Flyback Diode Needed?

[Diver300]

Just be certain i get this right I am to eliminate the relay and re-configure my pot (replacing R2 on my sch with the Pot) and my transistor (547)
EDIT. My current Pot is 500r but i do have 1K as well.

Yes, that's what I'm suggesting.

 

[ThomsCircuit]

Updated Schematic. I do not have 1K pots I hope this 500R will be ok. If needed I can add a resistor (470R) in series with U1
Voltage in 5.5v
Voltage at pin1 of ttp223 is 4.9v

 

[Beau Schwabe]

Your FAN will never come on with that schematic ... look at where your power is coming into Q1 specifically. And for what it is worth, a reverse biased diode across your fan might save you from a headache later.

 

[ThomsCircuit]

Your FAN will never come on with that schematic

Im going by the sample provided by a member. The pots connection looks odd but thats how its listed here. What did i get wrong? the resistor coming from the ttp223 is a pot. Im pretty new at this. thank you for pointing it out. what do i need to do?

a reverse biased diode across your fan might save you from a headache later.

We discussed the need for the diode earlier, the consensus was it was not needed because of the fans low power. you can see i left it in the schematic just in case. i can put an 1N4148 .

 

[ThomsCircuit]

Your FAN will never come on with that schematic

I see my mistake. i connected the collector to positive and added the flyback diode

 

[Diver300]

I don't think that the TTP223 will have enough power to run the LED.

This datasheet:- https://components101.com/sites/default/files/component_datasheet/TTP223-Datasheet.pdf
says that it's only rated to provide 4 mA (Output Port Source Current) which I why I suggested a 1 kOhm potentiometer. The 150 Ohm resistor and the LED will reduce the output voltage a lot.

You could either leave it out or have another transistor to drive the LED. An emitter-follower would work well for that.

 

[Diver300]

When you connect an inductive load to an emitter-follower, there is no need for a flyback diode. The transistor will conduct if the emitter goes negative, and that will stop the voltage getting too big.

 

[ThomsCircuit]

I don't think that the TTP223 will have enough power to run the LED.

is this because of the re-configuration of the 547? I currently have the prior schematic (post #1) running an led and it does work.

An emitter-follower would work well for that.

Ill add the above to my project. LMK if i got this right

 

[ThomsCircuit]

Ill be using this pot for the fan U1.
Specifications
Power: 0.06W
Resistance: 1K OHM OHMS
Type: Logarithmic
Type A
<--Datasheet-->

 

[ThomsCircuit]

datasheet says that it's only rated to provide 4 mA

somethings off about that. I have tried looking up how 5volts could = 4mA but i got no answers. I measure 4.9 volts at the output. Is there something i dont understand? There are sample diagrams showing an LED connected to pin 1 of ttp223.

 

[ThomsCircuit]

Output Port Source Current

Ok I looked this up and surprisingly I understand it. Here is a snip I read...
"...locate one that’s rated for 5 mA and has a forward voltage of 2 V. Ignoring the ON resistance of Q (usually very small), we can calculate Rsink to be (5-2)/0.005 = 600 Ohms to ensure that the port doesn’t sink more than 5 mA."
So now i understand that the 4mA you are referring to is Source/Sink current. I also understand that exceeding that will damage the port. (ttp223 /pin1) Im glad I ran into this because I like using these touch switches and do want to configure them properly. I found this formula and applied it to my project and came up with a 750R ohm or higher resistor is required in series with an LED that has a forward voltage of 2v. My question is since Im going to use a transistor to drive the LED do I still apply this formula?

 

[Diver300]

The source current is rated at 4 mA. I doubt if going beyond that would damage the TTP223. What would most likely happen is that the output voltage would reduce a lot as the current increased.

An LED in series with 150 Ohms will take about 20 mA from 5 V. A 500 Ohm potentiometer will take about 10 mA, so with both of those you would be taking 30 mA.

The best information that I had was that the output was rated at 4 mA. That doesn't mean that anything beyond 4 mA is impossible or harmful. It just means that the 4 mA is a guaranteed figure. The 4 mA rating was at 3.3 V supply, I think, so there could be a lot more current available at 5 V. There could be different versions of the TTP233.

If your circuit was working with the LED and the 500 Ohm potentiometer connected to the ttp233, you probably don't need to change anything.

On the formula, you still need to calculate the series resistance to get the current that you want in the LED. However, with an emitter-follower, there is an extra voltage drop of Vbe of the transistor, around 0.7 V, so you should use Rs = (Vs - Vbe - Vf) / i

The formula was ignoring the On-resistance, and therefore ignoring any other voltage drop than the LED, which is fine in many cases, but not where there is an emitter follower.

 

[Diver300]

I don't think that the TTP233 will have enough power to run the LED.

is this because of the re-configuration of the 547? I currently have the prior schematic (post #1) running an led and it does work..

In post #1, the LED is driven by the relay. The only power from the TTP233 is about 5 mA to operate that transistor.

If you attempt to run an LED at 20 mA from the TTP233 it may well not work.

 

[Diver300]

Ill be using this pot for the fan U1.
Specifications
Power: 0.06W
Resistance: 1K OHM OHMS
Type: Logarithmic
Type A
<--Datasheet-->
View attachment 133462

A logarithmic one isn't ideal for this application. You may find that the fan doesn't run until the potentiometer is turned nearly all the way up to full.

 

[ThomsCircuit]

Rs = (Vs - Vbe - Vf) / i

4.9-.07-2 = 2.83 / .0004 = 703R
I'll most likely use a 1k

A logarithmic one isn't ideal for this application

Thanks. Should I get a linear taper one instead?

 

[Diver300]

4.9-.07-2 = 2.83 / .0004 = 703R
I'll most likely use a 1k


Thanks. Should I get a linear taper one instead?

Yes, I think that a linear one would work better. Logarithmic potentiometers are almost only best for volume controls.

 

[ChrisP58]

You can still use your 500 Ohm pot if you insert a 510 ohm resistor between it and gnd.

That will make the total 1K. It will limit the voltage range of the pot though. Instead of 0 to 4.5 Volts, you'll have ~2 volts to 4.5 Volts.
That should be OK as the fan not likely to have much usable range below 2 Volts.

 

[ThomsCircuit]

You can still use your 500 Ohm pot if you insert a 510 ohm resistor between it and gnd.

Thank you. I thought that would work.
I have a few of these but for this project I want the adjustment available from outside the box.

Im going to get 500R like in post 29 and pair it with a 510R resistor

 

[ThomsCircuit]

Can you help me figure out what i have done wrong here? Did i mis-interpet the schematic you provided in #24?
It only works if i remove the GND from pin3 of the pot. But the the pots adjustment has no effect. If i do have the schematic right then my BB is wrong. I added a pic as well.

 

[rjenkinsgb]

I think the 510 Ohm resistor shoudl be between the pot and ground (in series with pin 3) not in series with pin 1 ??