually went down to 2.20 V).

So I assume the rectifier wasn't switching as fast as the signal, right?

What I'd like to know is how to read the datasheet so I would know the operating frequency of this rectifier, and perhaps "predict" the minimum voltage at a specific frequency. Is that possible?

I was using a Sine Wave, at 5 v P-P. Min voltage started to become non-zero at around 320Hz.

Here is the datasheet:

https://www.diodes.com/assets/Datasheets/ds21227.pdf

Adding a resistor on the output did the trick. Here's a before/after waveform. This one is 4V P-P Sine wave, at 3kHz. Interesting drop right after the crossing:

Before Resistor

After Resistor

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    \$\begingroup\$ Please edit your question with the frequencies where interesting things happened. It could be test set up, or it could be the diodes. \$\endgroup\$
    – TimWescott
    Feb 1, 2023 at 15:52
  • \$\begingroup\$ What waveform you used, sine, square, what frequencies, amplitudes and offsets, how was the rectifier connected to generator, scope, or elsewhere? \$\endgroup\$
    – Justme
    Feb 1, 2023 at 16:20
  • \$\begingroup\$ Question updated. As far as "how" it was connected, sine wave to input, -/+ to scope. \$\endgroup\$
    – LarryBud
    Feb 1, 2023 at 19:16
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    \$\begingroup\$ @LarryBud Yeah, but it leaves a lot of room for interpreting which combination of pins you used and how. Also it leaves your test setup hidden. If you are using a generator which has grounded output, and oscilloscope with grounded input, you have shorted away one diode, and another diode shorts the negative voltages to ground. Describing that with text is much harder than a simple diagram to show how it is connected. \$\endgroup\$
    – Justme
    Feb 1, 2023 at 19:40
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    \$\begingroup\$ (Interesting drop right after the crossing Reverse-recovery effect) \$\endgroup\$
    – greybeard
    Feb 2, 2023 at 17:55

3 Answers 3

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Perhaps parasitic capacitances are acting as a smoothing capacitor would, storing charge, in the classic bridge and cap setup:

schematic

simulate this circuit – Schematic created using CircuitLab

As source frequency increases, the time between peaks of the rectified output decreases, and whatever measuring equipment you have connected across that output has less time to discharge that capacitance.

In fact, the measurement device (an oscilloscope, with 40pF input capacitance?) may be the very capacitance preventing the output voltage from discharging to zero quickly enough.

Therefore you see the output never returning to zero, unlike the classic full-wave rectified signal we are shown in school.

A good way to test this idea is to connect a resistive load across the DC output of the bridge, which would discharge any capacitance there, but not so small as to overload the source signal generator. A 1kΩ across the output would be probably be fine.

In any case, I suspect the maximum frequency of operation of this setup will be more a function of load impedance (including any capacitance, explicit or parasitic) and source impedance, than the speed of the rectifier diodes themselves.

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  • \$\begingroup\$ This is the correct answer +1. Typical trr of that kind of rectifier is a few microseconds, negligible at 320Hz. \$\endgroup\$ Feb 1, 2023 at 19:35
  • \$\begingroup\$ Excellent! I did as you suggested and put a 1K resistor across the output and the bottom flattened right out at 0, with a bit of ringing after the sine wave crossed 0 going negative. Very interesting, didn't know the probe would have such an effect! \$\endgroup\$
    – LarryBud
    Feb 1, 2023 at 19:44
  • \$\begingroup\$ I've added before/after pics of the waveform with the resistor added. \$\endgroup\$
    – LarryBud
    Feb 1, 2023 at 20:01
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... how to read this datasheet so I would know the operating frequency ...

The data sheet isn't very detailed in this regard. The first two lines of the "Maximum Electrical Specifications" section says:

Single phase, 60Hz, resistive or inductive load.
For capacitive load, derate current by 20%.

That means that they're not going to give you more information -- and they're not going to guarantee suitable operation at higher than 60Hz.

Normally you'd use the diode's reverse recovery time, and you'd choose a \$t_{rr}\$ that's small enough that you can be confident that it'll never cause a problem. This is especially so in a power supply, because backflowing diodes tend to cause excess power consumption, overheating diodes, and either a loss of efficiency or escaping magic smoke.

