In this post I have explained how to make a simple transistor latch circuit using just two BJTs and a few resistors.
Introduction
A transistor latch is a circuit which latches ON with a permanent high output in response to a momentary input high signal, and continues to stay in this position as long as its in the powered condition, regardless of the input signal.
A latch circuit can be used to lock or latch the output of the circuit in response to an input signal and sustain the position even after the input signal is removed.
The output may be used to operate a load controlled through a relay, SCR, Triac or simply by the output transistor itself.
Working Description:
The simple latch circuit using transistors I have I have explained in this article can be made very cheaply using just a couple of transistors and some other passive component.
As shown in the figure transistor T1 and T2 are configured in such a manner that T2 follows T1 to either conduct and or to stop the conduction depending upon the trigger received at the input of T1.
T2 also acts as a buffer and produces better response even to very small signals.
When a small positive signal is applied at the input of T1, T1 instantly conducts and pulls the base of T2 to ground.
This initiates T2 which also starts conducting with the received negative biasing offered by the conduction of T1.
It must be noted here that T being NPN device responds to positive signals while T2 being a PNP responds to negative potential generated by the conduction of T1.
Uptill here the function looks pretty ordinary as we witness a very normal and obvious transistor functioning.
How the Feedback from R3 Works to Latch the Circuit
However the introduction of a feedback voltage through R3 makes a huge difference to the configuration and helps to generate the required feature in the circuit, that is the BJT circuit instantly latches or freezes its output with a constant positive supply.
If a relay is used here it would also operate and stay in that position even after the input trigger is completely removed.
The moment T2 follows T1, R3 connects or feeds back some voltage from the collector of T2 back to the base of T1 making it conduct virtually “for ever”.
C1 prevents the circuit from getting activated with false triggers generated from stray pick-ups, and during switch ON transients.
The situation can be restored back either by restarting power to the circuit or by grounding the base of T1 through a push button arrangement.
The circuit can be used for many important applications, especially in security systems and in alarm systems.
Calculations and Formulas:
Threshold Voltage for BJT Activation
The threshold voltage values for turning ON or OFF the NPN and PNP transistors remain the same:
- NPN Transistor Turn-On Voltage (
VBE(on)):VBE(on) ≈ 0.7V(for silicon BJTs) - PNP Transistor Turn-On Voltage (
VEB(on)):VEB(on) ≈ 0.7V(for silicon BJTs)
These values determine when the base-emitter junction of either transistor is forward biased, enabling current flow and turning the transistor ON.
Base Current (I_B) Calculation
For both NPN and PNP transistors, the base current is still calculated in relation to the collector current:
- Base Current for NPN Transistor (IB(NPN)):
IB(NPN) = IC / βNPN - Base Current for PNP Transistor (IB(PNP)):
IB(PNP) = IC / βPNP
Where:
ICis the collector currentβis the current gain of the transistor (typically 50–300)
This is relevant for understanding how the transistors maintain their ON or OFF states once latched.
Collector-Emitter Voltage (V_CE)
The voltage across the collector-emitter junction of each transistor is very important for ensuring the transistors remain in the active or saturation regions:
- NPN Transistor Saturation Voltage (VCE(sat)):
VCE(sat) ≈ 0.2V(when fully ON) - PNP Transistor Saturation Voltage (VEC(sat)):
VEC(sat) ≈ 0.2V(when fully ON)
These values are relevant when the latch is "set" or "reset," which ensures that both transistors are either fully conducting or completely turned off.
Latch Holding Condition
Once the latch is set or reset, the feedback ensures that the state is maintained regardless of the input signal:
- Feedback Current (IFB):
IFB > IB(required)
Where, the IB(required)
This feedback signal ensures that once the NPN or PNP transistor is turned on, the circuit remains latched in that state until forced to reset.
Resistor Calculations
Resistors will control the currents flowing through the transistor and define the behavior of the circuit:
- Base Resistor (RB):
RB = (Vinput - VBE(on)) / IB - Collector Resistor (R_C):
RC = (VCC - VCE(sat)) / IC
Where:
Vinputis the voltage applied to the base of the transistorVCCis the supply voltageIBandICare the base and collector currents, respectively
These resistor values help in controlling the current levels to properly switch and latch the circuit.
