Understanding arduino circuit and transfer to breadboard

Hey,
I am trying to connect my new Bluetooth module, HC-06 ZS-040, therefore I am using this tutorial:
http://www.martyncurrey.com/arduino-and-hc-06-zs-040/
but I ran into trouble when I saw the circuit:

first, I am using Arduino Uno, not such a problem,
second I am using breadboard and I am not so great reading circuits, so I created a Fritzing circuit, using breadboard and Arduino Uno and I would like to ask if the circuit I provided equals to the circuit I attached as picture. (link here)

Thank you :slight_smile:

NOTE: I forgot to attach jumper from ground to the breadboard ground line, note this as done.

You need to connect the gnd to the Uno as well.

Weedpharma

weedpharma:
You need to connect the gnd to the Uno as well.

Weedpharma

Yes I said I forgot, thank you :slight_smile: can I do it without 2k ohm resistor? I have only 1k, or can I put two of 1k?

davidbalas25:
Yes I said I forgot, thank you :slight_smile: can I do it without 2k ohm resistor? I have only 1k, or can I put two of 1k?

What that part of the circuit does is act as a voltage divider, to drop the output voltage of the data pin on the microcontroller (5V = HIGH) to a lower value (in this case, since the divider is configured in a 2:1 ratio - by half - to 2.5V) - because the device is a 3.3V TTL device (and 5 volt TTL would damage it).

So, no, you need that resistor.

As long as you keep the ratio the same, you could use other resistors (within reason) - so, if you had - say a 500 ohm (470 ohm would be ok) - and a 1K resistor, you could sub those in for the 1K and 2K resistors, respectively. Or, if you had a 2K and 4K (or whatever is close) that would work too. Just keep the ratio the same, and don't go too low or too high on the values.

You can use two 1K resistors - if you arrange them in series (NOT in parallel).

All of that said - a voltage divider is not the best way to do this kind of interface to these devices; you will be limited in speed (far lower than what you could really be able to use) - if this isn't a problem for your app, then don't worry about it - but if you need as much speed as possible, look into another solution that doesn't use resistors (rather, look into one that uses level translation via an IC or individual mosfets).

Lastly - your question indicates a lack of understanding or knowledge on your part about Ohm's Law - and how (and what) a voltage divider is for. These two items (and a whole host of others) are basic electronic building blocks and knowledge. What you are trying to do is build a house, without understanding what a hammer is, what nails are, or how to use a tape measure properly. But you have a large pile of wood, and you know how to saw it apart...

Get that knowledge. Also - pick up and learn to use a multimeter, if you haven't got one already.

cr0sh:
What that part of the circuit does is act as a voltage divider, to drop the output voltage of the data pin on the microcontroller (5V = HIGH) to a lower value (in this case, since the divider is configured in a 2:1 ratio - by half - to 2.5V) - because the device is a 3.3V TTL device (and 5 volt TTL would damage it).

So, no, you need that resistor.

As long as you keep the ratio the same, you could use other resistors (within reason) - so, if you had - say a 500 ohm (470 ohm would be ok) - and a 1K resistor, you could sub those in for the 1K and 2K resistors, respectively. Or, if you had a 2K and 4K (or whatever is close) that would work too. Just keep the ratio the same, and don't go too low or too high on the values.

You can use two 1K resistors - if you arrange them in series (NOT in parallel).

All of that said - a voltage divider is not the best way to do this kind of interface to these devices; you will be limited in speed (far lower than what you could really be able to use) - if this isn't a problem for your app, then don't worry about it - but if you need as much speed as possible, look into another solution that doesn't use resistors (rather, look into one that uses level translation via an IC or individual mosfets).

Lastly - your question indicates a lack of understanding or knowledge on your part about Ohm's Law - and how (and what) a voltage divider is for. These two items (and a whole host of others) are basic electronic building blocks and knowledge. What you are trying to do is build a house, without understanding what a hammer is, what nails are, or how to use a tape measure properly. But you have a large pile of wood, and you know how to saw it apart...

Get that knowledge. Also - pick up and learn to use a multimeter, if you haven't got one already.

I got one, thanks, and yes, I am not so good with Ohm's law, I learned the basic at eight grade and that's it, I will learn more, last question, I will use 5k and 10k, is it ok? and thank you very much dude!

I will use 5k and 10k, is it ok?

Probably but I would lower values. Have you got two of each? If so use two in parallel instead of just one.

