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Remote Control Circuit with CD4017 - counter decoder

KrisBlueNZ

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Receiver: 1) Why is C2 arranged in different way compare to original? (Originally, positive to ground but yours is positive to the source)
It's better that way. A transistor responds to the voltage between its base and emitter, and the purpose of C2 is to smooth Q1's base-emitter voltage. You can connect it from Q1's base to the 0V rail as in the original design, but then any noise or disturbances on the power supply will be coupled into Q1's base. Connecting C2 across the base-emitter of Q1 is the proper way to do it.
2) R6, 2K2 means 2002ohms?
2k2 = 2200 ohms. The k replaces the decimal point and gives the multiplier, so the value is 2.2 kilohms or 2200 ohms.
3) Q2 - I got BC548 instead of BC547B
That's fine. If you're getting a BC337-40 for Q1 in the transmitter, you could get a second one to use as Q2 in the receiver, but a BC548 is fine too.
U1 - I got HCF4017BE instead of CD4017BE
That's fine; it's just a different manufacturer, and some of them use different prefixes. The "standard" prefix is CD, but Philips/NXP use HCF; ON Semiconductor use MC1, and there are a few others.
4) Any idea on getting 5V of supply?
Sure. The power supply does need to be regulated, for the TSOP1738. If you have a DC power source of 8V or more, you can use a standard 5V three-terminal regulator such as the 7805 (http://www.digikey.com/product-detail/en/MC7805CTG/MC7805CTGOS-ND/919333) or 78L05 (http://www.digikey.com/product-detail/en/L78L05ACZ/497-2952-ND/634716). If you want to power it from a 6V battery (I recommend using at least C-cells), you'll need a "low dropout" regulator, aka "LDO". Your cheapest option from Digikey is a Microchip MCP1702: http://www.digikey.com/product-detail/en/MCP1702-5002E/TO/MCP1702-5002E/TO-ND/1098464 and another option is an LM2931: http://www.digikey.com/product-detail/en/LM2931AZ50R/497-4262-1-ND/725545 or an LP2950: http://www.digikey.com/product-detail/en/LP2950CZ-5.0/NOPB/LP2950CZ-5.0/NOPB-ND/148222
Transmitter: 1) My question is, at pin 1, is the signal 0? And is it then inverted and come out as "1" at pin 2? Then is inverted to "0" again then enter CLK??
Not exactly. U1 contains six inverters. The first one is connected to pins 1 and 2. It takes a voltage on pin 1 (its input) and drives its output (pin 2) to the logically opposite state. The ceramic resonator feeds this signal back to the input in a way that is frequency-dependent, and this causes the whole circuit to oscillate at a frequency determined by the ceramic resonator. The 1M resistor is needed to put the inverter into its "linear region", where it can detect and amplify small voltage variations at its input. The output is buffered by another inverter in U1.
2) The signal entering CLK of U2 (from pin 12 of U1) is 455 kHz while the signal entering pin 11 of U1 is 37.91667 kHz (after being divided by 1100 binary), am I right?
Right. U1 is used for both frequencies. It generates and buffers the 455 kHz clock frequency, and it buffers the 37.991667 kHz signal to boost it so it can drive the transistor that drives the LED.
3) And I'm wondering how the current flowing in the oscillator and U1.
I'm not sure what you're asking. The operation of a Pierce oscillator is not easy to explain, and you don't need to understand it to be able to make it work!
4) Is the current having phase angle of 38kHz? Or else how the IR LED emit IR with 28kHz?
"Phase angle" doesn't mean anything in that sentence. The IR LED is pulsed at 38 kHz. There is no 28 kHz frequency anywhere in that circuit.

No problem. Those were all reasonable questions except the last one.
 

Jay.E.Eh.S

Apr 11, 2014
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Haha. I'm sorry about the last question and the typo too. Ok, now I understand how both the circuits work, will start doing it :) Will post my questions at here again if I got any problem / once I completed the prototype. Once again, I thank you from the bottom of my heart :)
 

KrisBlueNZ

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You're welcome :)

If you want, you can "Like" some of my posts or leave a comment on my profile.
 

Jay.E.Eh.S

Apr 11, 2014
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Hi :)

Transmitter :
1) IC : Is it ok if 74LS92 instead of CD4024BE?

2) Ceramic Oscillator : Is it ok if CRV455E instead of ZTB455E? (both 455kHz, but alphabets different)

Receiver :
1) RELAY : Is it ok if YL303H-S-5VDC-Z instead of V23079A2001B301?


Above are the components which I think it's okay to be replaced with the one you mention.

The pictures showing the lists of electonic components that are being sold at a particular shop (the nearest, will look for another shop soon) and also a picture of the relay I found.

Current problem : Couldn't find TSOP1738 via any local shops (currently) and also no online buying allowed for my country :( And the hex inverter in the list (74LS04) is not unbuffered. Also, couldn't find a LDO voltage regulator in the lists (only "linear" right?)
 

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KrisBlueNZ

Sadly passed away in 2015
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Hi!
Transmitter :
1) IC : Is it ok if 74LS92 instead of CD4024BE?
The 74LS92 is a member of the LSTTL logic family. It is not CMOS. It requires a power supply voltage in the range 4.75~5.25V so you would need a 5V regulator in the transmitter as well as in the receiver. It draws more current than the CD4024. It could be made to work, but some changes would be needed.
2) Ceramic Oscillator : Is it ok if CRV455E instead of ZTB455E? (both 455kHz, but alphabets different)
I couldn't find much information by Googling that part number (CRV455E). What is the manufacturer's name? Can you give me a link to the data sheet?