Switching Time
The switching time for the latch circuit or the time it takes for the latch to change states is determined by the charging and discharging of junction capacitances:
- Rise Time (tr):
tr ≈ (RB * Cj) - Fall Time (tf):
tf ≈ (RC * Cj)
Where Cj is the junction capacitance of the transistor which determines how quickly the transistor can switch between states.
Hysteresis Voltage
Hysteresis ensures that once the circuit is latched in a particular state... it remains stable:
- Hysteresis Voltage (Vh):
Vh = IFB * Rfeedback
Where:
Rfeedbackis the feedback resistor value.
This feedback voltage creates a gap between the switching thresholds which helps to prevent oscillation and ensurs stable operation.
Testing procedure can seen in the following video tutorial:
Parts List
- R1, R2, R4 = 10K,
- R3 = 100K,
- T1 = BC547,
- T2 = BC557
- C1 = 1uF/25V
- D1 = 1N4007,
- Relay = As preferred.
PCB Design
With over 50,000 comments answered so far, this is the only electronics website dedicated to solving all your circuit-related problems. If you’re stuck on a circuit, please leave your question in the comment box, and I will try to solve it ASAP!
Hi. Hope you had a great Christmas.
I’ve used the latch circuit to latch after it detects the first activation ping from a burglar alarm. This works as expected. I’d hoped the output from the latch would be enough to run a 555 timer and 4017 counter that will run relays to dial a cell phone to call me, an idea from one of your other blogs, but the output is only 1.9v I s’pose all that’s left of the 12v after voltage drop. It’s not enough to run the 555.
So the latch output is now connected to a BC547 base that is acting as a switch passing 12v to the 555, 4017 and SRD-12v-SL-C relays. But I imagine the base needs resistor. The math and required understanding has got the better of me what size resistor does it need?
Seems to work alright with the 4017 driving LEDs without a resistor but I’ve had transistors cook on me before.
Hey Thanks, hope you too are enjoying the Holidays.
You do not need a BC547 at the output for powering an external circuit. You just have to replace the shown relay coil with your external circuit.
For adequate current you may have to replace the BC557 with a 2N2907 or an 8550 transistor.
So, the collector of this transistor now connects with the positive supply line of your 4017/555 circuit, and the negative to the common ground of the supply.
I’m learning electronics in my spare time. I already know some basics about electronics and still learning. I’m trying to make an alarm that uses 4011 NAND gates for checking if the an alarm wire is broken and then it should use a latch circuit to turn on a microcontroller that turns on the speaker, leds etc.
Then after a specific time the microcontroller should turn off the latch circuit, to preseve energy. This is because I want to make a small battery powered multipurpose alarm system that could be used in bikes to prevent thefts (with vibration activated sensor/switch) or as a door alarm (with NC switch) in my apartment.
I already understand quite well how NPN and PNP transistors works, but I don’t currently understand the purpose of the C1 capacitor how it works here in the circuit.
Thank you for your question, I appreciate your interest in electronics.
Could you please tell me where is this C1 capacitor situated?
Is it connected in series with a resistor or after the resistor and across ground?
If you can refer me to the schematic or provide the exact configuration details of the capacitor in the circuit, I can certainly help you to understand how the capacitor may work to fulfil its specific job.
Actually I mean the C1 capacitor in this example latching circuit you have made. To fully undertand this latching circuit, is this a decoupling capacitor or what exactly is it called in this setup? Or is this used for some kind of filtering or debouncing prevention?
But about the alarm circuit I’m trying to make, part that would activate the latch circuit would look something like this. I have 4011 ICs for this purpose, so one NAND gate would be used for checking if the circuit is closed and another NAND gate would be used to invert the output, so it could be used by a latch circuit.
I have not yet fully planned this, but my initial idea was to keep the alarm latched for a specific amount of time and it could probably be done using your latch circuit. In this example image I posted, the 1K resistor is just there because I quickly added something to demostrate.
Why I want to keep the alarm latched is that if someone would close the circuit again, the alarm so still be on and not turn off.
Ok, understood!
The C1 in the above latch circuit is used for filtering external noise and to prevent false activation of the latch circuit.
However, the best position of C1 is across the base/emitter of T1, so that the noise is filtered right at the beginning of the circuit.
