SparkyChem
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It's just Ohm's law.In the documentation referring to exclamation point there is a 3V pulse going in the input through the base of the transistor and it recommends that very specific voltage. However I don't have it available. Can 5V be used instead?. What modification should be made on that part?
Hi,
It's just Ohm's law.
Think about this:
1)
For a raw estimation you could say the 1.5kOhms resistor limits current at 3V to 2mA.
It should be no problem for you to calculate the resistor value that does the same for a 5V signal?
2)
For a more precise calculation you may consider that V_BE_ON is about 0.6V.
If you now use Ohm's law you will get a more close operating situation for the bjt.
3)
Use Ohm's law to calculate the current through the 1.5k Ohms resistor with 5V input signal.
Then read the bjt's datasheet whether the bjt can handle this base current.
Klaus
As Klaus mentions, it is only related to re-calculating base resistor values.
That particular transistor all that does is to fully discharge the capacitor that had been previously charged with the signal voltage.
After the base drive is released, the capacitor will recharge again.
The resulting display really grabs your attention.
For all that matters, the bipolar transistor could even be replaced with a small signal Mosfet, like the 2N7000
Your calculation.the collector current is 200 mA
Hi,
Your calculation.
I3 goes to the base, thus you need to look formax base current, but you said collector current.
but max base current is just to check if it can withstand the increased (5V instead of 3V) base current.
You don't go to the limit.
When you calculate with base current on a switching (saturated) bjt, then you should consider I_base = I_collector / 10.
Where I_collector is the max expectable load current and not the maximum limit.
Sadly I have no experience with LM3914, thus I can't help you with that.
Klaus
Hi,
I haven't thought this through fully but to give a rough idea, could you use a non-inverting OA (for the +5V) and an inverting OA (for -5V) type of circuit and feed their signals into a single supply device, maybe a non-inverting summing OA, which is used as the input device to the LM3914?
I recommend having a good browse through op amp application notes for similar circuits, maybe a pos/neg peak detector could work. As far as I can see, you'll need a level shifter on a dual supply for the negative range to make the output polarity positive.
I assume you did not read through bjt basics.Not sure why i_base=I_collector/10 ? Does it mean that the gain is just 10?. If so why?. I'm assuming that this comes from the fact that Ic=ib x beta
But in your calculation you used 200mA....Do you really expect 200mA collector current?Where I_collector is the max expectable load current and not the maximum limit.
Hi,
I assume you did not read through bjt basics.
h_fe or transistor gain is mainly relevant for analg bjt circuits....but in your case it is a switching bjt circuit.
Thus you "overdrive" the base (current) in a way that the bjt becomes saturated and V_CE becomes close to zero.
--> read through "linear" or "analog" ur "amplifier" bjt circuits
--> read through "switching" bjt circuits
The value "1/10" is a rule of thumb for usual bjts (no daringtons) in switching circuits.
But in your calculation you used 200mA....Do you really expect 200mA collector current?
True, the max base current is not given in the datasheet.
I_C/I_B = 10 is given several times. This gives 20mA, but I assume the base can withstand higher current, but it doesn't make sense to drive much higher current..
Klaus
Hi,
I think you want to combine two different circuits and that may be somewhat difficult. If I had the time I might try to square that circle but currently I don't, sorry. The CD40109 is only for positive voltage shifting. To give an idea of what I was suggesting - and this is a crude, rushed attempt that might have some issues in the real world outside a simulator - to invert the negative input and as you will see, use as many devices as I could think of squeezing in to the circuit (joke). It's a starting point that can be much improved. You would need e.g. an astable 555 to make the 1kHz flashing signal whose reset pin could be controlled by a comparator even.
As I said, this is a rushed circuit and needs reducing and refining, and the 555 added, and I'm not even sure it's what you were asking about beyond trying to combine two different LM3914 circuits, which as I said, at present I do not volunteer to wonder how on Earth to Frankenstein together, I'm afraid due to lack of free time.
Let me know where I've misunderstood you and I hope to dedicate more time to this at the weekend.
View attachment 149799
View attachment 149800
Btw, what software did you used for simulation?
I have some problems understanding the first op amp (referred to as 1/4 LM324) why there are three 5.1k resistors in series?. This set up have you used is it a difference amplifier?. The second op amp (the other 1/4 LM324) looks as a summing amp as you mentioned earlier. The second stage is more clear for the comparator. But why there are 5.1k and 100k in the output of the LM193?.
The rest an inverter will make +V into -V but the rest I don't get very clear the idea why an or gate will be needed?.
