Zero Drop ORing or Drop Less Diod

If this is required at (low) signal level use Op-Amp configured as diode:

Else use active rectification with Mosfets. (But you will need seperate higher voltage for gate control)

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You might look for a germanium diode instead of a silicon diode. Germanium has a voltage drop of around 0.3 volts as opposed to the 0.6 volts of silicon. I'm sure germanium diodes must still be manufactured.

What is your inverter used for? Is it for a solar power application? If so, I'd consider putting more solar cells in series to get a higher voltage to work with, and thus a lower power loss (as a percentage of the power developed) through the diode(s).

What is your actual power source capable of wattage wise and what power level is your inverter capable of and at what output voltage?

1.6 volts is very low for any serious power production levels.

You can make an "active diode" by using a comparator to sense when the "diode" should be conducting, and using the output to turn on a MOSFET switch. There are comparators, mosfets and other electronics that can run on as little as 1.6 volts (was it 1.6V max, or 1.6V minimum that you have?), but it's a more difficult task than working with higher voltages. Sometimes folks add a low-power dc-dc converter to step up their 0.9 to 1.8V, etc., to a higher voltage like 3.3 to 5V. They run all the low-current control electronics from the higher voltage, which greatly simplifies the task.

Back to power diode ORs, there are quite a few specialized ICs that have been created to do this task. They include the comparators, mosfet drivers, and sometimes the mosfets as well. For example, TI's TPS2102 and TPS2112, and Linear Technology's LTC4411, LTC4412, LTC4416 and LTC4357. The LTC4354 is a negative-voltage version.

These all require higher than 1.6 volts to operate, so you may have to make your own design, but they are useful to study and see how their schemes work. The LTC4357 is shown above. It uses N-type mosfets, so the substrate diode (not shown) conducts, and the mosfet is turned on to make a low-voltage conduction, as if it was a low-voltage diode.

TI's TPS2105 switch is pictured below. Unlike the other parts mentioned, this part requires an external signal to decide which input to activate. It uses an n-channel mosfet (top) for the ordinary high-current mode, and a p-channel mosfet (bottom) for low-current standby-power mode. The 0.25-ohm N-type mosfet requires a positive voltage wrt its source pin = the output, hence the need for a charge pump to create the gate-drive voltage. It takes 18uA to run this pump. In the standby mode with the 1.3-ohm P-type device on, only 0.75uA supply current is required. The minimum operating voltage is said to be 2.7 volts.

The TPS2105 switch uses internal mosfets. But if you roll your own circuit, you'll need to find low-voltage parts. Examples of N-type mosfets that'll work with 1.6 volt gate drive are the FDG327 and the Si2312. Low-gate-drive-voltage P-type mosfets are harder to find, but there are some 2.7-volt parts that may work with 1.6 volt drive at reduced currents. Perhaps a ZXM61P02 or an IRLML6402.

Many redundant power supply designs use power Schottky diodes with their nominal 0.4V forward bias voltage to OR sum the output voltages. This is not as low of a dropping voltage as a germanium diode but germaniums diodes are not fabricated today for large power handling capability. There are many large current Schottky devices available on the market today that will do this. I prefer to heat sink each diode separately, this will assure that only one power supply will be driving the circuit. This is because the diode handling the current will self heat and will then have a lower forward voltage drop. This will significantly reduce the thermal stress on the backup supply. Thus the remote possibility of both supplies failing at the same time will remain a remote possibility.

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several 1n270 germanium diodes in parallel should do it. They will handle 100volts. They usually have a voltage drop of .3 volts. I have found some batches vary.

they have a surge capacity of 0.5 amps for one second and an average rectified forward current capacity of 40 milliamps.

You did not mention your current usage. knowing that would help

Several of you have mentioned low-voltage-drop germanium diodes, etc., but let me point out that the O.P. mentioned "Zero Drop ORing or Drop Less Diode", which I take to mean zero-voltage-drop.

Of course, very few things will truly approach ZERO, but while we may have to live with some series resistance, one idea is to avoid any additional relatively-large semiconductor-diode threshold-voltages.

That's what "active diodes" made from switched FETs accomplish.

You mention the 100V capability of the 1n270 germanium diode. To match your example, the FDB3632 mosfets suggested by LTC for use with their LTC4357 chips, in the drawing I posted, are rated at 100V and have an impressive low 7.5 mΩ series resistance.

