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How to make a Transistor Latch – Testing and Working Explained

Last Updated on by 105 Comments

In this post we learn how to make a simple transistor latch circuit using just two BJTs and a few resistors.

Introduction

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 described in this article can be made very cheaply using just a couple of transistors and some other passive component.

 

simple transistor latch circuit
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.

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

PCB design for transistor latch circuit

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Comments

  1. 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.

    Reply

  2. 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.

    Reply

  3. 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.

    Reply

    • 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.

      Reply

  4. Hi Swagatam.

    I made a similar circuit which works quite well, but I noticed that in my own circuit, touching a wire or one of the components (the BC547) would turn on the circuit. That is not acceptable in my application.
    So I started looking for circuits on the net to see if I could find a solution.
    I came across your circuit, built it and tested it, but this circuit has the same problem; if I touch the base or the collector of the BC547, the circuit turns on. Even if I touch the emitter of the BC556, the circuit also turns on.

    Can this be prevented ?

    Reply

    • 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

      Reply

  5. 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..??

    Reply

    • 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.

      Reply

  6. 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

    Reply

    • 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

      Reply

  7. 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

    Reply

    • 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)

      Reply

  8. 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?

    Reply

    • 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

      Reply

  9. 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.

    Reply

  10. 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.

    Reply

    • 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://homemade-circuits.com/2014/06/how-to-convert-inverter-to-ups.html

      Reply

    • 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

      Reply

  13. 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

    Reply

    • 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://homemade-circuits.com/2011/12/automatic-white-led-emergency-light.html

      Reply

  14. 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

    Reply

    • Hi Moein,

      for that you will need the following designs, the above will not work:

      https://homemade-circuits.com/2011/12/build-these-simple-flip-flop-circuits.html

      Reply

  15. 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.

    Reply

  16. 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

    Reply

  17. 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………

    Reply

  19. 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.

    Reply

    • 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.

      Reply

  20. 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

    Reply

  23. 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

    Reply

  24. 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……….

    Reply

    • 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.

      Reply

    • 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??????

      Reply

  25. 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

    Reply

    • 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.

      Reply

  26. Latest news:-
    I hav tested the circuit by connecting an another similar relay used to give an external 12 V supply to the motor driving relay. The new relay is controlled by the voltage from the output transistor driver stage. Thereby overall the motor has been isolated from the driver. Now i got thr motor functioning. So what should i have to understand from this ? Is that the problem caused by the high back emf associated with thr motor as you said
    Or
    the problem due to 555 stages ?
    If the high back emf may be the real reason, i think the whole home wiring would have been get affected from this atleast the Refrigerator kept operated near it would have been affected. But ni such problems are there in connecting the motor as usual

    Reply

    • The 555 IC stage cannot be compared to a refrigerator, because the 555 is an electronic stage consisting of analogue circuitry which can be vulnerable to noise and transients.
      Try replacing the 555 with (7)555, these are same as 555 ICs but are CMOS versions which are entirely resistant to external noises and disturbances.

      Reply

  27. Exactly the problem what you said sir.
    The circuit is getting disturbed as soon as the motor is turned on automatically and not giving intented results. But the same works perfectly when a table fan is connected in place of the motor.

    Since the driver transistor drives so many stages such as CD4060, buzzer indicating motor running, also the relay i thought that there are some disturbances arising to these stages due to problems in relay stage since they are all under same section. Hence i decided to drive the relay on a separate stage thereby chose the MOSFET switching stage. On searching internet about flyback problems faced by relays, i found a config. ( as shown in the diagram as the combination of diode and zener in phase to phase ) which nest suites for relays in terms of avoiding transient suppressions and retaining long life. So i used it.
    And about the extra capacitors used :-
    When unexpected beaping sounds from the buzzer i just put a 100 uf capacitir across the driver terminals in theintention if avoiding any nuisance signals, by a luck i got succeeded. I have repeated the same in relay stage also. These are to say about the extra components.

