You seem to be confusing duty cycle with voltage. If you have a PWM signal of 3V at 60%, that does not make the voltage 1.8V. The voltage is either 3V or 0V in the two parts of the cycle.


A Voltmeter will read something like 1.8V because it is measuring an average.

But that is NOT the voltage to use in your LED resistor calculation. The voltage you should use for that is 3V minus the Vcesat of the transistor.

Assuming a saturation voltage of about 0.3, and 80 mA, the calculation looks like this:

3 - 0.3 - 1.3 = 0.08 R

R = (3 -0.3 - 1.3) / 0.8 = 17.5 Ohms

Bob

Thank you
I can see thing cleariest now...
I tought that voltage on the colector rises as duty cycle rise to 60% (0-1.8 V) in other words i tought that colector voltage reaches 60% only and then transistor shuts off.
I was wrong

 

The transistor does not saturate because its base current is too low.

The base current is low because the TLC555 has a very small output current when it sources current (when it goes high).
On the datasheet it shows a minimum output current of only 300uA (0.3mA) when its supply is 2V and its output voltage is only 1.5V. Its output current is a little higher when its supply voltage is higher and is probably 0.8mA when the supply is 3V and the output voltage is 2.2V.

With a 0.8mA base current then the transistor would have a saturated 8mA collector current but the collector current will be about because the transistor is not saturated and is amplifying the base current maybe 20 times, the collector resistor is not doing anything.
0.8mA in the 525 ohm base resistor produces a voltage across the resistor of 0.42V. Then the output voltage of the TLC555 will be limited to 3V - 0.7V (base voltage) - 0.42V (resistor voltage)= 1.88V which is less than the 2.2V with 1mA so its output current will be a little higher than 0.8mA or maybe 0.82mA. The datasheet for the Intersil ICM7555 which is the same as a TLC555 shows a "typical" output high current of about 1.5mA when the supply is 3V and the output is driving a dead short. The current is less when its is not so heavily loaded and where will you buy a "typical" IC instead of getting a "minimum" one? Also, will you replace the battery every day so its voltage is always 3V?

Thank you very much
As i said i learn more and more.
I thought that decoupling capacitor should help batery to last longer but obviously i was wrong.
Is it there any chance to get this thing work on 3V?? If it is,can you give me advice how?
Thank you

 

can't find the link i'm looking for . . . so couple instead :
https://www.analog.com/en/technical...essful-ultralow-light-signal-conversion.html#
d

Thank you

 

Is it there any chance to get this thing work on 3V??

Of course.
You were shown that the TLC555 has a puny output current going high but it has almost 10 times the current going low.
You were also shown that a transistor with a fairly high hFE saturates with a hFE of only 10, so use two transistors, one driving the other for a gain of 100 but with the output fully saturated.

1) Let the output TLC555 produce output pulses that go low.
2) The TLC555 drives the base of a PNP transistor with about 7mA and it saturates with an output of 70mA but you want more current so reduce this transistor's input and output to about 2mA input for a 20mA output, then use two transistors.
3) The 20mA output of the PNP transistor feeds into the base of an NPN transistor that has a saturated output of 200mA.
4) The 200mA output drives the LED.
5) both transistors need series base resistors to limit their base current.

You might need more than 200mA pulses so that your project still works when the battery voltage drops to 2V but watch out for the maximum allowed current in the LED when the battery is new.

 

T

Of course.
You were shown that the TLC555 has a puny output current going high but it has almost 10 times the current going low.
You were also shown that a transistor with a fairly high hFE saturates with a hFE of only 10, so use two transistors, one driving the other for a gain of 100 but with the output fully saturated.

1) Let the output TLC555 produce output pulses that go low.
2) The TLC555 drives the base of a PNP transistor with about 7mA and it saturates with an output of 70mA but you want more current so reduce this transistor's input and output to about 2mA input for a 20mA output, then use two transistors.
3) The 20mA output of the PNP transistor feeds into the base of an NPN transistor that has a saturated output of 200mA.
4) The 200mA output drives the LED.
5) both transistors need series base resistors to limit their base current.