At the end of the circuit there is only one output but how does this controls one side to remain off (between let's say 0 to +V) while the other on (-V to 0)?
You intend to include into this circuit a 555 as an astable (although any other IC could be used for this like CD4047) ? But since you said that the reset pin (on the 555) might be needed as to be controlled by the comparator you mentioned.
It looks like a good way to start but the design doesn't seem to get the idea.
Maybe this can be better illustrated by this flux diagram and it can help you to get an idea of what this riddle is about.
I hope this makes clear the idea, although the issue with making LM3914 to be as an exclamation point and alarm flasher, from the looks of it seems that is not possible?
Still there is some misunderstanding.Again, I was not sure if the collector current was 200 mA, since this was the only value that I could find in the datasheet.
For sure you can't find the value for the load current in the bjt datasheet.Where I_collector is the max expectable load current and not the maximum limit.
Hi,
Just to clarify:
I don't want rocket science here, either. I don't expect exact nodal analysis.
And in post#2 I gave two alternative solutions, which do not involve collector current...the OP is free to choose the one solution he likes the most.
I want the OP (and all future readers of this thread) to understand how to calculate a base resistor on his/her own, not only for this particular situation.
For sure one may use 20mA ... but if the OP does not know where this 20mA come from he/she will not learn anything and thus has to ask similar questions again and again.
Klaus
I didn´t take it as offense. Nothing to worry about.Definitely not aimed at you, quite the opposite.
Hi,
Still there is some misunderstanding.
For sure you can't find the value for the load current in the bjt datasheet.
If the load is a standard LED fir example you may set the LED current to 20mA... then the LED is the load
The LED current = the load current = the collectir current.
For a motor it may be 2A, for a relay it may be 70mA .... it simply depends on the circuit connected to the collector
Klaus
Hi SparkyChem,
Just to clarify:
The supply voltage you will have available for the LM3914 (and any additional circuitry needed) is +5V DC or +-5V DC or what?
What will the circuit be powered by, a wall wart/plug in power supply or a battery?
The input signal range maximum value is +-6V DC or what maximum value can it reach? Is the input signal AC or DC?
Otherwise, this feels rather like being asked to build a house but without knowing how many m2 are available for the foundations.
Thanks.
Hi SparkyChem,
Just grabbed an hour or so to fiddle a bit further with the idea. I can't help but feel that my version is parts-heavy and could surely be reduced if considered from another viewpoint. Anyway, here's the same version with a 555 added for the flashing functionality. I'm thinking about how to deal with the LM3914, personally, I'd suggest choosing either an exclamation mark version or a bargraph flasher but not try to combine both after looking at the schematics in the LM3914 datasheet more carefully...
Please bear in mind, this is just a circuit suggestion... The reason for the 4069 (or you can use the 4049, they're the same thing, basically) inverter IC is because otherwise the circuit has the undesirable quality/penalty of using more current doing nothing than when there are overvoltage events due to how comparators work, so it seems wiser to have the comparator inputs as in the schematic and add the inverter IC to use less current overall, milliamps or not, with a battery it would be not good to squander current...
The schematic:
View attachment 149821
The results of a simulation:
View attachment 149822
A close-up of what the 555 and NPN will do when the voltage is at or over (in this design) +6V or - 6V:
View attachment 149823
Hi,
Did a couple of sums for the LM3914...if I've understood what the datasheet requires...
First set of calculations were for a 3V full scale, but that is hard, i.e. weird-looking, to divide 6V into 10 LEDs and 3mA per LED. Also calculated for around 9mA per LED.
You can have the results if that scale appeals to you.
To use 6 LEDs to represent 6V input for the Ref pin rigmarole and the internal divider scale I would suggest dividing each input Volt down to 500mV. Four LEDs would be unused on both LM3914s unless you want to do the flashing bargraph version.
R1 = 4.9K
R2 = 10K
I LED = 3mA.
Not willing to spend another hour doing yet more iterative calculations for about 10mA LED current, I'll leave that pleasure for you. 3mA should be more than bright enough for average LEDs.
Ref Vout = (1.25V * (1 + (10000/4900))) + (0.00012A * 10000)
Ref Vout = 5.001V
Therefore each LED will turn on at 500mV increments from 0V to 5V, which represents a real world scale of 0V to 10V.
I LED = (5.001V/14900 Ohms) * 10
I LED = 0.0033A
For the above scale all you need to do is divide the input voltage by 2. That could be accomplished by a voltage divider made up of two 5.1K resistors. The LM3914 signal input must be limited to below 3mA.
