Design of a short circuit kit for DUTs (switch, opamp, LEDs)

In summary, the output of the OP-AMP is either high or low, depending on whether the circuit is open or short.
  • #1
RShan
11
0
Hi folks,

I would like to clarify few doubts on my mini-project as mentioned below:

I am designing a small kit using a transistor (2N3904) and an OP-AMP (LM741- Voltage follower) with few resistors connected to them. My power supply voltage value is 5V. Please refer to the below picture for your understanding. Am stuck in a situation where, I need the following logic as output but it is not the case.

  • When the circuit is OPEN, # The voltage at inverting input is half of the supply voltage (i.e., +Vcc / 2) # The voltage at non- inverting input is Zero # The output of the OP-AMP is either low or negative (i.e., LED 2 – status ON)
  • When the circuit is SHORT, # The voltage at inverting input is Zero # The voltage at non- inverting input is half of the supply voltage # The output of the OP-AMP is high (i.e., LED 1 – status ON)
I hereby attach the circuit diagram for your reference.

DSC_0232[1].JPG

I believe the above circuit will provide you a clear idea about my query. Hoping for a favorable reply from you folks.

Regards,
Shan
 
Engineering news on Phys.org
  • #2
You need some resistance from the base to ground, maybe 10K, so leakage won't turn the transistor on.

Once you do that, C will be VCC when DUT open, and 0.3 when DUT shorted.

You have the opamp configured as a unity gain follower such that the output will always be at the same voltage as the + input.

The 741 is a poor choice for an LED driver.

Is this a homework challenge? If so, it should be posted in the homework forum.
 
  • #3
meBigGuy said:
You need some resistance from the base to ground, maybe 10K, so leakage won't turn the transistor on.

Once you do that, C will be VCC when DUT open, and 0.3 when DUT shorted.

You have the opamp configured as a unity gain follower such that the output will always be at the same voltage as the + input.

The 741 is a poor choice for an LED driver.

Is this a homework challenge? If so, it should be posted in the homework forum.
I am doing this mini- project with my self interest in order to recall the fundamentals. It is not like homework. If this works well, then I will think further to advance it. Since am new to this forum, I thank you for sharing your views. Do you recommend me to use LM324?
 
  • #4
RShan said:
Do you recommend me to use LM324?
Yes, if you are using +5V and Ground only. Why do you show a split power supply to the opamp?

And in general, simple opamps will not be able to drive LEDs directly -- You should perhaps use a logic gate after the opamp as a buffer to drive the LEDs.
 
  • #5
Just some suggestions.
1. Find a 5V opamp with rail to rail drive capability.
2. You really don't need the input transistor. You can easily sense the open/short with the opamp directly.

3. Each Led should have a driver transistor. Opamps typically do not have adequate drive current to light an LED.
When the opamp is high, it can drive an NPN to conduct for 1 LED.
When the opamp is low it can turn on a PNP connected to Vcc to turn on the other LED.
You can also use small MOSFETs to do the same thing.

4. Or, instead of #3 You can use a CMOS driver like the 74LC244. I think there are 1 and 2 part packages like the 74HC2G34

5. You can easily do the whole thing with just the logic buffers (no op-amp)

6. Typically people use voltage comparators to drive analog threshold sensing to logic levels. (rather than an opamp)

I'll leave the details to you, but feel free to ask more questions.
 
  • Like
Likes berkeman
  • #6
berkeman said:
Yes, if you are using +5V and Ground only. Why do you show a split power supply to the opamp?

And in general, simple opamps will not be able to drive LEDs directly -- You should perhaps use a logic gate after the opamp as a buffer to drive the LEDs.
Thanks a bunch Berkeman.
 
  • #7
meBigGuy said:
Just some suggestions.
1. Find a 5V opamp with rail to rail drive capability.
2. You really don't need the input transistor. You can easily sense the open/short with the opamp directly.

3. Each Led should have a driver transistor. Opamps typically do not have adequate drive current to light an LED.
When the opamp is high, it can drive an NPN to conduct for 1 LED.
When the opamp is low it can turn on a PNP connected to Vcc to turn on the other LED.
You can also use small MOSFETs to do the same thing.