If you want to latch your alarm circuit for some specific amount of time, then I would recommend you using a 555 monostable circuit.
Alternatively, you can simply use your NAND gates to create a simple monostable for the same purpose.
Let me know if you have any further doubts or questions.
I want the latch to stay even if the input signal drops to 0. In my preferred solution the only way to break the latch is by switching off power. How do I achieve this?
The circuit will stay latched even if the input signal is removed. The circuit needs only a momentary input signal to get latched.
thanks for your quick reply. at the moment, the latch drops when the base resistor of T1 is connected to ground. I don’t want this to happen. instead I want the latch to stay until the power is turned off even if the input signal drops to 0. Can you suggest a modification?
For that, you can add a 1N4148 diode at the input side of R1, and feed the trigger signal through this diode.
it worked, thanks for your help
Jan
Glad it worked…
I have a circuit that takes less than 100usec to “get started” and my input signal is high during this startup period. How can I ignore the first 100usec of input before using the input to trigger? Within ~ 40usec my input is low and later (tens of seconds) the input goes high as expected and this is the transition I wish to react.
Controlling a microsecond signal looks difficult, I can’t figure out a configuration that would control the microsecond timings with such accuracy.
Amazing! Thank you so much!.
I simulated it and it works. Just the 2 questions I have are:
1) I have a 20us negative pulse that I need this transistor to switch on to and turn on the LED. Will this work?
2) There could be a lot of noise in the signal that will trigger the base of T2. I know you mentioned that C1 should help with the filtering, but now that i have switched the input to the base of T2 would this capaitor still help with noise or should I make modifications?
Thanks, Glad it helped.
C1 might help to eliminate false switch ON only if it is connected directly across base/emitter of T2. However if this done, then the 20us pulse will be too short and might get quickly absorbed by the capacitor, so that no signal reaches the T2 base.
Nevertheless, you can try smaller capacitor values such as 0.01uF across base/emitter of T2 and see if that helps to mitigate the noise and yet allows the 20us triggering.
I would like to know how to switch the circuit on a negative going pulse rather than positive. What changes do I need to make ?
Apply the negative pulse to the base of T2 via a resistor.
R1 can be removed, it is not required now.
I am giving triggering of 1 volt from ccd camera video signal ,how can we make it possible to disable the latching when the trigger voltage is deactivated
It means you don’t want the latching feature, for that you simply have to remove the R3 resistor link.
Thanks Sir
Thanks for the circuit, i made and it works fine, is there any way the the the latching will reset when the trigger circuit is removed if so please share diagram
Thank you for the update! Triggering is supposed to be done only once, then the circuit latches and remains latched. If the circuit resets on removing the triggering source then it won’t be a latch circuit.
In connection with previous ckt my inputs both are +ve for dis i need ckt sir.
Try this circuit then:
Pl give me SR Transistor lactch ctk with detail
here it is:
Built circuit but it latches on application of power. I have checked and rechecked the circuit and soldering but I am at a loss
Connect the C1 across base/emitter of T1. C1 can be a 1uF/25V. This will stop the self latching issue.
Hi
Thank you for the post. I tried replicating the circuit with BD139 and BD140 transistors. All seems to work well, except that there is a voltage drop across T2. The input voltage is 9V, the output is only 4.5V. Why would this be heppening?
Hi, that should not happen. The collector of the PNP must show the same voltage as its emitter supply. Did you check the voltage without a load or the relay? Please check it without a load.
Hi Swagatam,
I have what I suspect is an impossible circuit, but want to run it pass you to confirm.
I am sure you have knowledge of a simple moisture tester. This device uses dissimilar metals generally made up of copper and zinc. These metals when exposed to moisture (water) give a very low electrical current, this is the basis of my question.
I would like to know if there is any circuit and or device that could be powered by say a 3.7vdc lithium Ion battery or a 1.5VDC AAA battery that is basically in the off position, until two leads senses moisture, thereby powering up and indicating by a low volt LED that there is moisture across these 2 “LEADS” dissimilar metals but wouldn’t cause any, “shock” if applied to sensitive skin?
Hi Doug,
It is definitely possible, I have designed the circuit and have posted it at the end of the following article, you can check it out:
https://www.homemade-circuits.com/soil-moisture-tester-circuit/
For very less sensitive , what can we do as for water level sensing by 12 volt relay.
