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A simple and inexpensive ideal-diode MOSFET circuit

September 26, 2017 By Aimee Kalnoskas

by Crutschow

When you parallel a battery with another battery or other source, the batteries are often required to be back charged. This can be done with standard diodes, but that gives close to a half volt drop — even with Schottky diodes. This is especially problematic with low-voltage batteries, where that drop is a significant percentage of the battery voltage, noticeably reducing efficiency and battery life.

To minimize this forward drop you can configure a MOSFET as an ideal diode, which has a very low drop in the forward direction (equal to the current times the MOSFET’s ON resistance) while blocking the current in the reverse direction.

Below is the LTspice simulation of a simple ideal-diode MOSFET circuit. It uses inexpensive components consisting of a P-MOSFET (for use in the positive rail) with a dual PNP transistor and two resistors.
MOSFET circuit

Q1 and Q2 form a current mirror circuit. The indicated values of R1 and R2 cause Q2 to be on and thus M1 off (Vgs ≈0V), when there is no voltage difference between the drain and source of M1. The mirror has a gain of ≈130 from the voltage difference between the two emitters to Q2’s collector voltage change.

In the forward direction (output voltage lower than the battery voltage) the current mirror becomes unbalanced due to the difference in emitter voltages, such as to turn Q2 off, which puts the P-MOSFET gate near ground potential, turning it on. This allows current to flow from the battery to the output (left to right) with a low drop. (MOSFETs conduct equally well in either direction when on.)

When the output voltage becomes slightly higher than the battery voltage, this voltage reversal across the MOSFET unbalances the current mirror in the opposite direction, causing Q2 to turn on. This causes the MOSFET gate voltage to rise, reducing Vgs [V(G,Out) in plot], which turns it off and prevents reverse current flow.

This can be seen in the simulation, as the current only goes out of the V1 battery when the V2 output voltage is lower than the battery voltage, and doesn’t flow in the reverse direction when the output voltage is greater than the battery voltage. The maximum voltage drop, when the battery is providing 2A current is ≈32mV with the MOSFET shown, demonstrating the near ideal diode operation.

The current mirror operation is very sensitive to any offset between the two transistor base-emitter voltages, which could possibly allow some current conduction in the reverse direction. It is thus recommended that a matched transistor pair be used, such as the DMMT3906W shown on the schematic (basically two 2N3906’s in one package), which have their Vbe matched to within 2mV max and are thermally connected.

(The simulation was done with 2N3906‘s which are perfectly matched in the simulation, unlike real life.) The DMMT3906W pair are quite inexpensive, selling for U$0.37 here, for example.

The P-MOSFET selected should have an on-resistance small enough to give a low voltage drop when conducting the maximum battery load current. If the battery voltage is less than 10V then a logic-level type MOSFET should be used which have gate-source threshold voltages (Vgsth) of less than 2V.

One of these circuits can be used at the output of each battery; however, many are in parallel.

 

You can read more articles by Electro-Tech-Online “Well-known” member, Crutschow, here.

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  • What’s the difference between a MOSFET and a JFET?

Filed Under: Featured, Industry Experts, MOSFETS Tagged With: diodesincorporated, electrotechnoline.com, mouserelectronics

Reader Interactions

Comments

  1. Eric says

    April 17, 2018 at 1:38 pm

    Thank you for this article, I think it will suit my DC UPS project perfectly. I plan on having one of these ideal diode circuits after my battery, and one after my AC/DC power supply because I don’t want any reverse current into either supply. The outputs of the 2 mosfets will then be connected to the load.

    In the article you said: “When the output voltage becomes slightly higher than the battery voltage, this voltage reversal across the MOSFET unbalances the current mirror in the opposite direction, causing Q2 to turn on.”

    Do you have any idea of what that voltage difference is? So if the battery and AC/DC power supply are both at 19.5 volts, I assume they will each share the load and current will flow forwards through both ideal diodes. But as soon as the AC/DC power supply drops to 19.4 or 19.3 V then will the battery backup supply 100% of the current? Or does the voltage difference need to be higher?

    Thank you for your help

  2. Aimee Kalnoskas says

    April 18, 2018 at 8:23 pm

    Thanks for your feedback, Eric. Crutschow has asked that you post your comment to the original article as it appeared on our Electro-Tech-Online engineering community forum. Check it out as Crutschow writes quite frequently there. Here is the link to the article: https://www.electro-tech-online.com/articles/simple-inexpensive-ideal-diode-mosfet-circuit.817/
    cheers,
    Aimee

  3. David Albert says

    January 22, 2024 at 12:23 pm

    A nifty circuit; thank you!

    Unless I’ve mis-read it, something to be careful of here is the maximum voltage for V2 (which must not exceed roughly 5v7)…fine for the reference application of a 5V supply with 3v7 battery backup, but would be a problem if the main supply were 6V or higher.

    Vebo_max (emitter-base breakdown voltage) for Q1/Q2 is -5V for many dual-PNP transistors including the one shown in the reference circuit. When the battery (V1) is removed (V1=0V), then Veb of Q1 = V2 less Vbe_sat of Q2. This is OK if V2 is 5V, but if V2 were 6V, the EB breakdown voltage of Q1 would be exceeded.

  4. Aimee Kalnoskas says

    January 22, 2024 at 3:31 pm

    Thanks for your feedback, David. Crutschow has asked that you post your comment to the original article as it appeared on our Electro-Tech-Online engineering community forum. Check it out as Crutschow writes quite frequently there. Here is the link to the article: https://www.electro-tech-online.com/articles/simple-inexpensive-ideal-diode-mosfet-circuit.817/

  5. David Albert says

    January 23, 2024 at 12:15 am

    Hi Aimee, thanks for the link to the original article; I had not seen it before!

    The original article already discusses the Vbe limit and includes an addendum that addresses it elegantly with an op-amp. The op-amp solution is particularly nice because it has very low power consumption. There are lower-cost op-amps than the LT6003 mentioned e.g.: TLV2401IDBVR (880 nA) or TLV9301IDCKR (150 uA), but any op-amp is going to cost significantly more than the current mirror.

    A solution that supports higher voltages and costs nearly the same as the current mirror is Diodes Inc. DZDH0401DW-7 ($0.10 in qty 1K); current consumption is better than the current mirror (147uA) and it easily supports Vbat up to the FET’s Vgs_max (typically 12-20V).

Trackbacks

  1. Adding Backup Battery Circuitry to your Projects – Pranav's Tech Blog says:
    March 20, 2020 at 7:56 am

    […] If the voltage drop is a problem, ideal diodes with ~0 Voltage drop can be constructed using MOSFETs as explained here. […]

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