Help me understand this reverse-polarity circuit.

Other Hardware General Electronics
1

I am reading up on some reverse polarity options, since I think I busted some hardware recently when incidentally powering my Arduino Uno from the USB and from vin pin at the same time.

GreatScott had a youtube video which helped summarize some useful information which got me started in the right direction. He referenced this blog post and circuit which I am now dissecting and I wanted some confirmation whether or not I am fully understanding it correctly. I am not very good at reading circuits diagrams still. Plus his example uses a 5v BMS and a single cell battery, and in my project I am using a much larger 12v battery so the required mosfets might be subject to change.

The diagram image attached show the circuit in the NPN form. I am choosing this one over the PNP type simply because I already have some NPN mosfets lying around that I want to see if I can use. I have annotated this with how I think the mechanical behavior is functioning. I just wanted to confirm whether or not I got it right.

So after much thought, here is how I see the signal flow:

  1. When in the correct orientation, current would flow in parallel to the collector of the 2N2907A transistor as well as into the positive terminal of the battery cell.

  2. The battery cells (-) terminal would send a negative -3.7v in parallel into the gate of the 2N2907A and to the source of the 2N7000.

  3. The negative voltage would then open the gate of the 2N2907A transistor, allowing the positive voltage to flow in parallel to the base of the 2N7000 mosfet and to the TP4056 BMS.

  4. The positive voltage on the base of the 2N7000 would open the source/emitter path, allowing the negative voltage to flow back into the negative terminal of the TP4056 BMS, completing the circuit.

...and if the battery was the incorrect polarity:

  1. +3.7v would flow from the battery cell to the 2N2907A transistor. Since it is positive, it will remain closed.

  2. The diode prevents the +3.7v current of the TP4056 from flowing to the positive and negative lead of the battery cell, preventing a short from happening. The battery cell is now in a closed loop and there is no sink for the current from the TP4056 to complete the circuit.

I am skeptical that I got this correct. If so, then what is the interaction that happens at the 2N7000 stage as the -3.7v is allowed to pass through the collector/emitter of the 2N7000 while the positive +3.7v TP4056 is flowing the same way in parallel? How does the (-) of the TP4056 allow both a positive and negative current to flow back to the BMS?

Please help :disappointed_relieved:

battery_protection_plain_mosfet (1).png

battery_protection_plain_mosfet (1).png

2

when you connect the battery as per the circuit diagram, with correct polarity, it causes the transistor to turn ON and allows the current to flow through it.

3

since I think I busted some hardware recently when incidentally powering my Arduino Uno from the USB and from vin pin at the same time.

No you didn’t, there is a circuit in most Arduinos that switches the supply away from USB if there is a voltage on the Vin pin. What ever damaged your Arduino it was not that.

A transistor will work if you swap the emitter and collector over. The only difference you will see is that the gain is lower with them swapped over.

because I already have some NPN mosfets lying around

As ther is no such thing as a NPN mosfet I doubt you had one lying around. Maybe you had an n-channel mosfet instead?

When a chip is powered up with reverse power then it looks like a diode to ground, that is why the chip gets hot and lets out the magic smoke.

Your diagram makes little sense, you draw a transistor but you talk about a FET.

4

aspen1135:
The diagram image attached show the circuit in the NPN form. I am choosing this one over the PNP type simply because I already have some NPN mosfets lying around that I want to see if I can use. I have annotated this with how I think the mechanical behavior is functioning. I just wanted to confirm whether or not I got it right.

So after much thought, here is how I see the signal flow:

  1. When in the correct orientation, current would flow in parallel to the collector of the 2N2907A transistor as well as into the positive terminal of the battery cell.

yes

  1. The battery cells (-) terminal would send a negative -3.7v in parallel into the gate of the 2N2907A and to the source of the 2N7000.

The battery negative is connected to the drain of the 2N7000 and via R1 to the base of 2N2907 - BJTs don't have gates.

  1. The negative voltage would then open the gate of the 2N2907A transistor, allowing the positive voltage to flow in parallel to the base of the 2N7000 mosfet and to the TP4056 BMS.

the base current turns on the 2N2907, which then raises the gate voltage on the 2N7000
turning it on. "open" is not the right term here, and its the voltage on the gate than creates
the channel that connects source and drain - the gate itself is not carrying current.