    Now i think you could get my problem which is exactly same as you thought.
    Everything i needs now is a perfect solution for this.
    Expecting that soon with a good mind

    Reply

    • The problem is not due to relay coil back-emf, it's due to the pump motor winding back emf and transients that's causing the problem. The motor could be sending strong back-emfs over the entire housing wiring.

      As you have said that when you connect a fan or any other load, the circuit behaves fine….just switch ON the pump motor externally while the fan is connected, and check whether the fan gets rattled or not.

      If not it would mean that if the relay stage is isolated using a optocoupler could solve the problem….if the problem is witnessed then probably you'll have to get rid of the IC555 stages and replace them with other alternatives.

      Reply

  28. Sir, i forgot to say some last modifications done. I have again connected 1N4007 in between +12 V and the collector of output driver transistor also a 100 uf/25V capacitor has also been connected in b/w these terminals. These modifications eliminates the fast repeating sound of the buzzer BZ1 and made it functioning well. I have disconnected the relay feeding voltage terminals from this output driver stage and used a simple MOSFET( IRF540 n ) switching circuit to control the relay voltage. A 1N4007 and a 12 V , 1W zener diode are connected( in phase to phase ) in between the relay control voltage terminals ( MOSFET stage ) as a flyback. I will try to send this to your emails.

    Reply

    • Arun, I could not understand what's the exact problem with your circuit…I thought as soon as the motor is being switched ON by the relay the circuit is getting disturbed and not giving the required results…is it so??

      Reply

    • …I could not understand why you have so many parts, diode, zener diode, mosfet….all these have no relevance to the circuit functioning.
      Just one diode across the relay coil is all what is needed.

      Reply

  29. Sir it is nof the problem due to relay. I got my motor working when an external 12 V trigger has applied. So what will i do now ?
    I know i am disturbing you much. but what to do, i will have to complete this anyway

    Reply

  30. Sir,
    what must be the correct specification of the relay to be used with a 0.5 HP pump motor ?
    I have corrected all the problems and the entire circuit was found working perefectly on loading with a table fan, but when it was loaded with my 0.5 HP motor, water had not been pumped to the tank ( indicated by the drain motor section, also verified by physical examination ). Water flowed as there was an obstacle which restricts the pumping action also the entire circuit was seemed to be loaded…
    The written things on the relay i have used are shown:-

    50/60 Hz
    7A 250V ~ 10 A 125V AC
    12A 120V – 10 A 28V DC

    Reply

    • Arun, the relay is OK,
      try operating the circuit/relay through a 12V battery and see if the problem persists or not.

      If still it behaves the same way, we can try isolating the relay power supply by employing a separate external DC for the relay/4060 sections.

      Reply

  31. Atlast i got it sir. Since all the problems were seen to occur while connecting the relay stage, i thought that some types of backward emf is flowing from the relay to the preceding stages, so i have disconnected the wires feeding control voltage to the relay and connected two diodes (1N4007) as shown below which has solved the issue to an extend.

    Possitive wire …….-|>|………Relay possitive terminal
    Negative wire………|<|- ………….Relay negative terminal

    But my doubts are
    1). What is the cause for this back emf
    2). whether this could be taken as a permanent solution or whether i have to chose any other back emf protection circuits, if i have to could you please suggest me one

    Reply

    • Deva, in your circuit you already have one diode installed across the collector and positive of the transistor which is used for triggering the relay and the 4060 stage….so back emf cannot be the issue.

      The issue could be with the 555 IC1 and IC2, which needs to be identified patiently or better be replaced with transistorized set/reset circuit.

      Reply

    • I have checked the diagram, it looks similar to the previous one.
      Yes, the phase or the neutral either of the lines can be used with the relay for getting the cut offs, however connecting the phase to the relay or any switch is recommended.