Could not understand what is less sensitive, please explain properly!
Can latch be ‘broken’ by an input based transistor switch inserted in the connection with 110 k reactance?
If the answer is ‘yes’, can one use more than one input as ‘breakers’ ?
yes that’s possible, as given in the following figure:
Sir, I want to create a simple latching circuit,
which can turn on/off by using single push button only.
Sir,Please note that,i want to achieve this task without using IC and microcontroller.
Please help me to design this circuit
I eagerly waiting for yours response sir
Hi Yogesh,
You can refer to the second last circuit from the following article:
https://www.homemade-circuits.com/build-these-simple-flip-flop-circuits/
However, without using an IC, and using only transistors may not give reliable results….using IC will give extremely reliable results.
Thank you sir,☺️????
Sir, what is the use of resistor R4 in this circuit?
And what happens,if I won’t use R4 resistor present in this circuit?
Please clear my doubt sir
Yogesh, R4 keeps the T2 base to a proper switched OFF condition when T1 is not conducting, which ensures that the T2 base is always held at a well defined potential and never in the floating position.
Thank you sir, ???? now I understood
And again thanks for yours speedy response sir……
You are welcome Yogesh!
The R4 is called a pull-up resistor in this case. Just in case if someone else tries to figure out how the circuit works.
But about the R3 resistor, is it really needed, or could the feedback loop use only the R1 resistor? I have some difficulties understanding the function of the R3 resistor.
R4 is actually a pull-down resistor in this case, because T2 is a PNP transistor.
R3 is the heart of the latch circuit, without it the latching would never happen.
With a momentary base trigger when T1 is turned ON, T2 also is also turned On, which allows a feedback voltage to pass through R3 and reach the T1 base, so that now T1 and T2 lock on with each other ensuring that the circuit gets latched regardless of whether T1 base is getting any external trigger or not.
Hi, I simulated your circuit with 12V input but it doesn’t work as intended – can you please provide design guidelines for selection of resistors?
Hi, the circuit is fully tested, so you can be sure it will work if you build it practically.
I did build the circuit – is it mandatory to use BC547 and BC557? I’m trying to use BC807 and BC817 instead…
Any NPN/PNP combination should work. It is advisable to remove the capacitor C1 from the existing position and place it across the base/emitter of T1 for better response
Hi Swag,
It me again to bother you. Can I ask you a small question. I made a copy the latch circuitry as shown on the image below (this also a copy of your schematic).
+ The switch will turn on the LED1 at first press (Discharged C1 via Switch and Base of T1);
+ At the 2nd press LED1 will be off as voltage drop at Q1 then Q1 closed and turn off the LED.
In this two stages the current will run from different directions on the Switch.
Due to the nature of my project that require the load can be overridden by a touch button (in case of microcontroller failure). I need a way to control the latch using both Manual button and one I/O pin of microcontroller and can also handle state of the load (on or off) on that only one pin.
I am thinking of replacing the switch with something that allows the current from both direction (2 transistors or a triac). This will allow me to run this circuitry from a capacitive touch button or a PULL_UP pin from microcontroller. This is critical as I can
+ Turn off manually the load by using a touch button (TTP223 module) – without the use of Microcontroller and;
+ Turn off the load by using Microcontroller and can also monitor state of the load (on/off) using just single I/O pin.
Can you advice me on what part I can use to mimic the Momentary Switch button with thanks
Dang Dinh Ngoc
Vietnam
Hi Dang, you can try replacing the push button with a bidirectional switch as discussed in the following article:
https://www.homemade-circuits.com/bidirectional-switch/
Interesting circuit, thanks for sharing this. I was building one just like it for an evening class I’m following and without any capacitor I have no sensitivity issues, that’s nice I suppose. I do have a question though. It may be a stupid question, but isn’t your snubber/flyback diode forcing any current from the relay through T2 and damaging it??
Thanks, and glad you found the post interesting.
The sensitivity issue will not be there without the capacitor if the supply is from a battery.
If you are using an AC to dC adapter then removing the capacitor will simply latch the circuit each time power is switched ON.