You might need more than 200mA pulses so that your project still works when the battery voltage drops to 2V but watch out for the maximum allowed current in the LED when the battery is new.

Thank you for your advices so far.They worth a gold.I can se that you have great knowledge in electronic,and i am very thankful.My knowledge is not so rich so i must ask few things:
1)you said that TLC should produce low pulses.Does it mean that i must decrease duty cycle or let it stay on 60%?
2)in what configuration should i connect PNP? Base on pin 3,emiter on + and colector on -(i know that resistor goes to base region,but what with collector emiter region?should i put resistor on colector or emiter or leave witouth resistance?
The last part when you said that voltage drops to 2V... how i can menage saftey of a diode and on the same time ensure enough voltage/curent??
Should i leave decoupling capacitor?
Thanks

 

hi optus,
Is there any special reason that the project should work from just two 1.5V batteries, when 3 would solve most of the problems.?
E

 

hi optus,
Is there any special reason that the project should work from just two 1.5V batteries, when 3 would solve most of the problems.?
E

Hi
Unfortunately it is crucial to work on 2 bateries of 1.5 V

 

hi,
I am only asking these questions as I am interested in understanding the actual purpose of the project.

Can you give more details.?
E
BTW: Is it a college assignment.?

 

1) You said that TLC should produce low pulses. Does it mean that i must decrease duty cycle or let it stay on 60%?

Now the LED lights when pin 3 is low so turn your thinking about duty-cycle upside down.

2)in what configuration should i connect PNP? Base on pin 3,emiter on + and colector on -(i know that resistor goes to base region, but what with collector emiter region?should i put resistor on colector or emiter or leave witouth resistance?

Yes the PNP emitter connects to the +3V supply and the base has a series current-limiting resistor from pin 3 of the TLC555. The collector feeds a series current-limiting resistor into the base of the NPN. The emitter of the NPN connects to 0V like before and its collector has a series current-limiting resistor to theathode of the LED. The anode of the LED connects to +3V like before.

The last part when you said that voltage drops to 2V... how i can menage saftey of a diode and on the same time ensure enough voltage/curent??

What diode? A diode in series with the battery so that an idiot cannot connect the battery backwards? You need extra voltage to make up for the diode voltage loss.

Should i leave decoupling capacitor?

ALL electronic circuits, especially when powered by a battery, need a power supply decoupling capacitor.

 

Optus, I finally found time to breadboard your circuit to see what it is really doing. I used a 2N3904 NPN transistor. It has fairly similar characteristics to the transistor that you are using.
I just built the first 555 multivibrator with the reset connected to +.
I powered it with two NiIMH batteries which gave a 2.5 volts supply.
I changed the resistor from pin 3 to the transistor base to 220 ohms and the one from the transistor to the IR LED to 6.8 ohms (closest I had to 7.5 ohms).
When I connected my oscilloscope to pin 3 of the 555, with the ground clip to - I got a rectangular waveform switching between 1 volt and 0.
When I connected the scope probe across the 6.8 ohm resistor I got a waveform that was switching between 0.5 volts and 0.
From these results I can conclude that the transistor is switching correctly and when the IR LED is on, the current through it is 0.5 volts/6.8 ohms = 73.5 mA.

Try using your oscilloscope to see the dynamic waveforms and to observe the on and off levels. It takes all the guesswork out of it. Using the voltmeter to measure them can give you very misleading results.
Regards,
Keith

 

hi,
I am only asking these questions as I am interested in understanding the actual purpose of the project.

Can you give more details.?
E
BTW: Is it a college assignment.?

Hi
I apologize for non instant reply but i have so much obligation resulting in lack time for spending time online.
There is no some special reason for that voltage level.Since i have seen so many projects done with standard NE555(not Cmos version) powering with 5,6,9,12 V i foud out that there is 555 that consume less power and i decided to give it a try so i can see if there is a posibility to work around 3V.It would be a good teaching leason and challenge for me.

 

[QUOTE="Audioguru again, post: 1443654, member: 656595"


What diode? A diode in series with the battery so that an idiot cannot connect the battery backwards? You need extra voltage to make up for the diode voltage loss.