      Reply

  33. Sir i have sent to you a modified version of the previously presented circuit. Would you mind checking your both emails ? I have some doubts regarding it. Expecting a satisfactory reply soon

    Reply

  34. Sir,
    Could you please suggest me a circuit, the output of which should go high for programmable ON time and when the time lapses turns the output to zero permanently untill a next manual trigger is applied.
    I have searched many timer circuits, but majority of them have ON time as well as OFF time yielding the output to turm ON again when OFF time lapses

    Reply

    • Arun, you can try the circuits shown in the following article:

      https://homemade-circuits.com/2012/01/how-to-make-simple-versatile-timer.html

      The power to the circuit may be provided from any 12V dc adapter instead of the shown capacitive power supply.

      Reply

  35. No way sir. I am tired of checking my circuit many times after soldering…..
    Circuit is said to be fine, checked by you and me. No problems detected during in bread board. But i can't identify what is happening when it has been moved into the PCB.
    The IC3, IC4 stages and thereby relay are switched on when power reaches IC1. This didn't occur while testing in BB. According to my design these stages are switched on whenever the sensing nodes dipped in the lower tank ( A and A' ) are shorted by overflowing water, but what to say it is turned on even initially on powering the circuit. All the connections are examined, no errors found. Everything is perfect.
    The circuit is completely formed by me, thereby may error occurs. Can you identify anything after arranged in PCB SIR ?

    Reply

    • You can connect LEDs in series with the bases of the transistors, this will give you the indications regarding the relevant triggering.

      You can also try testing the stages separately, by cutting the PCB tracks between the stages and by checking the response of individual stages with manual triggering.

      Reply

  38. Sir a simple doubt,
    How this latch can be made disabled once it is activated ?
    I mean without turning of the supply..
    I couldn't disable it after activating the latch even connecting a low voltage as threshold

    Reply

    • Deva, I'll need a diagram to understand the problem, without the diagram it will be difficult to solve the issue.
      You can send the diagram to my email, if possible I'll check it.

      Reply

  39. Sir i have succeeded with ic 4060. A 555 ic has been used for generating the main logic. According to my design, when pin#2 of ic 555 gets low voltage output pin#3 becomes high which then switches on the transistor connected to this terminal. I have connected the timer stage in between +12 V and the collector terminal of this transistor, hence the timer stage is switched on when the pin#3 of 555 becomes high. pin#6 of 555 has been connected to pin#3 of 4066, so when timing lapses (100s ) makes this pin high, resulting in low value at the output pin of 555 also the timer stage….. The logics at these terminals ( 2 and 6 ) hence thereby used to establish the necessary turning on and off the motor.
    Now i am in dilemma, plz help me sir.
    According to ma diagram, motir switches on whenever the water overflows in the ground storied tank and switches of whenever time elapses or the overhead tank overflows…. But this continues till water stops in public water tap…. but, always my overhead tank overflows by second stage of pumping water from the ground storied one. So i want a slight change in this circuit that, after detecting an overflow in the overhead tank the motor shouldn't be turned on untill a next manual activation is detected

    Reply

  40. Hellow sir, can you please help me
    I want a simple latch circuit that holds the output signal for about 100 Seconds.
    I have tested timer circuit using CD4060 but didn't get my requirement, the result was ON time for 100 s and OFF time for the nearly same time, thereby output voltage again becomes high after 100 s.
    I am in the middle of constructing a simple water level controller. My intention is to fill my home water tank with water from a ground storied small water tank. The water from public water connection is being filled in the ground storied tank. So when the water level in that tank raises to its maximum i want to activate the motor automatically and then turn off the motor after 100 s again starting the motor when water level begin to overflow the lower tank.. This should be continued till the overhead water tank just begins to overflow…. So i want a latch circuit for 100 s

    Reply

    • Hello Deva, you can try the circuit which is explained in the above article.
      Just connect a 220uF series capacitor with R3, and make R3 = 2.2M
      Now whenever the input trigger is connected to positive via water, the relay will activate and the circuit will latch until the 220uF cap is fully charged after a predetermined delay.
      The delay will depend on the values of the 220 cap ad R3 which needs to be adjusted for getting the 100 seconds delay.