"INTERNAL VOLTAGE REFERENCE" and "CURRENT PROGRAMMING" descriptions are on page 9 of the datasheet.
"2) Pin 5 input current must be limited to ±3mA." Notes at the bottom of page 3 of the datasheet.
Using 6 LEDs in a straightforward way that is simple to interpret and lighting up all ten for the positive and negative overvoltage events seems okay to me. The exclamation mark version lights up the number of LEDs that correspond to the input voltage.
How have you been getting along with this circuit?
Hi SparkyChem,
You are still alive, and keen, are you?
Without wanting to sound too stroppy:
What's the limit at both ends for the exclamation mark (or the bargraph) to flash, +-5V or +-6V? Big difference.
It's neither clear from your first post nor from post #10. You keep saying +-5V and +-6V in the same thread posts...
It's pretty pointless for me to try to calculate this circuit when that aspect isn't clear to me, I'm afraid. I'm also having to guess that the supply voltage is +5V DC. I'm not doing theoretical homework for fun, I'm using my free time to try to help you. Some clear input from yourself would be appreciated if you are still planning on making this circuit.
My conclusion so far to make this as simple as possible is to separate this off into two LM3914 exclamation mark circuits. One will display negative voltages and the exclamation mark at -5V (or -6V) while the other will have no LEDs lit up and vice versa for +5V (or +6V). Each will use up to six LEDs/outputs and the "dot" one that is not an output but only goes on when the limit voltage is reached to flash the "exclamation mark". The exclamation mark version lights up one LED per Volt input signal according to the description in the datasheet.
How much current do you want for the LEDs? 3mA is more than enough for hand-held devices or something in the same room as the viewer. If you plan to illuminate a football stadium and want to max. out the output current to ~20mA per LED (with all 10 LEDs on) then it would help to know. The device can dissipate 1365mW. 10 * 25mA = 0.25A. 0.25A * 5V = 1.25W. Adding for housekeeping current the LM3914 will use, I'd guess the max. per LED is realistically about 20mA to 25mA, this amount of current is OTT for most LEDs and a waste of energy IMHO, and the LEDs will wear out sooner. Ant the LM3914 will get nice and hot, and wear out sooner. "Less is more" in the long run, I would say. Maybe 10mA if needs must, but as I said, 3mA to 5mA is ample LED current. Always derate and avoid several undesired issues.
Your explanatory diagram in post#10, translating +-6V signal to +-5V with 10 LEDs - as I said above somewhere - is pretty grim. Think about it, the numbers don't fit: 20 outputs and representing +-12V with them - that's not intuitive to the eye/brain. I'd want something that at a glance the lit-up LEDs told me the actual voltage, not making me do mental arithmetic. If I'm wrong about this assumption, let me know and my apologies.
The LM3914 datasheet points out the issue of:
"DOT OR BAR MODE SELECTION
The voltage at pin 9 is sensed by comparator C1, nominally referenced to (V+ − 100mV). The chip is in bar mode
when pin 9 is above this level; otherwise it's in dot mode. The comparator is designed so that pin 9 can be left
open circuit for dot mode.
Taking into account comparator gain and variation in the 100mV reference level, pin 9 should be no more than
20mV below V+ for bar mode and more than 200mV below V+ (or open circuit) for dot mode. In most
applications, pin 9 is either open (dot mode) or tied to V+ (bar mode). In bar mode, pin 9 should be connected
directly to pin 3. Large currents drawn from the power supply (LED current, for example) should not share this
path so that large IR drops are avoided."
In my opinion, unless I'm missing something, a miserably tiny 20mV max from V+ makes trying to alternate between dot mode and bar mode unlikely with any blocking/pass device I can think of due to inevitable voltage drop across a BJT or a MOSFET or a relay or anything else on this planet I can think of as a pass device apart from thin air so as to alternate between the two modes... I strongly recommend selecting one circuit or another but not hoping to combine three different ones from the datasheet as though this were a pick-n-mix sweet shop.
I'm trying to help, even if it may not always sound like it.
- - - Updated - - -
...Whoops, extremely bad mental lapse there...
5V LED supply - 1.7V LED Vf = 3.3V into LM3914 "outputs".
3.3V * 0.025A LED current = 0.0825W.
0.0825W * 10 "outputs" on = 825mW.
Supposedly the LM3914 comsumes about 10mA or so = 50mW
Total PD at all LEDs on presumably around 0.9W...
I'd still avoid that level of power dissipation unless it's momentary and sporadic.
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