If the replacement part has three pins, it's almost certain to be OK.
Receiver : 1) RELAY : Is it ok if YL303H-S-5VDC-Z instead of V23079A2001B301?
Again, I can't find data on that part number. It is probably suitable. Can you measure the coil resistance? If it's more than 50 ohms, it should be OK.
Current problem : Couldn't find TSOP1738 via any local shops (currently) and also no online buying allowed for my country :(
Similar units are available from several manufacturers. Try searching by description - infra-red receiver-demodulator.
And the hex inverter in the list (74LS04) is not unbuffered.
The 74LS04 is also an LSTTL device and is not suitable.
Also, couldn't find a LDO voltage regulator in the lists (only "linear" right?)
What country are you in?

Can you post links to the web sites of any electronic component suppliers in your area.
 

Jay.E.Eh.S

Apr 11, 2014
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Noted, will hunt for CD4024BE instead of making any changes to the circuit.

I too couldn't find the datasheet about CRV455E through Google yesterday. It's from a 10 year-old+ VCD player of mine, not a branded VCD player. Already not using it for like 7 years.

Yes, the relay it is more than 50ohms - 73.7ohms :)

I'm from Malaysia. I found this website just now (http://malaysia.rs-online.com/web/) when I key in "infrared receiver demodulator malaysia" in Google. I've checked, they are selling TSOP4038.
I found many electronic components at that website but they are selling they are selling in the quantity of at least 5 or 10 o_O
 

Jay.E.Eh.S

Apr 11, 2014
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Here's the datasheet of TSOP4038, supply voltage 2.7V - 5.5V, hmm, should be ok?
 

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KrisBlueNZ

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Yes, the TSOP4038 looks suitable.

Does that CRV455E have three pins? If so, it's very likely that it will work.

OK, I've added your location to your profile. It's often useful for us to know where you are.

So RS Components have the CD4069UBE and CD4024BE, but only in quantities of 25 and 10. Is it too expensive to buy that many?
 

Jay.E.Eh.S

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I just checked just now. Has only 2 pins. BTW, based on the schematic, the ceramic oscillator has only 2 connections, means it has one pin left unconnected to anything?

Ok. Thanks :)

Hmm, it is quite expensive, some more I don't need that many for this project. Haha. I will discuss with my group members regarding the bajet. Or else, I think I will pay for the remaining IC on my own because I am sure I will be using it in the future. Electronic is interesting :)
 

KrisBlueNZ

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Sorry, I'm getting this thread confused with a different one! Right, it should be a 2-pin resonator. The one you have will probably work.
 

KrisBlueNZ

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The VS1838B will probably work. I found a data sheet but it's in Chinese and I can't read it, so I can't be 100% sure. If it's not too expensive, it's worth trying.
 

Jay.E.Eh.S

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I found it too. I can read it. Any specific specification that I need to take note? Maybe I can check on my own? :)
 

KrisBlueNZ

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Yes. In the block diagram, there is a resistor connected between the OUT pin and VCC. Can you find out anything about this resistor? Either its resistance, or the output current when the output is high.
 

Jay.E.Eh.S

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Hmm. The datasheet written, when there's NO SIGNAL, the Icc is 1.5mA (Max).
That's what I found from the the table. The block diagram shows no reading.
 

KrisBlueNZ

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No, that's not the information I need. I don't think the data sheet has that information. It will probably be fine.
 

Jay.E.Eh.S

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Oh, okay. I'm sorry about that.

I've just uploaded 2 datasheet, one TSOP1738 and another one is VS1838B, I'm looking at their pins... The middle pin of TSOP1738 is "Vs" while for VS1838B is "GND". My question is :

1) Vs and Vcc are the same right?

2) I'm having the VS1838B in my hand now and I tested it with my digital multimeter. I found out that current is not able to flow from "Vcc" to "OUT". I thought, now it receive no signal, and so voltage is present in "OUT" pin, means current can flow from "Vcc" to "OUT" to cause reverse biasing in Q1(Receiver Circuit)?
 

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KrisBlueNZ

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Yes, in this case, Vs and Vcc mean the same thing.

"Current is not able to flow from Vcc to OUT"? I'm not sure what you mean. Do you mean it measures open circuit between those pins when you measure its resistance? That isn't necessarily a problem.

You can test it in the circuit. Measure the voltage on Q1's collector (relative to the 0V rail along the bottom). It will be close to 0V. Push the button on the transmitter and hold it. The voltage should jump up to about 5V. Then release the button on the transmitter. The voltage should drop to 0V within half a second or sooner. If it doesn't, add a 47k resistor between Vcc and OUT on the VS1838B. That will fix it.
 

Jay.E.Eh.S

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Yes that's what I mean :) I connect my multimeter, one with "Vcc" pin and another one with "OUT" pin, and I get infinity resistance, which means it's an open circuit. I thought it is a closed circuit?

Noted. I just almost completed the receiver circuit, left getting the voltage regulator and C-cells by tomorrow :)
 

KrisBlueNZ

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It's OK that you don't measure any continuity between Vcc and OUT. The "resistor" inside the module is probably implemented as a current source made from semiconductor components, which are easy to manufacture as part of a silicon chip, instead of a real resistor. So it won't have any continuity if the device is not powered up, and/or it will need at least 0.6V across it to cause any current to flow, and your multimeter probably only uses 0.2V for resistance measurement. That's why you don't measure any continuity.
 
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