      Reply

    • Sir i saw your reply at morning. I have changed the logic of filling the upper tank in another way compared to that said at last night. I have told you last night to help me in placing a logic for counting the number of times the lower tank is being filled with public water and thereby inhibiting the condition of overflow of the uppper one. Now i want to place a different logic that, whenever the overflow condition is detected once in the upper tank, i have to inhibit any future activation of the motor even the pin#2 of 555 gets lower threshold ( during overflow in lower tank )….. Expecting your suggestion soon sir

      Reply

    • Hi Vasilis,

      I think you won't need this additional circuit, because your pellet burner first 555 stage is itself a latch. You can add another BC547/relay stage with this existing monostable and invert the response. The base of this new BC547 stage can be connected with collector of the monostable BC547 via a 10k resistor. As soon as the monostable time lapses, the new stage relay would get activated for the intended actions.

      Reply

  42. Hi Swagatam

    The INPUT of this latch circuit is very very sensitive.

    1) Sometimes the latch triggers when the input is touched with finger.

    2) Sometime it triggers when the circuit is powered on thru a mechanical switch. It seems that when the switch is switched on, some signal is induced within the circuit which causes unexpected triggering of the latch.

    How can we sort out these problems?

    Reply

    • You must have a pull-down resistor on the input, pulling the base down to ground when there is no positive signal, else stray voltages (such as those from your hand) can trigger it.

      Reply

  43. Hi Swagatam

    I have used this latch circuit in my projects.

    Recently, I came across more simpler but identical circuit which does not has C1, R2 and R4. Instead a 1K resistor is there in place of C1.

    Just want to know what is the difference between the two circuits in terms of function/performance.

    Reply

    • Hi Abu-Hafss,

      No difference at all, the above circuit can be simplified to much greater extents, I have just provided the basic or the standard design here.

      Reply

  44. If i replace the realy with say a buzzer do i still need the diode? If so how would i connect it all?

    Reply

  45. Hi All,

    I need help please, I need a circuit that will allow between 2.5 +v and 10+v to pass through it at up to 4amp. I need the circuit tobe able to switch from voltage supply (a) to voltage supply (b) once teh voltage in supply (a) drops to say 3.5 +v.

    I can solder really well and understand basic electronics.

    Many thanks,

    Ian

    Reply

  46. hi swagatam,

    i need help from you. I configured the circuit posted at https://homemade-circuits.com/2011/12/how-to-make-simple-low-battery-voltage.html to work as high voltage indicator.

    Now my question is how to integrate both the circuits so that the above circuit posted here latches, when the output of opamp 741 in the link (configured as high voltage indicator) is high.

    Please help me.

    Reply

    • Hi Anonymous,

      There's no need to use the above circuit for latching, just connect a 10K resistor across the output of the opamp and the preset input of the IC.

      By the way for making the circuit work as hi level detector, i hope you've interchanged the input pin number positions….

      Reply

    • Thanks for replying.

      yes i have interchanged the pin positions for hi level indicator.

      What i am trying to do is reduce the output voltage of LM317 when particular voltage say 13v is reached. i.e instead of completely cutting off charging when the threshold is reached, i am trying to reduce the voltage.

      The output voltage of lm317 is reduced by connecting a parallel resistor across the resistor R2(the one connected from adj pin to ground).

      i thought of achieving this by using the op-amp output to latch the above transistor circuit posted here. Then connect a resistor across the LM317's resistor R2 using the relay shown above.

      Reply

  47. replacing 100K resistor by 15K the circuit worked fine….. why it didn't work with 100K resistor…and why worked with 15K resistor…pls help me again

    Reply

  48. i used load(led) instead of relay. but i dont know where exactly my work went wrong. I tried twice and there is correct voltages according to my calculations but circuit dont latch. i used all components of given specification. pls help me

    Reply

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