  1. The positive voltage on the base of the 2N7000 would open the source/emitter path, allowing the negative voltage to flow back into the negative terminal of the TP4056 BMS, completing the circuit.

gate of the 2N7000 - FETs don't have bases. source/drain path, not source/emitter path.
Voltage doesn't flow. Current flows. Voltage difference is what drives current flow.

...and if the battery was the incorrect polarity:

  1. +3.7v would flow from the battery cell to the 2N2907A transistor. Since it is positive, it will remain closed.

It remains open (i.e. off). Open means no current flow. open-circuit is off, short-circuit is on, - this is completely different to water flowing through a tap, note, see below.

  1. The diode prevents the +3.7v current of the TP4056 from flowing to the positive and negative lead of the battery cell, preventing a short from happening. The battery cell is now in a closed loop and there is no sink for the current from the TP4056 to complete the circuit.

The diode prevents the 2N2907 from being destroyed by backwards bias across base-emitter junction

BTW nomenclature in English for electric switches:

open - no connection - no current
closed - connection, current flows (also called "shorted")

And for a water tap:
open - water flows
closed - no flow

So you have to be careful if English is a second language - the water flow analogy is
great, but "open" and "closed" are reversed!

5

Current flows like traffic on an opening bridge. If the bridge is open, no traffic flows.

Water flows under the bridge. If the canal lock is closed, no water flows. :grinning:

6

All your comments about "-3.7V flows etc etc" are incorrect for the direction of current flow.
The voltage across R1 is a positive voltage, not negative because the end connected to the diode is biased
more positive than the end connected to the battery negative terminal and the voltage from Q1 source to
the BMS negative terminal is biased more positive at Q1 source than at the BMS negative terminal so it is
not negative either. With the exception of a small voltage drop across Q1, the BMS negative terminal, Q1
source, Q1 drain and the battery negative terminal are all at the same potential (all connected together)
so there is NO potential difference (no 3.7V). In order to have a potential difference, (a voltage drop) you
must have an R (resistance) and an I (current). The only R is the Rds(on) of the mosfet, and the the voltage
drop across that is negligible (approaches zero). (it is likely in the order of mV). For all practical purposes
we can consider the resistance of the wire to be 0 ohms.
so the current is flowing from the battery negative terminal to the BMS negative terminal.
To ascertain whether a voltage is positive or negative you must first identify the voltage
at either end. Perhaps some basic electronics courses would help.

7

Very grateful for all the responses. Last night I was staring at the schematic for like an hour or two, looking at the parts a bit closer and I eventually figured it out. I feel so dumb now.

For awhile, I was getting lost by thinking that the ON/OFF logic state of each switch types were inverted. I was thinking that "N" would mean negative closes the gate/base, and "P" would be counterpart of a switch that does the same but requiring positive voltage on the gate/base. I was missing a key characteristic of these switches which answered a lot of my questions.

However, as far terminology is concerned, why do they call them "N" channel mosfets if it takes a positive voltage to close the switch? Correct me if I'm wrong but since they are most commonly used for switching tasks, wouldn't it make sense to name them accordingly to the required signal type at the gate/base to turn them on?

Also: Many thanks to the people who helped clarify the terminology for me. My brain wants to think that these things are inverted for some reason and I feel I can communicate this issue a lot better now.

As far as the schematic goes, I think I got it. Maybe someone can verify:

The Q2 switch turns ON when the negative side on the battery grounds the gate on Q2 through R1. The diode (D2) prevents current from going to the negative side of cell when in the correct orientation. During this state, Q1 and Q2 are closed, connecting ground of the BMS to the negative terminal of the cell, letting it charge. When the polarity is inverted on the cell, Q2 switch opens up and the battery is left in closed protected loop. The diode protects it from shorting in reversed polarity again, and all the switches are left in an off state-- preventing any current from the BMS reaching the cell. R2 ensures that Q1 switch is pulled down and opened by default. And R1 does the same for Q2 switch, except it ensures the switch is left on if it was ever thrown into an off state, which happens when the cell is inverted.

Yes?

