# Can we use Zener diode as Battery low voltage cut off...?

#### ihsan.duet11

Joined Aug 10, 2021
11
Case# we have 13.5v Battery and we want the low voltage cut off point to be set at 10v. Can we achieve this simply by using zener diode...? If yes, then how...?

#### crutschow

Joined Mar 14, 2008
27,718
Zener diodes have a wide voltage tolerance which makes them problematic if you want an accurate cutoff voltage.
Better to use a precision reference, such as a TL431.
But 10V is generally to low for a 12V battery cutoff.

How much current is the battery providing?

#### click_here

Joined Sep 22, 2020
397
A zener usually takes a lot of current to be stable, but let's assume that it is not a problem with your application.

You would get a lower voltage than the cut off point, let's say 5.1V (1N4733A) and put a resistor in series so that it is 49mA at the voltage 10V

(10V-5.1V)/49mA = 100R
and...
P = (10V-5.1V) * 49mA = 240mW

 Error: This is not the maximum power - I'll redo this in another post below [/edit]

You would then make a resistor divider that is 5.1V when the voltage being applied is 10V
Ratio 4.9:5.1
47k+2k:51k

You would then put the output voltage into a comparator and monitor the output

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#### click_here

Joined Sep 22, 2020
397
You could also change the 2k to a 5k trimpot so you could calibrate the trigger point

#### ihsan.duet11

Joined Aug 10, 2021
11
Zener diodes have a wide voltage tolerance which makes them problematic if you want an accurate cutoff voltage.
Better to use a precision reference, such as a TL431.
But 10V is generally to low for a 12V battery cutoff.

How much current is the battery providing?
Thanks appreciates your response. Battery current is around 20A to 30A, depending on load. we have portable PV system with 50Ah battery.

#### click_here

Joined Sep 22, 2020
397
The maximum power in the 100R resistor would be
P = (13.6V-5.1V) * 49mV = 417mW

You should be able to get away with a 1/2W, but I'd play it safe and get something larger.

A good rule of thumb is add a third -> 417mW * 133% = 554mW

... So a .6W or a 1W would be a good choice.

#### Ramussons

Joined May 3, 2013
1,060
Case# we have 13.5v Battery and we want the low voltage cut off point to be set at 10v. Can we achieve this simply by using zener diode...? If yes, then how...?
No, this wont work.
You can set a reference voltage with a Zener. And you will need to compare the battery voltage with this reference. In your case, the reference itself is the battery voltage at the cutoff.

The right way is to set a lower reference voltage, say 9 volts and compare the 9/10ths battery voltage with this.
Comparing will need a Comparator with Hysteresis to avoid the "Grey range" when the comparator cannot make up its mind and keeps "chattering".

You can put the zener in series with a relay to do the job, but the Grey Range will increase and the relay will "chatter"

#### click_here

Joined Sep 22, 2020
397
No, this wont work.
You can set a reference voltage with a Zener. And you will need to compare the battery voltage with this reference. In your case, the reference itself is the battery voltage at the cutoff.

The right way is to set a lower reference voltage, say 9 volts and compare the 9/10ths battery voltage with this.
Comparing will need a Comparator with Hysteresis to avoid the "Grey range" when the comparator cannot make up its mind and keeps "chattering".

You can put the zener in series with a relay to do the job, but the Grey Range will increase and the relay will "chatter"
Just asking, and it's okay to say "no", but did you read through any of the other responses?

#### LowQCab

Joined Nov 6, 2012
1,096
If You want to build your own Circuit,
this one will allow You to adjust the Cut-Off-Voltage anywhere You like.
And, it has a delayed action to prevent a false shut-down when
a heavy Current Load starts up.
It also doubles as a Main-Power-Switch.
.
.
.

#### DickCappels

Joined Aug 21, 2008
7,704
It seems best that you consider a TL431-based solution or similar.

• Zeners, especially around 10 V have considerable thermal drift.
• The humble TL431 drifts at about 50 ppm/°C
• Most Zeners will require 10ma to 20 ma to get you a predictable voltage but
• only the accuracy will be 5% to 10% except for exotic Zeners
• TL431 needs less current
• If you try to use a 10 volt Zener, where will you get the bias current?
• If you don't have a TL431 I can send you some, but you have to wait for mailing from S.E. Asia

#### Juhahoo

Joined Jun 3, 2019
175
Case# we have 13.5v Battery and we want the low voltage cut off point to be set at 10v. Can we achieve this simply by using zener diode...? If yes, then how...?
Better use voltage reference and op amp circuit. It will be highly accurate and you can build hysteresis into it. Internet is full of circuit examples. Like this: http://www.high-voltage-lab.com/263/low-voltage-alarm

#### ihsan.duet11

Joined Aug 10, 2021
11
If You want to build your own Circuit,
this one will allow You to adjust the Cut-Off-Voltage anywhere You like.
And, it has a delayed action to prevent a false shut-down when
a heavy Current Load starts up.
It also doubles as a Main-Power-Switch.
.
.
.
View attachment 245375
In our case, the portable PV system is 300Watt and Battery = 13.5V (50AH) and I want cut off point to be set at 10V.
For achieving this what should I do...? which values I need to change in above circuit?
& can I use IRF4905 instead of above (IXTR170P10P-ND) FET ?

#### Andrew A. Cain

Joined Aug 16, 2021
2
Yes, you can use Zener diode but apply it in forward biased condition

#### LowQCab

Joined Nov 6, 2012
1,096
The 1K Resistor on the Gate of the FET is in the wrong position,
it should connected directly to the TL431.

There is no "Connection-Dot" on the Wires between C1 and R3.
It's not clear whether those wires simply crossover each other, or are connected.