The back EMF is developed across the relay coil in the reverse direction such that the diode orientation allows a forward path for this EMF to get short circuited through its terminals….so it is fine, no problem with the diode orientation.
Thanks for the speedy reply, very helpful
Hello Swagatam,
Thanks for the circuit and description which I enjoyed making.
Just like Abu-Hafss I found it to be very sensitive to finger touch and power ups.
So I increased C1 as you first suggest and eventually moved it to between T1 base and collector which worked great. Please would explain how C1 works in preventing these problems? as I want to understand why and not just assemble things without the knowledge of their workings.
Hello Paul, that is correct, shifting C1 across the base/emitter of the NPN is more effective for controlling the switch ON triggering of the circuit.
When power is switched ON, a voltage spike across the power line is normally developed which can cause small voltage to get through the base/emitter of the NPN and trigger it ON. The capacitor acts like a momentary short circuit which grounds this small voltage spike, preventing it from reaching the base of the NPN.
Thanks Swagatam – now I understand.
Hi, I’m not sure if my link went trough, I don’t see my comment. But here it is:
https://studentuml-my.sharepoint.com/:u:/g/personal/nuno_ekoyedesenamartins_student_uml_edu/EdukUKJEoQ5Phr5k8zYqvHMBzGlW6BGYIzEtVFY-pf5iLg?e=gMbATj
C1 and C2 corresponds to your circuit diagram above. Thank you!
Hi, Please show it in a schematic format, so that I can see where the transistor base, collector emitters are going, a rough block diagram will not help.
Or you can tell me what exactly are you trying to achieve in your application, I’ll try to design it for you..
It doesn’t allow me to post a picture here. But to make it simple, imagine 2 coils in series, and where these 2 coils connect is considered the positive side. Lets call the positive side A3 and the negative ends A2 and A1. I connected your diagram to one side of this coil connection. From A3 to A2 (set). This will be triggered by a 3.3v pulse and stay open until power is disconnected.
Now, I want to use the other coil to disconnect this circuit. When I send a pulse to A3 and A1 (reset) the relay will go back to Close state. Currently, i have connected your circuit to the one coil and it works as intended. I also did another same circuit connected to the other coil. I observed that Only one side works properly at a time, but not at the same time. For example if I have one side connected, and I pulse the other side to disconnect it doesn’t work. I guess it’s because it’s triggered already and it needs to be disconnected from power? What changes can I make to the reset side (second coil) so that when I pulse a signal the relay will go to Close state? I hope this is more clear, thank you!
Please upload your schematic to any free image hosting site, and provide the link here, if it is possible I will try to solve it for you.
Hello there, I’m new to electronics and I’m trying to connect this circuit to a bistable latching relay(2 coils relayRT314F05). The relay has 3 control pins (A3+ to A2- is to set/ and A1- to A3+ is to reset). So when i Connect one coil to the latch circuit works it fine. But when I add the same latch circuit on the other side to reset the circuit it doesn’t work. I found that when I trigger 1 gpio, both sides of the circuit are active therefore it doesn’t work… how can I set/reset using 2 different gpio pins without one circuit interfering the other? Operating 1 at the time. Thank you!
Hello, without seeing the diagram it can be difficult for me to understand the fault. However, in order to reset the above latch circuit or break the latch, you will need to add another transistor circuit which can ground the T1 base.
Please swagtam can i use this circuit in a 433mhz receiver circuit which has output of about 2V from ht12d output (pin 10,11,12,13).
Hi Sam, yes you can, the circuit requires hardly 0.3V to activate
I just want to ask I made the circuit but instead of the relay, i connected an LED to light when it will latch. The problem that I have is the input is 0V which mean the LED is off because of the latch is not working but mine when the input is 0V then the LED is working and to the base of the transistor of BC547 I have 0.7V which I don’t get it.
I will appreciate if you answer me to this question.
did you connect C1? Alternatively connect another capacitor (10uF/25V) across base emitter of T1
thanks a lot I now understand, its the reverse bias process from the pnp that confused me. now I got the concept. thanks again ones more. truth be told I followed a lot many website but this one is different and unique. one of the factors that made it unique is the quick response to your questions asked. I want to use this medium to say, a big thank you on behalf of all of your entire followers. thanks ones more.
thanks abba, glad you are finding my site useful…please keep posting your thoughts, and keep up the good work!!
hi Swagatam, my question is, in the theoretical concept if pnp base is connected to collector of the npn transistor the circuit to me wouldn’t work because to me the base of the pnp should be grounded and the npn base should be connected to the postitve of the power supply as well as ts collector. am confuse pls explain how the biasing will take place when pnp and npn are connected in the above manner.