4. Or, instead of #3 You can use a CMOS driver like the 74LC244. I think there are 1 and 2 part packages like the 74HC2G34

5. You can easily do the whole thing with just the logic buffers (no op-amp)

6. Typically people use voltage comparators to drive analog threshold sensing to logic levels. (rather than an opamp)

I'll leave the details to you, but feel free to ask more questions.
Let me check this week and update you as soon as possible with the observed result. Thanks for your kind help.
 
  • #8
Whether these part numbers- 74LCX244 and 74LC244 (both from FAIRCHILD) are same?
You mentioned about using a transistor as the driving circuit for LED, Do you suggest me to connect 2N3904 in between OP- AMP and LED?
 
  • #9
It depends a lot on what you want to do with this learning exercise. As pointed out by meBigGuy, you can use a single logic gate to drive LEDs from a switch input. If you are wanting to experiment with opamps and comparators and transistors, then there are other schematics that we can suggest...
 
  • #10
berkeman said:
It depends a lot on what you want to do with this learning exercise. As pointed out by meBigGuy, you can use a single logic gate to drive LEDs from a switch input. If you are wanting to experiment with opamps and comparators and transistors, then there are other schematics that we can suggest...

I want to learn and understand the relationship between them. Could you help me come up with new simplified circuits or modifying the above attached picture for my understanding
 
  • #11
Not sure what to tell you. There are lots of ways to do lots of things given lots of components to choose from.

For example, google "pnp LED driver" and look at images.

Here is an LED driver solution using a logic buffer:
An LC244 type buffer can easily drive 10ma High, or sink 10ma Low so it can drive the led directly (through a resistor).

If you connected two 74LC244 (or 74HC2G34) like buffer inputs together you can then wire 1 output to drive 10ma into an LED, and the other output to sink 10ma from an LED.
(through a current limiting resistor)
Then, one LED would light when the inputs are high, the other when the inputs are low.
 
  • #12
Yes I will check with the above mentioned data.

Meanwhile, is there any other simpler solution available for LED driver? If yes, please let me know.
 
  • #13
RShan said:
Meanwhile, is there any other simpler solution available for LED driver? If yes, please let me know.

Depends on your definition of simpler. Less discrete components? Can't get much simpler than 1 dual buffer, two resistors and 2 LED's.
 
  • #14
meBigGuy said:
Depends on your definition of simpler. Less discrete components? Can't get much simpler than 1 dual buffer, two resistors and 2 LED's.
I wish to make it simple circuitry as I am a beginner. So, as you mentioned here, could you please suggest me if I can use LC244 buffer to drive the LED, and
to add up further before going to hands- on, let me recheck with you about the circuit in the form of block diagram based upon my understanding and mentioned below:

1. Power supply ---> 2. LM324 ---> 3. 74LC244 ---> 4. LED's

Please forgive me if I am unclear with you.
 
  • #15
You can draw a schematic using those components and we can comment on it. I won't draw one for you.

One note --- The LC244 has 8 buffers and you might only use two. All LC244 unused inputs need to be connected to VCC or Ground.
The LM324 has 4 opamps. The unused opamps should have a connection from output to - input (unity gain buffer) and the + input should be grounded.

Generally no component signal inputs should be left floating. For example, in yopur first drawing, the base of the transistor is floating. If it is floating, then leakage current from the Collector/Base junction will act like base current and partially turn the device on. A high value resistor to ground (like 47K) will provide a path for the leakage.

Also, opamps and logic components require bypass capacitors from Vcc pins to ground. Generally 0.1uF ceramic capacitors are used. Also, there should be a 1uF or larger capacitor near the power supply input.

Ground wiring should be short and direct.
 
  • #16
I believe the attached datasheets for the buffer circuit spec LC224 are one and the same. Shall I proceed with this datasheet? Need your advice
 

Attachments

  • 74LCX244.pdf
    767.3 KB · Views: 299
  • DSA0049310_LC224.pdf
    432.3 KB · Views: 279
  • #17
Those parts are different logic families.
The LVC and LCX parts are recommended for 3.6V operation. LVC looks good to 5V.
You can read about the various logic families here
http://www.ti.com/lit/sg/scyt129f/scyt129f.pdf
https://www.fairchildsemi.com/collateral/Logic-Selection-Guide.pdf

You should look at recommended operating voltage, and what voltage the outputs are when driving high currents.
You also need to think about what packages are available for the part. The Fairchild parts are surface mount only.