You don't need to change anything.
When the Voltage at the wiper of the Trim-Pot drops below 2.75-Volts,
the FET will turn Off.

The P-Channel MOSFET that You select must be capable of withstanding a
"worst-case" scenario.
You are dealing with potentially extremely high, and continuous Currents.
2- Fuses, one on the Input, and one on the Output, would be a very good idea.
The Fuses should be sized such that they blow-out well before the
maximum rated Current of the FET(s) that You choose.

Most Electronics Supply Houses have a Parametric-Search feature on their website.

The IRF4905 is rated for ~50-Amps @ ~100C.
You say that You have a ~300-Watt System,
300-Watts divided by 12-Volts is ~25-Amps,
so You could probably get away with using that Transistor,
but I would recommend using a FET that has a large "Package" such as a TO-247
which will enable it to more readily dissipate any Heat that it may generate,
and find one with the lowest "Rds-On" possible, to reduce the amount of
Power that will be lost to Heat.
N-Channel FETs are better suited to this type of application,
because they generally will be cheaper,
for a usually lower "Rds-On" rating, than a similar P-Channel FET.
But certain applications demand that a "High-Side" Switch be used,
so a P-Chan FET compromise may have to be used.

As with the FET that I suggested, You can easily parallel 2,3,or 4 FETs
for extra "Blue-Smoke" insurance, and higher overall efficiency.
This does not remove the need for them to be mounted to some sort of Heat-Sink.
This is not a high speed "Switching" application,
so they will all play-nice together and

Paralleling FETs reduces the total "Rds-On" of the Switch,
and reduces the total amount of Heat-Dissipation,
and spreads-out the Heat that does get generated, over a larger area.
.
.
.

#### LowQCab

Joined Nov 6, 2012
1,096
It occurred to me that You didn't state exactly how or where this Switch would be used.
You stated that You have a "300-Watt-System",
but You did not say what the Battery was powering,
and how much Current may be demanded from the Battery, or
whether You are using the Switch to separate the Battery from the PV-Panels.

This Switch will not prevent reverse Current flow.
.
.
.

#### ihsan.duet11

Joined Aug 10, 2021
11
It occurred to me that You didn't state exactly how or where this Switch would be used.
You stated that You have a "300-Watt-System",
but You did not say what the Battery was powering,
and how much Current may be demanded from the Battery, or
whether You are using the Switch to separate the Battery from the PV-Panels.

This Switch will not prevent reverse Current flow.
.
.
.
Battery will be powering 12v DC fans + 12v DC bulbs. Current drawn from battery is upto 20A. This switch will be used between load and battery, to protect battery from completely discharging.
And regarding the circuit diagram I shared, its connection is not clear but it's the same circuit that you shared.same connection.

#### LowQCab

Joined Nov 6, 2012
1,096
"" but it's the same circuit that you shared.same connection. ""
The 1K Resistor is not in the right spot.

OK, the Switch is between the Battery and Load, it's probably all good.

The Fans could have a substantial Current-Spike when starting-up,
and Brushed-Motors can create super High-Frequency RFI / EMI, so lots of
differently sized Capacitors are a good idea, both Electrolytic and Ceramic,
on the Motor Terminals, and the Switch Output to Ground,
just a few things to keep in mind.

Incandescent Light Bulbs definitely have a huge Current-Spike when turning On,
but it's so short that it will probably not be an issue to the Switch.
.
.
.

#### ihsan.duet11

Joined Aug 10, 2021
11
Regarding above circuit,,,
0.5 watt or 1watt of R1/2/3/4/6 will be suitable...?
what should be the wattage rating for RV1 (Potentiometer) ?

#### LowQCab

Joined Nov 6, 2012
1,096
If You think that there may be some crazy chance that both the On and Off Buttons
could be accidentally depressed for an extended period of time,
then they should be rated 1/2-Watt or higher.
Under normally expected circumstances 1/4-Watt will be fine since the
chances of both Buttons being held down at the same time for more than
1-Second is extremely remote.

All other Resistors may be 1/8th-Watt or even less,
as none of them carry any significant amount of Current.

The amount of Current wasted by the entire Circuit when in the "Off" state
is so low that it is very difficult to measure.

The amount of Current used by the Circuit in the "On" state
is much less than 1ma.
The Green LED, (if You choose to include it), draws an additional ~20ma.

The amount of Power-loss when in the "On" state is determined by
the Load-Current, multiplied by the "Rds-On" Specs of the FET(s) that You choose,
this will result in the Voltage-Drop measurement across the FET.
The Voltage-Drop across the FET multiplied by the Load-Current,
equals the number of Watts of Heat that will be dissipated by the FET(s).

All of these are simple "Ohms-Law" calculations, nothing special.
.
.
.

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#### ihsan.duet11

Joined Aug 10, 2021
11
If You want to build your own Circuit,
this one will allow You to adjust the Cut-Off-Voltage anywhere You like.
And, it has a delayed action to prevent a false shut-down when
a heavy Current Load starts up.
It also doubles as a Main-Power-Switch.
.
.
.
View attachment 245375
I Tried to simulate this one. Its cut off voltage can't be adjusted by the 10k variable resistor as it cut off every input voltage below 13.5V. It work normally at and above 13.5v.
I have used IRF4905 and TL431 instead, due to unavailability of tl430clp and IXTR170P10P in the local market.
IRF4905 minimum gate threshold voltage is 2V while ixtr170p10p has 4V and tl431 VO adj (Min) (V)=2.495 compare to tl430 which is 2.75V.
Will using irf4905 & tl431 need some change in the resistors value, especially Potentiometer value?