Hi Abba, the exact reply to your question is already provided in the article, please read the “circuit Description” section and let me know whether you could grasp it or not, or which portion you couldn’t understand.
Hi, did you connect C1? C1 is specifically introduced to prevent this issue.
alternatively you can shift C1 across the base/emitter of the BC547, and check the response, the issue will be completely gone.
preferably use a 1uF for C1 or above
Hello, thanks you for the post. I do have a question however. I want to activate a relay with an arduino signal (less than 3 volts). The realy needs a power supply of 9V. So When I made the connections the signal from the arduino would not trigger the latch. I tried to use a positively charged wire coming form the +9V side to activate the latch and it worked. I realized that the arduino and the battery dont have common ground so the arduino signal doesnt really mean anything to the transistor. I thought of connecting the ground of the +9V and the arduino to have common reference but then I started thinking? WHat if the ground levels of the arduino and the battery are not in the same state(which probaly they are not) That means that either the battery ground will increase in voltage therefore reducing the batteries volatage difference and then the battery will not be able to activate the relay..OR worse the battery negative will be higher than the arduino so current wills tart flowing backwards in the arduino board…and now I am compeltelly confused..how do you go around that? how does commono ground works in cases where you have different intependet power circuit conected in parallel..??
hello thanks!
as you said it yourself, the common line is referred to as "ground" which is not "negative" rather it's a zero volt line (0V), therefore any voltage reaching this line would ultimately become "zero" therefore it is completely safe and also mandatory to connect the two grounds together so that the two units can correspond and work together.
sir for this circuit design can i connect "IR SENSOR "was this works. because it looks simple to me to construct. was load gets "ON" and "OFF" by opearating remote . and if how should i connect "ir sensor" ( sensor's out to output trigger . ground to T1 D1 junction , and positive to C1 R4 junction with resistor ) was this right sir
thanks a lot
Manjunath, no this circuit cannot be used like a flip flop, you will have to opt for an IC based circuit such as a 4017 IC for getting the intended results
hellow sir… this circuit idea is brilliant… can you please suggest an another simple circuit of same kind to implement -ve latch. i.e, output is first ON and when a signal applies output becomes zero and latches in that state…
my intention is to make an Emergency bank (7 W , 7 high brightness LEDs) lights up when a signal is applied as the input trigger to this above circuit… the problem is that. i cant connect the lamp in the output portion as shown in this circuit ( between collector of BC557 and ground ) since the current through this path is very less… so i have to add an extra bc547 stage following the output stage of this circuit but if i am doing so i will have to take output for the bank in between the possitive supply and the collector of BC547.. MY ISSUE IS THAT i dont want to take output like that.. i want to make the LED bank hellow sir… this circuit idea is brilliant… can you please suggest an another simple circuit of same kind to implement -ve latch. i.e, output is first ON and when a signal applies output becomes zero and latches in that state…
my intention is to make an Emergency bank (7 W , 7 high brightness LEDs) lights up when a signal is applied as the input trigger to this above circuit… the problem is that. i cant connect the lamp in the output portion as shown in this circuit ( between collector of BC557 and ground ) since the current through this path is very less… so i have to add an extra bc547 stage following the output stage of this circuit but if i am doing so i will have to take output for the bank in between the possitive supply and the collector of BC547.. MY ISSUE IS THAT i dont want to take output like that.. i want to make the LED bank connection with respect to the GROUND terminal only… so please help me sir with respect to the GROUND terminal only… so please help me sir
Hello, you can do it by changing T2 with another BC547 (emitter to ground and collector to positive via a 10K resistor)…after this you can connect a TIP147 transistor base with the collector of this T2, and then connect the LED across its collector and ground (emitter to positive)
make sure to remove R4 in the new design…and connect C1 across the base and ground of T2
Hello I made the above circuit without diode or relay and when triggered I get 9v (from pp3 battery) at base of T2 as expected. The problem is if I put a load (9v lamp) at base of T2 the circuit resets ie. the 9v is no longer present. When no load my meter shows constant 9v when triggered as expected. Can you explain to me please?
hello, sorry i did not understand your requirement.
why would you want to connect the load at the base of T2?…are you referring to the collector of T2?? please clarify
Can you reccomend a modification that will allow the circuit to unlatch after the input voltage becomes lower than a lower threshold value and then the output latches to the off condition? This would be like having hysterisis. Thank you.