Grumpy_Mike:
Probably but I would lower values. Have you got two of each? If so use two in parallel instead of just one.

Yes, I have 2 of 1k ohm, I created a new circuit, my English is not so good so I am not sure what is parallel so here is my new circuit, please tell me if it is what you meant. http://i60.tinypic.com/2hmd91f.png

Sorry totally wrong in almost every respect.
See attached:-

Grumpy_Mike:
Sorry totally wrong in almost every respect.
See attached:-

Ohh.. I hope I will understand it, thanks.

Grumpy_Mike:
Sorry totally wrong in almost every respect.
See attached:-

Ok dude, I had some time to work on it, I saw your image, it is like one resistor over the other, I used the fritzing bandable legs and I hope this will be the last image I will create, I really think this one better then the previous one :stuck_out_tongue:
here - http://i60.tinypic.com/qz5bgx.png

thank you for everything.

No, the single resistor is the one that should go to ground and the double resistor goes through the signal.

There is no need to use bendable legs, just put the two resistors side by side horizontally and wire up to that.

Grumpy_Mike:
No, the single resistor is the one that should go to ground and the double resistor goes through the signal.

There is no need to use bendable legs, just put the two resistors side by side horizontally and wire up to that.

Sure, just switched between the resistors, thanks a lot!!!!

Lastly - your question indicates a lack of understanding or knowledge on your part about Ohm's Law - and how (and what) a voltage divider is for. These two items (and a whole host of others) are basic electronic building blocks and knowledge. What you are trying to do is build a house, without understanding what a hammer is, what nails are, or how to use a tape measure properly. But you have a large pile of wood, and you know how to saw it apart...

The penalty for forgetting Ohm's Law is a PIE in the EIR (P=IE, E=IR)

I learned the basic at eight grade and that's it

And you're in what grade now ?, (9th ?)

I am not sure what is parallel

And this means what ? You have never heard of a dictionary or you have never heard of Google Translate or you are lazy , or all of the above ?

Google serarch

raschemmel:
The penalty for forgetting Ohm's Law is a PIE in the EIR (P=IE, E=IR)

And you're in what grade now ?, (9th ?)

And this means what ? You have never heard of a dictionary or you have never heard of Google Translate or you are lazy , or all of the above ?

Google serarch

YES I am. you know you can behave normally if you want . lol

YES I am. you know you can behave normally if you want . lol

Well, let's not go there because if we did then we would have to start by defining "normally", wouldn't we ?
Then we would eventually get to the fact that apparently it is "normal" for a 9th grader to skip all the research and just come right out and ask a question that other people would "normally" answer themselves with a little effort and the help of Google :

But since you brought it up, let's go there:

Hey,
I am trying to connect my new Bluetooth module, HC-06 ZS-040, therefore I am using this tutorial:
Arduino and HC-06 (ZS-040) | Martyn Currey
but I ran into trouble when I saw the circuit:
http://www.martyncurrey.com/wp-content/uploads/2015/03/HC-06_01-297x300.jpg
first, I am using Arduino Uno, not such a problem,
second I am using breadboard and I am not so great reading circuits, so I created a Fritzing circuit, using breadboard and Arduino Uno and I would like to ask if the circuit I provided equals to the circuit I attached as picture. (link here)

#1- Your greeting:

Hey,

Is that how you greet your teachers at school ? Come on, really ? Is that appropriate here on an international forum ? You may not be called on the carpet by the Global Moderator but is it really appropriate ?

#2-

I am trying to connect my new Bluetooth module, HC-06 ZS-040, therefore I am using this tutorial:
Arduino and HC-06 (ZS-040) | Martyn Currey

See all those toolbuttons at the top of the screen ? What do you suppose those are for , decoration ?
If you look to the right of the symbol resembling a computer screen, you'll see a symbol resembling a chain link (get it ? LINK , as in URL

After copying your link, if you click on the LINK toolbutton, a window will pop up at the top of the screen.
If you paste your link into the FIRST window that pops up, and click "OK", a SECOND window will pop up showing the link highlighted in the window. At this point you need to decide what label you want to use for the link. It doesn't really make much sense to use the URL itself as the label because it doesn't describe anything, so you can simply press the DELETE key and remove the link from the second window and type in a proper label, (like URL, or "this" or "here" or whatever).

#3 (question)
So , before we get into your question :

I would like to ask if the circuit I provided equals to the circuit I attached as picture.

