Doubt regarding schmitt triggers

[Shikhar]

I have a very basic doubt in schmmit triggers...suppose if I connect the required power supplies to the op-amp (ie. V+ and V-) and make the desired connections(see attached circuit). Now, what would be the output when Vin=0, ie. before the signal is applied.

 

[sophiecentaur]

It could be on or off as the initial input condition isn't specified. It's normal to use a pull-up or pull-down resistor (either explicitly or by virtue of what's in the previous stage) to define the output state at switch on.

 

[elect_eng]

I have a very basic doubt in schmmit triggers...suppose if I connect the required power supplies to the op-amp (ie. V+ and V-) and make the desired connections(see attached circuit). Now, what would be the output when Vin=0, ie. before the signal is applied.

The starting output condition is indeterminate in this case. It will depend on the exact OPAMP used and the initial transient conditions when starting. The output may be near V+ or near V-.

 

[Shikhar]

The starting output condition is indeterminate in this case. It will depend on the exact OPAMP used and the initial transient conditions when starting. The output may be near V+ or near V-.

so how would i know what would be the output at starting?
actually, i want the comparator to give a high signal only when the input signal (which is always positive but of very small magnitude) is applied. Before the application of input signal, the output must be low..
can you please suggest me a circuit or modifications for the same?

 

[sophiecentaur]

By defining the input volts at startup. But what is your input? What range of Volts are we dealing with? If that is well defined then the output state of the Schmitt is defined also. The Hysteresis of the circuit needs to be appropriate for the range of the input signal - enough to commit it in one direction or the other yet no so big that the input signal can't change the output. (This is set by the values of R1 and R2) The '+' input needs to have two states - separated by less than the '-' input signal swing but by more than the 'irrelevant' variations in the input signal, which you want to suppress.
Does the circuit use V+ and V- supplies? If not, then you may need to shift the input signal volts to somewhere around half the supply volts.
Your question is too undefined to make any concrete suggestions. Have you any experience with these things or is it your first involvement with Electronics?

 

[Phrak]

Like all electrical circuits, it could be metastable. The output could be high, it could low, or it could balance at zero for a while (metastable).

Place a pencil with an eraser on a table. Now hit the pointed end so it ends up standing on the eraser for a while. It doesn't happen very often, but if you do in enough times, the pencil will stand on the eraser. Do it enough times and you will see it balanced for a long, long time. This is the metastable state.

Say you have an operational amplifier or comparitor with two independ signals into the + and - inputs. Eventually the two will arrive an nearly identical times, and the output of the device will not switch fully from high to low, or low to high, but hang in the middle. This is metastable hang.

Now look at the specific question you asked "suppose if I connect the required power supplies to the op-amp (ie. V+ and V-) and make the desired connections(see attached circuit). Now, what would be the output when Vin=0, ie. before the signal is applied."

First, there is always a signal applied whether it be external or feedback. What was it before it was Vin=0. Was it an open circuit?


What is the specific sequence of events, including application of power to the OpAmp, that you are interested in?

 

[Shikhar]

First, there is always a signal applied whether it be external or feedback. What was it before it was Vin=0. Was it an open circuit?


What is the specific sequence of events, including application of power to the OpAmp, that you are interested in?

The signal Vin is initially open circuited. I want an output of 5V when the signal is applied(Vin would be the range of 5mv)

 

[elect_eng]

The signal Vin is initially open circuited. I want an output of 5V when the signal is applied(Vin would be the range of 5mv)

Now you are contradicting yourself. First you said Vin=0, now you say Vin is open circuited. There is a big difference between the two. You don't want to leave an input of an opamp open circuited, because a DC path must be maintained. Even though there is very high input impedance and the input current is negligible, the DC path must be there. You can place a 100K or 1M resistor from the input to ground or to one of the power supplies (if your opamp is capable of being driven to the rails on the inputs). If you tie to one of the supplies then your startup problem will no longer be there.