As you can see from the Characteristic Curves for the part, one suffers less than 8mV of drop at a current of 1A and under 40mV at 5A.

The 1N270 germanium diodes are impressive, but now with a little extra electronics, like an LTC4357 ideal diode controller, and a '3632 (right), for about $5 a pair, we can do much much better.

OTOH, MicroSemi 1n270 diodes (left) cost only 99¢ each at Jameco, or $1.13 at Newark, with over 1000 in stock at each distributor, so maybe they still have a place.

I give you a good answer on this one. I was unaware of this part. Definately better especially at higher currents.

thanks,

and sorry not to provide details,

i want to OR current and voltage analogue signals to feedback to KA3525 ERR amplifier, so the current is only few miliampers. Both entities need tobe controlled saperately.

deepak

Excuse me but giving us more confusing details doesn't really help. Which undefined entities need to be controlled. The KA3525 is a SMPS controller chip. With this you can control the voltage OR the current. As the load impedance (X) changes one parameter or the other (V or I) must change. Remember V=IX. You cannot have two constants in this equation and one changing variable for it to be a true equation.

Obviously he is trying a 'shortcut' by ORing both current and voltage signals into KA3525 error amp so both adjust voltage and limit current

Nothing at all is obvious here. Well unless you count that the obfuscation is obvious.

I made the following drawing to show you an "almost zero voltage drop diode" by using a mosfet (i.e. IRLM6401).

When a positive supply voltage is applied on Vcc, the npn transistor is turned "on" forcing the mosfet to be turned "on" too. Mind that the 'drain' (and not the 'source') must be connected to the Vcc, due to the internal parasitic diode of the mosfet (as shown in the figure). In this way, a negative voltage on Vcc will not pass through the 'diode' of the mosfet.

The 'load' can be a whole circuit. In some cases, when Vcc is not present, a small voltage may exist on point A due to other signals or dc voltages applied on the 'load' circuit. So this voltage should not be passed on the Vcc track (because this voltage could cause problems to other circuits 'hanging' on the Vcc track). In this circuit, when Vcc is not present and such a voltage exists on A, this voltage will not pass to the Vcc track because the mosfet is turned "off" (as the 'gate' and 'source' of the mosfet have the same voltage). When Vcc is present (i.e. a positive supply voltage) then the mosfet is turned "on" and the Vcc is passed to the 'load' circuit.

You have a significant typo there. Your MOSFET is backwards. The body drain source diode junction will always be ON.

No, Redfred, there is no error. This circuit acts as a 'diode' (having a very low dropout voltage). It accomplishes two tasks:

1) If Vcc is positive, it will pass to the 'load circuit'. If Vcc is negative, it will not pass.

2) If there is no voltage on Vcc and there is another small positive voltage on point A (due to -e.g.- signals or other dc voltages on the 'load circuit'), this voltage will not pass to the Vcc track.

Agree with that 'patrida' but still I see a flaw. This will start conducting at about 0.6V (one diode drop)and will fully conduct (load will see only Rdson) at Vcc about 0.6V+ gate threshold. It will also self destruct at about 30V. I think Redffred didn't understand at start that the Mosfet's junction actual current will be reverse than normal. You just short the diode. How about using the cirquit in my last post (THAT NO GENIOUS SEEMS TO NOTICE THAT IT DOES ELIMINATE Vdrop) driving an other open-loop gain op-amp with -In at ground, (fed with less than 20V external so not to destroy Mosfet gate)

So: Theoretically zero Vdrop, Problem: Op-amps nead low-power seperate Voltage:

Your op-amp circuit does do an excellent job of removing all of the voltage drop problem. It also suffers from the problem you point out that it must have a separate higher supply source to function. Your circuit works elegantly though for instrumentation to measure a signal above zero.

This is really a fundamental engineering trade off problem that requires knowing the whole system instead of just the goal of making an ideal diode. One can design a circuit that mimics the pertinent properties of an ideal diode to work in a design. But one cannot make one circuit design to replicate all aspects of a real ideal diode.

Yes, Simple Mind, at first Vcc must be more than 0,7V. In this way, the npn transistor will be turned "on" and, also, the voltage will pass through the parasitic 'diode' (so that point A will get a small positive voltage). But, after this, the mosfet will be, also, turned "on" bypassing its parasitic 'diode' (and giving a really small voltage drop).