Not being "normal", I decided take 5 seconds to Google your bluetooth module and found myself here:
See the wiring list right under the diagram ?

HC-06 Vin to 5V (can be from the +5V out from the Arduino)
HC-06 GND to common GND
HC-06 RX to Arduino pin D3 (TX) via a voltage divider
HC-06 TX to Arduino pin D2 (RX) connect directly

#4 (purpose of voltage divider)
See the phase "voltage divider"

Again, not being "normal", I decided to spend a few seconds and I found this.

Why do we need a voltage divider ?

Well, not being normal, I decided to actually read the information at the start of the tutorial and it says this :

Connections
The Bluetooth module the ZS-040 is based on is a 3.3V device. The HC-06 has regulators that allow a larger input voltage to be used, in the range of 3.6 to 6 volts. The RX pin can still only accept 3.3V though, which means a voltage divider is required to connect to a 5V Arduino. The Arduino will read 3.3V as a HIGH so the HC-06 TX pin can be connected directly to the Arduino. A simple voltage divider can be created using 2 resistors. I am using a 1K ohn resistor and a 2K ohm resistor.

So it seems that the whole point of the voltage divider is to reduce the 5V output of the arduino to 3.3V for the HC-06. Knowing this, let's go back to your question:

I would like to ask if the circuit I provided equals to the circuit I attached as picture.

#5 (missing ground connection)
Comparing the two diagrams, the one I linked (with ground symbol going into "space" (which was supposed to represent the breadboard you see in the photo)) , and your Fritzing, we see that your Fritzing is missing the ground connection from the arduino to the ground bus of the breadboard. If we
look at the wiring list again, it is clearly there;

HC-06 Vin to 5V (can be from the +5V out from the Arduino)
HC-06 GND to common GND
HC-06 RX to Arduino pin D3 (TX) via a voltage divider
HC-06 TX to Arduino pin D2 (RX) connect directly

Summary
Four connections (5V,GND, Tx, & Rx)
Special Notes: 5V to 3.3 V level converter required for arduino TxOUT to HC-06-RxIN. (voltage divider simplest method)

Are there OTHER ways to reduce a 5V logic signal to 3.3V ? (maybe)

In conclusion, I find that the answer to your question could have been obtained by making a wiring list and comparing it to your fritzing and the linked diagram, and researching voltage divider to understand how to construct one to convert 5V to 3.3V. I think if you took the time to research those items you would have found your answer and have a better understanding of electronics. Once you understood the voltage divider and how the resistor values wound up as they were and why they were necessary, your original question about whether your Fritzing matched the other one becomes a moot point. If you had used a wiring list , it would have been obvious your ground connection was missing from your Fritzing.

Ok dude, I had some time to work on it, I saw your image, it is like one resistor over the other, I used the fritzing bandable legs and I hope this will be the last image I will create, I really think this one better then the previous one

Really ? (you're a high school freshman ? Do your High School teachers let you call them "Dude" ? Regardless of what you think of me, you should keep in mind that , like me, most of the experts on this forum are or were working professionals in the electronics or technology field and have between 20 and 40 yrs experience on average. (I have 30, and both Mike and Russelz have 40 yrs experience) . Out of respect for all the years they (or we) put in learning what we know, do us the courtesy of showing some respect. I don't expect you to address us as "Sir' (although in your case, it would probably be appropriate due to your age), but at least don't address us as "Hey" , or "Dude". Ok ?

This poster was more tactful than I, but the message was pretty much the same...

Lastly - your question indicates a lack of understanding or knowledge on your part about Ohm's Law - and how (and what) a voltage divider is for. These two items (and a whole host of others) are basic electronic building blocks and knowledge. What you are trying to do is build a house, without understanding what a hammer is, what nails are, or how to use a tape measure properly. But you have a large pile of wood, and you know how to saw it apart...

Get that knowledge. Also - pick up and learn to use a multimeter, if you haven't got one already.

HC-06 input current (CMOS) Typical = 0.1 mA
Let Delta V = (5V-3.3V) = 1.7V
Let I min = 0.1mA = 0.0001 A
Solve for R:
V = IR ==> R = V/I (by dividing both sides by "I")

R = V/I = 1.7 V/0.0001 A = 17000 ohms.

For the resistor in series with the arduino output, 17k is the maximum value can use. Almost anything less than that (but not less than 500 ohms) is ok.