EDIT:
I have a question regarding when the cell is in a inverted state. The switches would be open, essentially disconnecting the ground of the BMS to the battery, but the positive terminal of the BMS would still allow current to flow to the negative side of the cell. Does this current not pose any harm to the battery because it cannot flow to the BMS or to the negative side of the cell?

8

why do they call them "N" channel mosfets if it takes a positive voltage to close the switch?

There are two types of material used in the construction of bipolar transistors and FETs, N and P.

These are both mainly silicon but are "doped" or mixed with traces of another element so that in an N type material there is an excess of negative type charge carriers ( electrons ).

Where as in a P type material there is an excess of positive type charge carriers ( holes or absence of electrons ).

Aluminum or Gallium, or Indium is added to make silicon into a p-type semiconductor, while in the n-type semiconductor impurity like Arsenic or Antimony or Phosphorus or Bismuth are added.

Also with respect to FETs there are two types, enhancement and depletion. What you think of as a FET is more correctly called an enhancement FET, but depletion FETs work the other way round.

So they are named after how they are made and have nothing to do with how you will perceive their operation.

9

One way to remember it is POLES rthymes (spelling?) wth HOLES and POLES starts with a P.

10

The voltage between gate and channel for an n-channel FET is positive on the gate
because its negative on the channel, and since the channel is what conducts the current
between source and drain it is N-type when the gate is positive.

In other words the voltage seen by channel should be thought of as Vsg, not Vgs, and Vsg
is negative to turn on an n-channel FET - the negative charge is what creates the channel
by inverting the p-doped region of the channel to n-type. Since source and drain are both
n-type they are isolated from each other until an n-type channel forms.

If you think about the gate, the polarity is the other way of course, but the current isn't
carried in the gate, the gate merely forms the opposite plate of a capacitor with the channel.

The positive charge on the gate balances the negative charge than makes up the induced
n-type region of the channel and holds it in place while the device is on.

11

One other question came to mind as an after-thought when I was buying components?

Why could you not just switch the pnp BJT for a P-channel mosfet? Wouldn't they accomplish the same thing?

12

Why could you not just switch the pnp BJT for a P-channel mosfet? Wouldn't they accomplish the same thing?

When mosfet Q1 is OFF, the voltage across R1 is 3.7V-0.7V = 3V.
When Q1 is ON, the diode prevents current from flowing in the reverse direction so the voltage on the
anode of the diode is approximately 0V. (which turns on the pnp transistor).
A p-channel mosfet would turn on with 0V but whether or not it turns OFF depends on the mosfet.
There's probably some that would. I couldn't say which one but I suppose it could work.

13

Lets say the battery is completely drained and you inserted it into this circuit with the incorrect polarities. Wouldn't the high internal resistance of the battery and R1 combined give you little to no current, switch ON Q2, and damage the battery?

14

I understand that you are asking a simple question .
Dare to try to answer yourself:
Before you attempt to answer these questions , you have to first remove the labels you put on the schematic
about "-3.7V etc..."
Why ?
For starters "-3.7V flows..etc" is an invalid statement because 3.7V is a voltage, and voltage doesn't flow. Only
current flows.
Second, "-3.7V opens gate" is an invalid statement because all voltage measurements are made with reference
to the B- output of the BMS, since that CANNOT be reversed .
If on the other hand you specify the meter ground reference by saying "Vbe Q2 is -3.7V, turning on Q2, which
conducts, applying +3.7V on the gate of Q1", then that would be a totally valid statement.
That being said, what is Q2 Vbe if you put the battery in backwards ?

Q1: Which device in this circuit determines whether current flows between the BMS and the battery ?
Q2: Given that Q1 is an N-channel mosfet, what polarity voltage would be required for it to turn ON and allow
current to flow between the BMS and the battery ?
Is that too difficult to figure out ? (that's a question, not a wisecrack) Either it is or it isn't. Once we know that,
we can go from there, but we need to know if you are on the same page or somewhere else.
It's an open book test.
Here's the datasheet for Q2
Do you know exactly HOW you would measure VBE ? (where you would place the black lead and where you would
place the red lead ?)

15

raschemmel:
That being said, what is Q2 Vbe if you put the battery in backwards ?