Also, Vin of 5 mV is very small and it is not easy to get a comparitor working in this range. Great care must be taken.

 

[sophiecentaur]

If you have as little as 5mV between the two states of the input signal then you need very little positive feedback or the Schmitt will never be pulled from its original state. Say R2 = 100KR and R1 = less than 20R. (depending on your output voltage swing).
Why not amplify your input signal by a factor of 100 before you go into the Schmitt?

 

[Shikhar]

If you have as little as 5mV between the two states of the input signal then you need very little positive feedback or the Schmitt will never be pulled from its original state. Say R2 = 100KR and R1 = less than 20R. (depending on your output voltage swing).
Why not amplify your input signal by a factor of 100 before you go into the Schmitt?

i can't understand why the schmitt won't work...

if i assume that initially the output is at +V(sat)...the triggering point would be at
+V(sat)*R1/(R1+R2). So as u said if R2=100Kohm nd R1=20 Kohm, then triggering voltage would be around 1mv...this is lower than my signal...so why won't the schmitt work?
is there any other consideration also here?

 

[sophiecentaur]

The absolute level of your 5mV, perhaps?
Try a range of input voltages. Can you get it to change polarity with any input signal? If not, there's something really wrong, I think. Check that both rails are working. What voltage is on the + input?. Are both supplies connected? Are the right pins connected to the right signals? (I'm clutching it straws here but I've done my share of daft things when building circuits so don't be offended)

 

[elect_eng]

is there any other consideration also here?

Well yes! Most definitely noise is the other consideration. Your signal level is not much larger than typical noise levels. Noise is one of the reasons why Schmit triggers are used in the first place. You want your trip-point voltage difference due to hysteresis to be larger than your noise level to prevel oscillations and false triggering. Like I said above, getting a comparitor to work at these voltage levels is not easy. The suggestion above to amplify your signal first is worth a try.

Another thing to keep in mind is that offset voltages on some opamps can be greater than 1 mV, some even as high as 10 mV.

 

[Shikhar]

The absolute level of your 5mV, perhaps?
Try a range of input voltages. Can you get it to change polarity with any input signal? If not, there's something really wrong, I think. Check that both rails are working. What voltage is on the + input?. Are both supplies connected? Are the right pins connected to the right signals? (I'm clutching it straws here but I've done my share of daft things when building circuits so don't be offended)

its all right sophiecentaur!
but i think u misinterpreted my question...i still haven't started working on the hardware...
jst designing the circuit right now!
...u told in your earlier post that i should amplify the signal by a factor of 100 before passing it through the schmitt...so i ws asking what's the need of it when jst the schmitt trigger can do the work alone by using the values of r1 and r2 as specified in the last post?

 

[sophiecentaur]

elect eng is so right when he talks of noise as being a possible problem. If you only need a bandwidth of a few tens of Hz then take your original signal pass it through an op amp filter with a gain of a few hundred at DC and put a capacitor in the feedback loop to bring the gain down after, say 50Hz. Then you will have a sensible level, low frequency, clean' signal for your Schmitt to look at. You can use a chip with two op amps on it so it won't take up much room.
Why use an amp plus the Schmitt? The Schmitt, being non linear, has a hard job to do the amplification and the decision making all in one go. The op amp can do some tailored filtering at the same time, too. Ideally, I guess you could say that the one circuit could do it all but the noise consideration (plus actually getting it to work) makes adding an op amp a better bet.
What is the source of your 5mV signal, btw?

 

[uart]

Shikhar, you still haven't specified enough about your input source.

Why is it open circuit when inactive. Is it swiched, if so what type of switch? What impedence is the input signal, where does it come from etc?

You don't seem to understand that you cannot design this system without more info about the nature of the input signal.