\It would be a good idea to read through and study a data sheet carefully (take 20 minutes or more) at least one time and study what sorts of things are specified and at what operating conditions. Subsequent data sheets can be digested quicker.

You need to do that for the 244 and the LM324. Understanding the data sheet is critical to designing with the part. Every spec has a reason for being there.
Also, you should look at application notes for the logic family or for the part, especially the LM324
(actually, the LM324 is an old opamp. Look for something that drives rail to rail and is stable at unity gain)

Seems like a lot of work before you get to do anything, but either you do it in the beginning, or you do it later while trying to figure out what went wrong.
 
  • #18
Sorry I was busy with my exams these days. But for a trial, I completed a short-open circuit using LM741. Please post your comments if this has any corrections or not.
Updated LED Driver.jpg


where, R1 = R2 = 10K ohms and R3 = R4 = 100 ohms.

Shall I replace the two LEDs in the circuit with a Bi-color LED? Need your advice.
 
  • #19
1. need a pulldown on + input. Open is not equal to zero (because of input leakage).
2. 741 won't run on 5V. Needs something like +-5V.
3. 741 won't drive rail to rail.
4. Opamp output drive is generally not enough to drive LED's, and it certainly won't drive to the rail when driving a big load.

An LM7301 is a better 5V opamp, but it also won't drive much current (most opamps won't)

You need LED drivers.
 
  • #20
Thank you @meBigGuy and @berkeman. I have completed that mini- task.
Sorry for my delayed response. I wish to be in touch with you guys for future reference if any.
 
  • #21
Be cool to see your completed schematic.
 
  • Like
Likes RShan
  • #22
I think i'd use a plain old 555 . Tie TRIG and THR together and it's a great Schmitt trigger buffer with reasonable Zin, works on 5 volts, plenty of output drive.
http://www.ti.com/lit/ds/symlink/ne555.pdf
 
  • Like
Likes RShan

1. What is a short circuit kit for DUTs?

A short circuit kit for DUTs (devices under test) is a circuit designed to test the functionality and performance of electronic components such as switches, operational amplifiers (opamps), and light-emitting diodes (LEDs). It allows for quick and easy testing of these components under different conditions.

2. How does the short circuit kit work?

The short circuit kit consists of various components, including a power source, resistors, switches, and LEDs. The DUT is connected to the kit, and the relevant components are selected and connected to the DUT. By manipulating the switches and varying the resistance, the DUT's behavior can be observed. The LEDs also provide a visual indication of the DUT's response.

3. What are the benefits of using a short circuit kit for DUTs?

Using a short circuit kit for DUTs can save time and effort in the testing process, as it eliminates the need for manually connecting and disconnecting components. It also allows for a more controlled and consistent testing environment, making it easier to identify and troubleshoot any issues with the DUT.

4. Can the short circuit kit be customized for different DUTs?

Yes, the short circuit kit can be customized for different DUTs by using different components and adjusting the circuit design. For example, the resistance values can be changed to accommodate DUTs with different power requirements or sensitivity levels.

5. Is a short circuit kit necessary for testing electronic components?

While it is not necessary, a short circuit kit can greatly simplify and streamline the testing process for electronic components. It can also help to ensure more accurate and reliable results, making it a valuable tool for scientists and engineers working with DUTs.

Similar threads

Replies
2
Views
426
Replies
55
Views
3K
  • Electrical Engineering
Replies
32
Views
2K
  • Electrical Engineering
Replies
6
Views
1K
Replies
9
Views
4K
  • Electrical Engineering
Replies
9
Views
3K
  • Electrical Engineering
Replies
12
Views
1K
  • Electrical Engineering
Replies
3
Views
1K
  • Electrical Engineering
Replies
22
Views
2K
  • Electrical Engineering
Replies
20
Views
4K
Back
Top