If you're really trying to cut corners , this tutorial probably has everything you need to eliminate most of the work.

raschemmel:
Well, let’s not go there because if we did then we would have to start by defining “normally”, wouldn’t we ?
Then we would eventually get to the fact that apparently it is “normal” for a 9th grader to skip all the research and just come right out and ask a question that other people would “normally” answer themselves with a little effort and the help of Google :

But since you brought it up, let’s go there:

#1- Your greeting:
Is that how you greet your teachers at school ? Come on, really ? Is that appropriate here on an international forum ? You may not be called on the carpet by the Global Moderator but is it really appropriate ?

#2-
See all those toolbuttons at the top of the screen ? What do you suppose those are for , decoration ?
If you look to the right of the symbol resembling a computer screen, you’ll see a symbol resembling a chain link (get it ? LINK , as in URL

After copying your link, if you click on the LINK toolbutton, a window will pop up at the top of the screen.
If you paste your link into the FIRST window that pops up, and click “OK”, a SECOND window will pop up showing the link highlighted in the window. At this point you need to decide what label you want to use for the link. It doesn’t really make much sense to use the URL itself as the label because it doesn’t describe anything, so you can simply press the DELETE key and remove the link from the second window and type in a proper label, (like URL, or “this” or “here” or whatever).

#3 (question)
So , before we get into your question :
Not being “normal”, I decided take 5 seconds to Google your bluetooth module and found myself here:
See the wiring list right under the diagram ?

#4 (purpose of voltage divider)
See the phase “voltage divider”

Again, not being “normal”, I decided to spend a few seconds and I found this.

Why do we need a voltage divider ?

Well, not being normal, I decided to actually read the information at the start of the tutorial and it says this :
So it seems that the whole point of the voltage divider is to reduce the 5V output of the arduino to 3.3V for the HC-06. Knowing this, let’s go back to your question:
#5 (missing ground connection)
Comparing the two diagrams, the one I linked (with ground symbol going into “space” (which was supposed to represent the breadboard you see in the photo)) , and your Fritzing, we see that your Fritzing is missing the ground connection from the arduino to the ground bus of the breadboard. If we
look at the wiring list again, it is clearly there;

Summary
Four connections (5V,GND, Tx, & Rx)
Special Notes: 5V to 3.3 V level converter required for arduino TxOUT to HC-06-RxIN. (voltage divider simplest method)

Are there OTHER ways to reduce a 5V logic signal to 3.3V ? (maybe)

In conclusion, I find that the answer to your question could have been obtained by making a wiring list and comparing it to your fritzing and the linked diagram, and researching voltage divider to understand how to construct one to convert 5V to 3.3V. I think if you took the time to research those items you would have found your answer and have a better understanding of electronics. Once you understood the voltage divider and how the resistor values wound up as they were and why they were necessary, your original question about whether your Fritzing matched the other one becomes a moot point. If you had used a wiring list , it would have been obvious your ground connection was missing from your Fritzing.

Really ? (you’re a high school freshman ? Do your High School teachers let you call them “Dude” ? Regardless of what you think of me, you should keep in mind that , like me, most of the experts on this forum are or were working professionals in the electronics or technology field and have between 20 and 40 yrs experience on average. (I have 30, and both Mike and Russelz have 40 yrs experience) . Out of respect for all the years they (or we) put in learning what we know, do us the courtesy of showing some respect. I don’t expect you to address us as "Sir’ (although in your case, it would probably be appropriate due to your age), but at least don’t address us as “Hey” , or “Dude”. Ok ?

This poster was more tactful than I, but the message was pretty much the same…

HC-06 input current (CMOS) Typical = 0.1 mA
Let Delta V = (5V-3.3V) = 1.7V
Let I min = 0.1mA = 0.0001 A
Solve for R:
V = IR ==> R = V/I (by dividing both sides by “I”)

R = V/I = 1.7 V/0.0001 A = 17000 ohms.

For the resistor in series with the arduino output, 17k is the maximum value can use. Almost anything less than that (but not less than 500 ohms) is ok.

If you’re really trying to cut corners , this tutorial probably has everything you need to eliminate most of the work.

DIDNT READ LOL .

DIDNT READ LOL

I think you mean CAN'T READ, LOL

Bad hair day! :astonished:

Ever had an OP that just won't do any work ?

You can take a horse to water but you can not make it drink.

You can take a whore to culture but you can not make her think.