 

[Shikhar]

elect eng is so right when he talks of noise as being a possible problem. If you only need a bandwidth of a few tens of Hz then take your original signal pass it through an op amp filter with a gain of a few hundred at DC and put a capacitor in the feedback loop to bring the gain down after, say 50Hz. Then you will have a sensible level, low frequency, clean' signal for your Schmitt to look at. You can use a chip with two op amps on it so it won't take up much room.
Why use an amp plus the Schmitt? The Schmitt, being non linear, has a hard job to do the amplification and the decision making all in one go. The op amp can do some tailored filtering at the same time, too. Ideally, I guess you could say that the one circuit could do it all but the noise consideration (plus actually getting it to work) makes adding an op amp a better bet.
What is the source of your 5mV signal, btw?

thanks to all there...im really developing a good understanding of my circuit!

actually the 5mV signal is from a photodiode which i have used to detect the light during a bomb explosion. When the explosion takes place, the photodiode gives a current which when passed through a small resistance gives a voltage of about 5mV. I want this voltage to trigger a switch for which i require 5V. Thus, frequency is not much of a problem for me. I jst need a fast response circuit.

 

[Bob S]

i can't understand why the schmitt won't work...

if i assume that initially the output is at +V(sat)...the triggering point would be at
+V(sat)*R1/(R1+R2). So as u said if R2=100Kohm nd R1=20 Kohm, then triggering voltage would be around 1mv...this is lower than my signal...so why won't the schmitt work?
is there any other consideration also here?

Why don't you try R2=100 kohm, and R1=20 ohms, rather than R1= 20 kohms.?

Bob S

 

[Shikhar]

Why don't you try R2=100 kohm, and R1=20 ohms, rather than R1= 20 kohms.?

Bob S

sorry to all..actually i wanted to write 20ohms not 20Kohms...

 

[Adjuster]

Hello Shikhar,

I see your problem, but I also agree with the other posters that 5mV is uncomfortably low, as it is the same sort of order of magnitude as the random offsets associated with many op-amps. You could increase the input voltage simply by increasing the photodiode load resistor, so try this first. This will however increase the RC time constant of the input circuit, possibly giving too slow a response.

If this is the case, one possible solution might be to use a TRANSIMPEDANCE circuit around the photodiode. Try doing a Web search on this subject, and see this link to some notes by TI: - http://focus.ti.com/lit/an/sboa055a/sboa055a.pdf Note that TIs example shows the photodiode with zero bias, which is correct if you want to minimise the effect of the dark current. On the other hand, a faster response should be possible with some bias applied - in Fig.1 of the example, the side of the photodiode shown grounded would be biased negatively.

In a transimpedance receiver, an amplifier is used to provide negative feedback around the load so that the apparent time constant is shorter for a given load resistance. Note that the degree of possible improvement depends on the gain and bandwidth of the amplifier. The offset performance of the amplifier is also key (the offsets must be small, both the inherent input voltage offset, and any added offset generated by the input bias currents).

 

[Phrak]

The signal Vin is initially open circuited. I want an output of 5V when the signal is applied(Vin would be the range of 5mv)

First of all I want to know why you want to detect bomb explosions. Before that I can't help any further.

 

[Shikhar]

First of all I want to know why you want to detect bomb explosions. Before that I can't help any further.

Don't think it in the other way man...actually I am doing my training (a part of our course curriculum) in a defence organisation..we are working on the impact of explosions on a building for which I am using various sensors. Now, in order to trigger my circuitry, i wish to use the photodiode...

hoping for a fruitful reply this time

 

[Adjuster]

Don't think it in the other way man...actually I am doing my training (a part of our course curriculum) in a defence organisation..we are working on the impact of explosions on a building for which I am using various sensors. Now, in order to trigger my circuitry, i wish to use the photodiode...

hoping for a fruitful reply this time

I'm not happy about this, and now wish that I had not made my earlier suggestions. Incidentally, I think that if you really are training with a defence organisation, you should never disclose anything about it.
Posting such subjects on a public website really does not seem like a good idea, and in some circumstances could have extremely unpleasant consequences.

