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Since When Are Diodes Active Electronic Components?

I provided some engineering assistance a couple years ago for a company designing photo-voltaic array (PVA) based power inverters. These are sometimes referred to as micro-inverters since they are rated at a few hundred to a few thousand watts (rather than tens or hundreds of kilowatts). At first glance, I assumed all the high technology was in the inverter. The PVA portion seemed pretty low-tech. I was mistaken.

In a PVA there are long series strings of solar cells. The individual solar cells are matched to the extent that is practical (output voltage vs. light intensity). Unfortunately, that's not good enough.

When a section of the panel gets shaded (by trees or clouds), its output drops. So you've lost the matching you tried to achieve. With the shaded section producing less voltage, current gets pushed through from the other cells. This results in heating of the darker cells. You're wasting the energy you're trying to collect. The fix has been to set up sections in the PVA and add bypass diodes like this:

(Source: Texas Instruments)

(Source: Texas Instruments)

That helps, but it's not perfect — it reduces the heating some and moves the location of the heat source. Here are some typical numbers: With 10A flowing through the panel, a standard silicon diode (forward diode drop of 0.6V) would waste 6W. Even a Schottky diode with a forward drop of 0.4V will waste 4W.

When I cranked through these numbers, I decided what we needed was some sort of electronic switch that acted like a diode but had active control circuitry. I thought of a BJT or an FET, but I would need a separate isolated power source for base current or gate drive for each device. What had seemed like a good idea for a few minutes quickly looked rather impractical.

Apparently, someone else went through the same analysis and concluded almost the same thing — except for the part about it being impractical. Someone came up with a simple way to put the needed power supply circuitry right where it was needed, like this:

(Source: Texas Instruments)

(Source: Texas Instruments)

This is the SM74611, a “smart bypass diode” from Texas Instruments. It's intended for use with PVAs. It has a low-power switched capacitor boost power supply plus an N-channel power FET. The power supply generates sufficient voltage to drive the gate and fully enhance the conduction channel. Compared to those two diodes cited above, this device drops 26mV at 8A forward current, so that's a power dissipation of just over 200mW. The devices come in a D2PAK much like the standard Schottky diodes. You would normally mount these on a small PC board with quick disconnect terminals, in a totally sealed junction box. Low power dissipation means low temperature rise. I'll leave it as an exercise for the student to estimate temperature rise in a sealed box with a 6W load inside.

I wish I'd had these two years ago.

14 comments on “Since When Are Diodes Active Electronic Components?

  1. Michael Dunn
    March 21, 2013

    Wow, nice part. I was wondering how it could work once turned on. Depletion FET? Cycling? The datasheet confirms the latter 🙂

  2. Brad Albing
    March 21, 2013

    Well, not such a good application for a depletion mode FET, tho' I like them for some apps (hey! blog idea). Enhance mode makes perfect sense – as long as you've got the juice to enhance the gate region. Hence, the charge pump chugging along in the background.

  3. Vishal Prajapati
    March 22, 2013

    I have done a bit of calculation for the deployment of the PVAs. But never thought of wastage of watts from the reverse protection diode. I will definitely try to simulate this solution built entirely with discrete componts. This is really interesting.

  4. DEREK.KOONCE
    March 22, 2013

    Very elegant solution. The trick is to balance the capacitor size and charging to maintain the FET on. Depletion FETs will not work since they require negative voltage to turn off. I suspect the cap is silicon based to ensure optimum parameters; otherwise, it may be discrete to keep silicon costs to a minimum. Be interesting to see the interior.

    Diodes, in my book, are considered active since they are silicon based and have an on/off state of some form. But this is not a typical diode, only labeled as such for simplicity.

    Second musing, FET selection or design is important. It could be the FET they purchased from Ciclone so that there is minimal loss on the gate drive. But I believe Ciclone FETs did not get up above 20 or 30 V.

  5. David Maciel Silva
    March 25, 2013

    Brad,

    We have here a solution with cost less than $ 5!

    How many diodes would be required for the final application?

  6. Brad Albing
    March 26, 2013

    It's simply a function of the number of sections in the panel. The companies that make the PVAs have already set them up in sections (probably vertical sections with the same number of cells per section or column).

  7. Gregst
    March 27, 2013

    The upper scheme, with the diodes, would cause a great variability of the output voltage, from the full voltage to 1/3 of it, as shown. Hence, it needs to be followed by some voltage stabilizer. You can imagine how much energy would be wasted in such a stabilizer.

    Now, the TI part does not provide much of unidirectionality that characterizes the diodes. The current could flow from source to drain and from drain to source. Yes, as it is N-mos, it would conduct better in one direction, yet the net effect is that it effectively shunts the solar cells, allowing both forward and reverse currents. Thus, any light falling on the shunted cells would be used to heat these cells. Not only it is wasteful, it might even destroy the cell, or diminish its lifetime.

    If I'm mistaken in anything, please correct.

  8. Brad Albing
    March 27, 2013

    The assumption here is that there is current flow from the remaining panel sections that are still brightly illuminated, so the current flow (as far as the “diode” is concerned) is in what we will call the “forward” direction (observeing the network as if the diodes were regular PN junction diodes; conventional current flow). So the dark panels are producing low potential and appear to be rather high-resistance, so the current flow favors the diode path.

  9. Brad Albing
    March 27, 2013

    Oh – current always flows in the same direction, so that while the MOSFET could allow curent flow in the reverse direction, that consideration is moot.