... perhaps "predict" the min voltage at a specific frequency ...

That's not something you want to do.

There are only three places where I know of that you want to do anything other than deliberately select diodes for faster reverse recovery than your application demands.

The first, and unhappiest, is when you're designing a switching supply at the limits of the available diodes, and you're trying to figure out just how fast you can switch. In this case you're being squeezed between wanting the highest frequency (magnetics are bigger and more expensive the lower your switching frequency), and not wanting your catch diodes (or the intrinsic diodes in your FETs) to blow off the board from overheating.

If you have good diode (or FET) models, you can simulate this effect. It's complicated enough that you can't really predict it on paper.

The second is driving LEDs or BJT transistor bases, in which case you can speed up the "turn off" by deliberately applying reverse voltage to the junction to "suck out" the carriers. I've had two occasions to try this: neither design actually made it to production.

The third is in the use of PIN switching diodes in RF applications, where you use a diode whose junction is deliberately made with an intrinsic layer between the P-type and N-type layers to increase \$t_{rr}\$ in a controlled way. Treated right, such diodes will act as low-value resistors when they have average forward current on them, even when transients cause the current to reverse for parts of the RF cycle, while still acting as high-value resistors when they're reverse biased.

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    \$\begingroup\$ Sucking charge out of a BJT base by taking it a volt or so the wrong side of the emitter is a fairly standard thing in class AB audio power amps where the speedup in a large output stage is entirely worth it. \$\endgroup\$
    – Dan Mills
    Feb 2, 2023 at 14:57
  • \$\begingroup\$ @DanMills: thanks! I've never done serious audio design, so haven't encountered that. \$\endgroup\$
    – TimWescott
    Feb 2, 2023 at 16:41
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    \$\begingroup\$ It buys you two things, less cross conduction at high frequency (We have some by design to maintain gm at crossover, but you don't want the bottom device conducting much current when the top one is pulling up by maybe 100V or so, makes much heat and does the SOA and second breakdown no favours), but it also reduces the tendency of an output (or driver) transistor to stick to a rail when you clip the thing, and improving clipping behaviour can very much improve the sound at high volume. \$\endgroup\$
    – Dan Mills
    Feb 2, 2023 at 22:36
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Your standard pn junction diodes can't turn off instantaneously.

In simple terms, when a diode is conducting in the forward direction, minority carriers are injected across the junction, and the presence of these minority carriers allows current to flow through the diode.

Even if you stop applying a forward voltage, those carriers are still there. Given time, they'll eventually hook up with a majority carrier and recombine, emitting a photon in the process. But this isn't an instantaneous process--it can take tens or even hundreds of microseconds before all the carriers recombine. And while they're still there, the diode can conduct in either direction, not just forward.

The time a diode takes to turn off is called the reverse recovery time, and can usually be found on datasheets. For high frequency operation, you need diodes with low recovery time; there are two main options for this.

First, you can get what's called a FRED (Fast-Recovery Epitaxial Diode). This is a pn diode with additional doping of usually either gold or platinum, elements that interact with the electronic structure of silicon in ways that make minority carriers recombine faster (for advanced readers: they produce traps with states in the middle of the bandgap). These typically have reverse recovery times on the order of tens of nanoseconds, three to four orders of magnitude better than your typical diode. As a side effect, this also increases the diode's forward voltage.

Alternatively, you can use Schottky diodes. Schottky diodes use a metal-semiconductor junction instead of a semiconductor-semiconductor junction, and as a result of this are majority-carrier devices; there are few if any minority carriers in Schottky diodes even while conducting. As there are no minority carriers, this means there's no need for the minority carriers to recombine, so a Schottky diode has a theoretical reverse recovery time of zero. As a bonus, Schottky diodes have a lower forward voltage than pn diodes. The downsides to Schottky diodes are that they have high reverse leakage and low breakdown voltages; you can't use a silicon Schottky for anything beyond a hundred volts or so. Silicon carbide Schottkys exist, and can work at very high voltages (I've used a 3.3 kV one before), but they have a higher forward voltage even than silicon pn diodes.

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