 

[elect_eng]

I have no prior or independent knowledge of the OP and what he is trying to do, but there is no indication that he is doing anything wrong. Unless he is working on a classified project, he is under no constraint to not mention the topic of his investigation. Also, there is no indication that he is doing anything harmfull. He is asking about making a detector for explosions, not about making explosives themselves. There is a big difference.

People who legally blow up buildings are interested in doing it in a controlled way and would want to study the process scientifically. Those doing so for the wrong reasons would not be trying to detect and analyze the process. Also, if he were doing something wrong, I think he would have avoided mentioning the application altogether. He's just trying to make a light detector and comparitor circuit. Give the guy a break.

 

[Adjuster]

I have no prior or independent knowledge of the OP and what he is trying to do, but there is no indication that he is doing anything wrong. Unless he is working on a classified project, he is under no constraint to not mention the topic of his investigation. Also, there is no indication that he is doing anything harmfull. He is asking about making a detector for explosions, not about making explosives themselves. There is a big difference.

People who legally blow up buildings are interested in doing it in a controlled way and would want to study the process scientifically. Those doing so for the wrong reasons would not be trying to detect and analyze the process. Also, if he were doing something wrong, I think he would have avoided mentioning the application altogether. He's just trying to make a light detector and comparitor circuit. Give the guy a break.

I'm not trying to be nasty to anyone, but where I come from almost anything connected with military technology is de facto classified under the Official Secrets Acts. The penalties for contravening this are very severe, and are absolutely rigorously enforced.

I concede that some countries have less of a secrecy culture, but I would hate to think of anyone talking carelessly and getting themselves into very serious trouble.

 

[berkeman]

thanks to all there...im really developing a good understanding of my circuit!

actually the 5mV signal is from a photodiode which i have used to detect the light during a bomb explosion. When the explosion takes place, the photodiode gives a current which when passed through a small resistance gives a voltage of about 5mV. I want this voltage to trigger a switch for which i require 5V. Thus, frequency is not much of a problem for me. I jst need a fast response circuit.

You contradict yourself with that last sentence. Fast response time means wide bandwidth.

As others have said, 5mV is not a practical voltage to work with, and shows that you haven't learned how to use photodiodes yet. For example, how does the reverse bias voltage of the photodiode affect its response bandwidth?

Are you receiving your light from an optical fiber? I would think that if you are trying to get timing information on explosions in a building (presumably multiple explosions as part of a controlled demolition), then getting that timing information as accurate as possible would be important. What are the response time and skew time requirements that you have been given?

It does seem a bit strange that they are using beginning university students for developing this instrumentation. From your questions, it does sound like you are just beginning to learn about electronics. Is your instrumentation going to be used in parallel with professional instrumentation?


EDIT -- I don't mean to be insulting. It's just that the story seems a bit off base so far...

 

[Shikhar]

THanks to all...finally i have come up with the circuit...
Please have a look and let me know of any errors or improvements...


...and for all those who think that i have bad intentions in my mind...i would like to point out that I am a final yr engg student doing my 2 months industrial training..
and if the matter was of such a secrecy..i won't be given this work!
I'm jst trying to learn things..in a practical way...and such comments are really discouraging for me!

 

[berkeman]

THanks to all...finally i have come up with the circuit...
Please have a look and let me know of any errors or improvements...


...and for all those who think that i have bad intentions in my mind...i would like to point out that I am a final yr engg student doing my 2 months industrial training..
and if the matter was of such a secrecy..i won't be given this work!
I'm jst trying to learn things..in a practical way...and such comments are really discouraging for me!

Why are R2 and R3 separate 10k resistors?

 

[Shikhar]

Why are R2 and R3 separate 10k resistors?

It was mentioned in the Application note of Photodiode that such a configuration helps in minimizing the parallel stray capacitance of the feedback circuit. I don't have any idea why is it so?