[In general, a power supply voltage (i.e. Vcc) is usually more than 0,7V and less than 8V (which is the max V_GS voltage for this mosfet). (E.g. it can be 1,8V or 2,5V or 3,3V or 5V for digital circuits). So this circuit is okay for the most cases.]

Hi guys - my first post on this forum and my electronics knowledge is a bit rusty (mostly from my childhood). So please be gentle if I'm talking nonsense ;-)

I currently have a tiny pager motor which I'm driving directly via a 1.5V AAA battery - so far so good. Now what I'm trying to do is to limit the polarity going to that motor (for certain mysterious reasons - LOL) - meaning I only want it to spin if input A is negative and input B is positive.

I can accomplish this by simply putting a regular junction diode in series - done. You guys already covered that above. But as expected it'll eat up half my current! So, I started to look at a Schottky diode and that one will work but still eat up ~0.3V - thus affecting the rpm of my motor.

Will the circuit shown above work for my purpose? Can I simply use this circuit and replace the 'load' with my tiny pager motor?

Also, I looked up 'BL847B' and that transistor does not seem to exist. Is that supposed to be a 'BC847B'?

Any input would be greatly welcome.

Also, I forgot to ask - the IRLM6401 MOSFET appears to be tough to get over here in the U.S. - can someone propose a design based on a more accessible component?

Thanks!

Thank you for such comprehensive insights! Truth to be told much of it is a bit over my head but I do understand that you are using simple PNPs to create a threshold switch (I think ;-))

Funny you should suggest a ZTX1149A - some fellow I'm thinking of hiring to help me out with this proposed a ZXTN19020DFF, which I assume is similar to yours.

Is there any forward voltage drop with your design? Recall that I am trying to pull all the current available in my AAA battery to spin my motor.

The ZXTN19020 transistors are NPN, which means they would have to switch the low side. It takes a PNP type to switch in a high side BJT circuit. The MOSFET circuit you put up (first posted by G.K.) is high-side, but for floating loads, like motors, LEDs, heating elements, etc., low-side switching circuits are fine. In fact they can work a bit better than high-side circuits, since NPN and nMOS parts are superior to PNP and pMOS parts.

To compare the ZXTN19020 and the ZTX1149A, we look at the switch voltage drop, which is the V_CE(sat) spec in the datasheets. The '19020 datasheet says 30mV (max) at 1A with 100mA drive, or 65mV with 10mA drive. It has a V_CE(sat) vs I_C plot (left) that says 35mV at 1A with 20mA drive (I_C/I_B=50) The '1149A has a similar plot that says about 80mV under the same conditions. So we see that the NPN part is better than the PNP part. But the PNP part will pass 95% of the battery voltage on to the motor at 1A, compared to nearly 98% for the NPN, so both of them will do very well.

Question: what's the maximum current your motor can draw? Also, do you have other voltages available (positive or negative)? Maybe some higher voltages?

If the switch choices are sufficiently good, and work at low voltages, your real choice should involve the most effective circuit for your needs. The original thread was for a simplified OR'ing switch. But you want something different, so there may be an even more suitable circuit we can suggest, if we know your entire goals. You said you were "trying to limit the polarity going to that motor (for certain mysterious reasons - LOL) - meaning I only want it to spin if input A is negative and input B is positive." Can you please tell us more about your negative A and positive B, and, for that matter, maybe something about the "mysterious reasons"?

Another question, I assume this will be a little prototype you want to solder together and use? If so, you may prefer "through-hole parts" like the TO-92 package with its three convenient wire leads, as opposed to the SOT-23 package, which really wants to be mounted on a designed and manufactured printed-circuit board (PCB). We call that SMT, or surface-mount technology. The '19020 is in an SOT-23, but the same die that's inside it is probably available in a TO-92 package, or certainly a similar-performing die will be.

Hi molecool. As I wrote in my post#18, this circuit is okay for the usual Vcc of digital circuits (i.e. from 1,8V to 5V). I'm not sure about the lower voltages. (Perhaps you should try it and see.) Probably, the solution that Mr Hill suggested is better for your application.

OK gents always a pleasure talking with you. S.M.