I meant IR LED, not a diode...my bad

 

Optus, I finally found time to breadboard your circuit to see what it is really doing. I used a 2N3904 NPN transistor. It has fairly similar characteristics to the transistor that you are using.
I just built the first 555 multivibrator with the reset connected to +.
I powered it with two NiIMH batteries which gave a 2.5 volts supply.
I changed the resistor from pin 3 to the transistor base to 220 ohms and the one from the transistor to the IR LED to 6.8 ohms (closest I had to 7.5 ohms).
When I connected my oscilloscope to pin 3 of the 555, with the ground clip to - I got a rectangular waveform switching between 1 volt and 0.
When I connected the scope probe across the 6.8 ohm resistor I got a waveform that was switching between 0.5 volts and 0.
From these results I can conclude that the transistor is switching correctly and when the IR LED is on, the current through it is 0.5 volts/6.8 ohms = 73.5 mA.

Try using your oscilloscope to see the dynamic waveforms and to observe the on and off levels. It takes all the guesswork out of it. Using the voltmeter to measure them can give you very misleading results.
Regards,
Keith

Thank you..
I will try when i get home

 

Try using your oscilloscope to see the dynamic waveforms and to observe the on and off levels. It takes all the guesswork out of it. Using the voltmeter to measure them can give you very misleading results.
Regards,
Keith

So where to adjust my switch on handheld osciloscope? On the voltage place and than take osciloscope measurements or in HZ/DUTY place and then take the osciloscope measurements?

 

The most difficult problem is that the circuit can be designed (probably with a second transistor) to work with new battery cells and a 3V supply but when the battery voltage drops then the output power and range are reduced. It stops working soon when the battery voltage becomes maybe 2.4V.
Transistors and ICs have minimum, typical and maximum spec's. Keith Walker might be lucky that his parts have typical or maximum spec's. Your parts might have minimum spec's and the circuit will fail unless it is designed to work with minimum specs and a low battery voltage but still must be designed to survive much higher spec's and battery voltage. A higher battery voltage and a voltage regulator will help.

 

So where to adjust my switch on handheld osciloscope? On the voltage place and than take osciloscope measurements or in HZ/DUTY place and then take the osciloscope measurements?

I am not familiar with that instrument but you need to select an input range that will display all of the vertical waveform and then set the timebase to display the shape of the waveform. Check your user manual.
I agree with the other comments about using 3 volts for power. It means your circuit is running at it's limits. You would get much more reliable results if you added one more cell to your battery and you would be able to drive much more current through the IR LED.

I tried powering the same circuit with three alkaline cells which gave 4.8 volts. That gave a current of 300 mA through the IR LED when it is turned on.

 

AAA battery cells are small so they do not have much "juice" to deliver 300mA pulses for longer than an hour which can be extended with a short duty cycle:

 

AAA battery cells are small so they do not have much "juice" to deliver 300mA pulses for longer than an hour which can be extended with a short duty cycle:

If I was building it to be portable I would use an 18650 Lithium ion cell with a 5V booster.

 

Hello guys...
Unfortunately failure again.I tried only with one NPN tr.just to see what is hapening.I was powering my system(both IC1 and IC2) with 3 V suply.Then i measured curent between pin 1 and 3 of a IC1(30 kHZ) witouth conecting transistor.It was 2.2 mA.Then i conected transistor to pin 3 of IC1 witouth base resistor(i asumed that there is no need for resistor since curent is only 2.2 mA).But then i measured voltage on pin 3 of first IC and i get 0.65 V,same between base and emiter of transistor.Colector-emiter voltage was around 1.3 V.So 3 V-0.7 V =2.3V should be on pin 3 but i get onky 0,65 V??Transistor didnt go to saturation.Maybe transistor is on his failure point???

 

.But then i measured voltage on pin 3 of first IC and i get 0.65 V

hi,
You would get only 0.65V because you do not have a series resistor, you are shorting pin 3 to 0v via the transistor BE junction [ Not recommended]
E