  10. Brad Albing
    March 27, 2013

    And yes – output of the PVA goes to a voltage stabilizer, a.k.a. a switching power supply; buck, boost, or buck-boost as the sytem needs dictate.

  11. Gregst
    March 27, 2013

    Brad, dear, if you ever done a classic DC net analysis, you surely know that current is not always flowing in the same direction, and on a same wire shared by 2 closed circuits you could have 2 currents flowing in the opposit directions. Use Kirchhoff's current and voltage laws. As an experiment, take a battery and connect it to an LED. Next, take another battery, and connect both electrodes of it to one of the wires hooked to the LED. You can position your battery in any direction + closer to LED, or + farther from LED. Let's see  if your new circuit is going to affect glowing of LED, and what would it do to your second battery. Just stay away far enough, it might exlode. There would be 2 currents flowing in the same wire: one, to LED, second, from the short-circuited second battery.

    It appears that your MOSFET  (almost) short-circuits a number of solar cells. I'm saying “almost” because the current would be limited by the MOSFET resistance, and interlal resistance of the solar cells. Nevertheless, all light falling on the shunted cell would be used for their heating.

    The interesting question is: does the solar battery itself increase the absorption of light ? I.e., get a solar battery not connected to anything and radiate light on it from a powerful source. A part of that light would be reflected, a part of it absorbed by the cell and converted into the thermal energy. Now, short-circuit the cell. Would it cause an increase in the cell's temperature ? If yes, then why ? It is presumed that only a part of the absorbed light energy would be converted into electricity. So, if the cell is not connected to anything, all the absorbed energy is converted into heat. If it is short-circuited, then a part of energy is converted into heat, another part is converted into electricity, and then into heat, so the net should be the same. I wonder, if it is indeed.

  12. Brad Albing
    March 27, 2013

    >Brad, dear, if you ever done a classic DC net analysis, you surely know that current is not always flowing in the same direction, and on a same wire shared by 2 closed circuits you could have 2 currents flowing in the opposit directions. Use Kirchhoff's current and voltage laws.

    Greg, dear, just to keep this discussion easier to follow, let's consider a long string of PV cells connected in series and let's go back to using the old-fashioned real PN diode connected across cells. Let's take the extreme case and put a diode across each PV cell with its cathode on the positive terminal of each PV cell. As the output of any one cell drops sufficiently low and it starts to appear more and more resistive, current caused by the other cells' EMF will start flowing thru the diode rather than thru the low output cell. With the active diode IC, the result is the same, except that the forward drop is a lot lower.

    >As an experiment, take a battery and connect it to an LED. Next, take another battery, and connect both electrodes of it to one of the wires hooked to the LED. You can position your battery in any direction + closer to LED, or + farther from LED. Let's see  if your new circuit is going to affect glowing of LED, and what would it do to your second battery. Just stay away far enough, it might exlode. There would be 2 currents flowing in the same wire: one, to LED, second, from the short-circuited second battery.

    An interesting experiment, which might show some interesting insights if you modeled it properly by introducing some resistance in different legs so as to mimic real world conditions. Then you could write the loop equations for the different portions. Absent those resistances, all curents would be infinite. A nice Electricity 101 problem. But not the one under discussion. Per my paragraph above, the current flow is very easy to follow.

    >It appears that your MOSFET  (almost) short-circuits a number of solar cells. I'm saying “almost” because the current would be limited by the MOSFET resistance, and interlal resistance of the solar cells. Nevertheless, all light falling on the shunted cell would be used for their heating.

    The MOSFET short circuits the PV cell in the same way that a diode short-circuits a cell – i.e., not at all. Sketch out a series circuit with any number of cells with a shunt diode across each cell. Assume the cell is a voltage source whose output varies with the incident light. Add a series resistor to each cell. Assume that the resistor varies inversely with the incident light. Should be pretty clear where the current flow (or flow of electrons, if you prefer) is going and why.

  13. Gregst
    March 27, 2013

    Dear Brad, there was no argument about a regular PN diode. You are making a definition trick labeling a power mosfet with the control circuits as an “active diode”, implying that it has the I-V characteristics of an ordinary diode and stopping there. Could you please explain how would a power n-mosfet prevent the effective short-circuiting of the solar cell ? Yes, Nmos would conduct much better in one direction than in another direction. Yet its directionality is not equivalent to that of a PN diode. For example, there are no pmos transistors working at high currents (~30A), therefore nmos transistors are routinely used in the high-current regulated power supplies (say, those for inductive heating) for conduction in both directions.

    It is mosfet, not bipolar. Drain could function as source, and source as drain. The electron channel is there, albeit with the pinch-off that might be a wrong side. Why that channel would not shunt the solar cells and allow the current to flow through the shunted solar cell loop ?

    Again, I'm not claiming that it is bound not to work, but that if it works, it is due to the physical principle that the light energy is either converted in heat or in electricity. Hence, it might be possible to safely short-circuit a photoelectric cell. I've never tried though.

  14. Brad Albing
    March 28, 2013

    I think I see where our misunderstanding is coming from. The FET is only turned on as needed. Specifically, when the FET's body diode starts conducting (meaning the “dark cell or cells” condition has occurred), that produces the 0.7V drop (approx.) in the diode's forward direction. That small voltage is enought to start the charge pump circuitry working. That produces enough voltage to turn on the gate (enhance channel conduction) for a while until the capacitor in the charge pump circuitry bleeds down. Then the FET turns off and the process repeats. I should have pointed to that explanation in TI's data sheet.

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