Q1: Which device in this circuit determines whether current flows between the BMS and the battery ?
Q2: Given that Q1 is an N-channel mosfet, what polarity voltage would be required for it to turn ON and allow
current to flow between the BMS and the battery ?
Is that too difficult to figure out ? (that's a question, not a wisecrack) Either it is or it isn't. Once we know that,
we can go from there, but we need to know if you are on the same page or somewhere else.
It's an open book test.
Here's the datasheet for Q2
Do you know exactly HOW you would measure VBE ? (where you would place the black lead and where you would
place the red lead ?)

Now you all are making very skeptic about my understanding of circuits :slight_smile: But that's okay, I am humbly here to learn. I am also always open to tests and challenges :wink:

What you said about the relation to positive and negative voltage according to the ground signal path makes sense.

I did have to look up what Vbe was again. It is the voltage drop across the base-emitter path of the transistor. I am familiar with what this is just not the term abbreviations. The voltage of Vbe (when cell is backwards) would be +0.7v. In normal orientation, it'd be -0.7v

Q1: The transistor and thus, the battery component, are what determine what turns ON/OFF state of the Q1 mosfet.
Q2: The N-channel mosfet would require a positive polarity above the rated gate threshold voltage specified on the mosfets datasheet to turn ON. When the pnp transistor is ON, the switch closes allowing +3.7v from the BMS to turn on the mosfet (not accounting for any voltage drop across the collector-emitter path).

I believe I am understanding this correctly, yes?

What I don't understand, is that if a transistor is a switch that's state is determined by the amount of current on the base, and if a pnp transistor specifically requires the base to be grounded, then if the batteries polarities were flipped and the battery were completely drained, then wouldn't there be too little current on Q2's base for it to turn OFF?

That is why I was wondering if a logic-level P-channel mosfet may work better here instead of a pnp transistor, since the battery should still hold some sort of positive voltage after it is drained assuming that the cell itself is not already damaged.

EDIT:
I looked up some additional information about pnp-based transistors and it may have enlightened me a bit. It seems pnp transistors emmiter's are "on top" of the schematic symbol instead of the collector (How in the world does that make sense?). So as we are discussing the base-emitter voltage, we are discussing the voltage difference of the polarities between Vb3 to Vb4 and Vb3 to VQ2b.

I may not be using the node abbreviations correctly, but what I meant is the voltages between the points of the BMS source terminals +/-, and the points between the BMS to the pnp transistor's base +/-. The base voltage has to be less than 0.7v of the Emitter's to turn ON.

Before we go further, maybe someone could clarify a few things for me:

  1. Why is the node at which current is received called the "Emitter". How is it 'emitting' anything. Isn't it technically 'collecting' current? I see this as an input rather than an output node..

  2. Why are pnp transistors considered to be characterized by their current if it is the voltage difference between the emitter-base path that makes them switch ON/OFF? Is it because can amplify current across the emitter-collector path?

Also the question still remains: If the battery were drained and inverted the voltage delta between Vbe could be < 0.7 still which would open the switch, allow current to be drain, and short the battery since a discharged battery still contains a voltage lower than when it is charged?

What am I not getting here? I hope you can appreciate my efforts and attempts to researching and understanding this. Maybe someone can explain it to me like I am 5 lol :sweat_smile:

I apologize for the crappy MS paint visual. The original reference photo is pretty low res >.<

battery_protection_plain_mosfet.png

battery_protection_plain_mosfet.png

battery_protection_plain_mosfet.png

battery_protection_plain_mosfet.png

16

I can't follow what you said because there are contradictions.
Let's keep it simple :
"what do you expect to measure with the BLK LEAD on Q2 emitter and the RED LEAD on Q2 base for normal and reversed 3.7V batteries ?

17

What I don't understand, is that if a transistor is a switch that's state is determined by the amount of current on the base, and if a pnp transistor specifically requires the base to be grounded, then if the batteries polarities were flipped *and the battery were *__*completely drained, *__then wouldn't there be too little current on Q2's base for it to turn OFF?

Prove it.
First of all, if you are going to discuss current you have to FIRST calculate it.
SHOW YOUR WORK !

18

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