It could be probably done by tweaking the value of R4…however the results might not be as effective as could be with an opamnp circuit
sir,i want to use this circuit as a switch of relay from 230 volt ac line in the point to "input trigger"to the perpose of inverter switch on or off in the absent or present of 230 volt ac line.please, give a solution for the condition of this circuit.
Sudip, you won't reacquire such an elaborate circuit for your purpose, you can simply use a relay operated from a mains adapter, and use its contacts for the changeovers, as shown in this example article:
https://www.homemade-circuits.com/2014/06/how-to-convert-inverter-to-ups.html
I tried, it using connecting not gate IC in place or relay, and then output of NOT gate IC goes to realy to turn it off.
but it was not working
Hello sarang, you can use the above circuit for the said application, just do the following modifications to it:
remove C1 and connect a push ON switch across the base and ground of T1
hello sir,
what changes should I make, If I want to build a latch circuit to Turn OFF relay instead of ON. input should be NON RE-TRIGGERABLE..
Sir,
please provide me switching circuit using transistor in which, when to AC fails it turn on the Leds(powered by battery) and when it recovers the load is off. I have made this circuit using relay but it is bulky
Thanks sir,
I have made this automatic circuit and it is working perfectly.
Sir can you please explain the value of resistor for the base of the BD140. I am using a 6v transfo for charging in this circuit.
https://www.homemade-circuits.com/2011/12/automatic-white-led-emergency-light.html
that's great Syed,
anything between 1k and 4k7 1/4watt would be good enough.
hi sir. i need the on off latching circuit, did it work like this? i mean this circuit just latching on signal i need to triggre input again and relay cut off, please help me out
Hi Moein,
for that you will need the following designs, the above will not work:
https://www.homemade-circuits.com/2011/12/build-these-simple-flip-flop-circuits.html
Sir, i have found an anothet mistake too.
In the low voltage protection circuit i have chosen a low cut off DC IN voltage of 10.5 V, below which the relay activates and the supply to the logic section will cut off. But i forgot to think that if the voltage is too low such that( producing a 0 to 9V DC ) it will cause the relay to remain in idle state and willn't deactivate the logic section, thereby causing the motor ON/ OFF carrying out even in pow voltages ( say in 5-9 V ) . So whqt will i do. May i ask you onething tht whether it is possible to revert the logic at 741 opamp by just interchanging the 2 and 3 pin connections, which will solve this issue.
……Just swap the input pins of the opamp, meaning for pin2 use pin3…. and vice versa.
…change the relay contact connections also accordingly.
Hai sir,
I found a serious mistake in my current design when i just took a look on its working today. In my previous design, there was a provision to disable the circuit permanently after achieving the condition of overflow in the overhead tank. It will disable all proceeding circuitry unless it is switched on again, while doesn't allow the circuit to activate again when the downstoried tank is overflowing
Now considering the current circuit, the motor will be disabled as soon as the overhead tank is overflowed but it will again activate in the nearby future if the downstoried one is overflowed.
So could you plz suggest a slight modification in my current design in which, the latching stage once deactivated shouldn't be activated automatically unless the power is turned off and on again.
I will sen u the current design
Hi Arun,
use another latch circuit for the lower stage in place of the single BC547 transistor.
Thank u sir. I found the problem was due to the triggering of the circuit by ic 741 which was used to avoid activation of the circuit under low voltages. Isolating this stage by a 12 V relay with the proceeding portion solved the issue. Using the relay i could enable the 12 V supply to the logic section iff the reliable ac voltqge level is present. I got a decent 11.5 V something at the output while the o/p relay is working. The transients are still remains present. I could understand it by hearing
some nuisance signals in the dtain water indicator buzzer. But it doesn't affect the functioning of the ckt. much. I fear whether it will be a problem to the timer stage and the associated buffer. Anyway thanks for ur valuable supports sir………
That's great Arun, I appreciate your efforts.
Sir i am waiting to hear your valuable suggestion about the problem of reduction in voltage by connecting the relay
Due to lack of time i am not able to check the circuit that you have sent, I'll come back to you soon…
Sir when i am connecting the relay as shown in the latch circuit given in this article, the voltage at the output is getting reduced to 6 or 8 V. I have also tried by driving the output stage with a transistor. But no change. What 2 do ?
I have used an another NPN transistor for reset purpose.
Arun,
since the relay is the only load here which is consuming most of the amps, the voltage could be dropping due to it.
It could be happening because of the lower resistance of the relay coil compared to the input amps capacity. You can check the current consumption by connecting a mA meter in series with the positive supply….if it matches with the relay coil current would confirm the cause…and the problem could be rectified by increasing the input amps.
If this is not the case then it could some kind of fault or short in your circuit design which could be causing the issue.
The above circuit is perfect and should work immediately….
Hellow sir deva again. I was busy with some other works for the last 3 days, thereby couldn't update the results. Now i am entirely changing my circuit replacing the 555 stages by transistir set/reset circuits as you said earlier. I have skipped all unnecessary components such as mosfets, zener etc. from the previous design. Now i am here to seek for any options available for storing a single state even after power failure. I asking you such an option, coz i need to modify the circuit by adding the facility that it SHOULD PROVIDE US THE CURRENT STATE OF ACTION i.e, it should indicate a low iff the overhead tank had overflowed and should show a high state if it is under processing ( tank not overflowed ). This state shouldn' t fluctuate with any external disturbances such as power failure etc. That is, if it had shutt off it should remain shut off even aftr power reaches after a failure. I have designed such a one. But it is less effective. Couls you please put a look at that. I have sent it it your mails
Thanks sir. I got it. Now can you provide me a ' ON after 2 s delay ' circuit using transistor
please refer to the following design:
https://www.homemade-circuits.com/2013/02/make-this-simple-delay-on-circuit.html
Sir, in the circuit i requested before, the output should go low permanently when a trigger is applied
Usha, You can try the above circuit….take the output from the collector of the NPN transistor for getting the specified results.
……the relay may be removed and replaced with a 100k resistor.
Sir,
Could you plz suggest a transistor or other circuit that holds its output at high when power is applied and turns its output to low when a trigger is applied
Sir also, if i am gonna use a MOV or TVS diode or suitable RC network to suppress high transient spikes, how they can be connected to the present circuit ? I could understand that one terminal connected to the ac L line. Where the other end to be connected ? To Neutral or Ground ????? OR to the N/O terminal of the relay?????
requesting not to forget to say the the specification of TVS or MOV suitable for this circuit if is gonna used……….
According to me MOVs or any such device won't help, as mentioned earlier you can try replacing the ICs with (7)555 ICs or other CMOS alternatives.
In the latch circuit given in this article, the output voltage (latched voltage) reduced when the feedback resistance 100k is connected. I have used 12.4 V to supply the circuit, but got only 11.3 V something as the output voltage. When it has been checked keeping the feedback resistance disconnected i got full voltage. Tried with different valuee of resistances, but same was the results. In the circuit i requested before, i need full voltage as output and should have a default on stqte and should latch in low state when a trigger is applied. Could u plz help me sir??????
it could be happening due to high consumption by the relay coil…..try using a relay with 400 ohm resistance,
Now i have understood that all those problems were due to high transients from the motor as you said earlier, coz the circuit isfunctioning properly even on a mixer grinder…. So nothing to worry about 555 stages.
Now my doubts are:
1). Why the mixer grinder didn't produce transients eventhough it is basically an induction motor ?
2). How to chose an optocoupler or
Whether a bidirectional Transient Voltage Suppression (TVS) diode will solve it ? If it can, what specification of it to be chosen???
3). Is there are any other simple transient suppression circuits?
Expecting your reply soon
1) mixer grinder is too small compared to a 0.5 HP motor.
2) The switch ON transients from the water pump motor could travel throug the air also and could be easily picked by the opamps inside the 555, so I don't think it can be stopped by any means.