Hello!

I'm obviously a rookie on this subject, so, I'm asking for basic help!

I have bought an IR sensor, and I was trying to amplify it's output current.

I checked the output current for normal conditions, I mean:

I connected +5V DC to the IR sensor's + terminal, and GND to the IR sensor's - terminal. Then I measured the current at the IR sensor's out terminal! I measured only 1.8mA.

So I want to amplify this current! I tried to use a single bjt but I didn't get much of an amplified output current! So, I tried a Darlington transistor as in the attached circuit.

But when drawing an equivalent circuit where I replace the IR sensor by a current source of 1.8mA, I'm not sure, if I'm doing things correctly to make some calcs and try to predict the output current.

current_amp.asc is the real circuit I was thinking to build.



current_amp_1.asc is the equivalent circuit I'm trying to analyse!


I get that the current through R3 = 1kΩ and D1 will be about 8.4888A if the input current is 1.8mA. (R1 = 1kΩ)
I have assumed that hFE1 of Q1 is 130 and hFE of Q2 is 35...

I found that Ie2 = Ib1 * (hFE1 + 1) * (hFE2 + 1)

Is there any change this is correct?

PS:
I have a +12V DC voltage to feed the circuit and a separate 5V DC to feed the sensor!
So, that +Vcc in the circuit are +12V DC and the sensor + terminal is connected to a +5V DC supply.

 

The output current of your module is not specified, but the output voltage is. That is only 3 volts, too little to drive your darlington emitter follower.

Aside: see if this still puts out 3 volts when it sees movement. By measuring the current as you did you effectively shorted out the output and that may cause damage... Or not, depends on how the output is wired internally.

Try grounding the emitter, and put your led and resistor up where R1 is. I would add a 10k or so resistor 1k series with the base as good practice.

 

hi Psy,
Your PIR is a module, it outputs a +3V voltage signal when motion is detected in front of the fresnel lens and close to 0V for no activation.
Eric

 

The output current of your module is not specified, but the output voltage is. That is only 3 volts, too little to drive your darlington emitter follower.

Aside: see if this still puts out 3 volts when it sees movement. By measuring the current as you did you effectively shorted out the output and that may cause damage... Or not, depends on how the output is wired internally.

Try grounding the emitter, and put your led and resistor up where R1 is. I would add a 10k or so resistor 1k series with the base as good practice.

Hum, those are details I completely am unaware.

But why are 3 V not enough to drive my Darlington transistor? If I consider both transistors at the active region, both Vbe will add 1.4V. So it will yet remain 1.6V.

I couldn't understand what you meant with the underlined part of your post! The resistors positions - 10kΩ and 1kΩ.

Also, isn't the output current of a transistor at the emitter of Q2? Why I should move the load to the Q2 collector?

hi Psy,
Your PIR is a module, it outputs a +3V voltage signal when motion is detected in front of the fresnel lens and close to 0V for no activation.
Eric

Yeah, that info I was aware...

 

hi,
Why are you requiring to amplify the current from a fixed 3V voltage level.?
Add a 1K resistor in series between the PIR output and transistor Base, the Collector current will be sufficient to drive a small relay or LED circuit.
E

 

hi,
Why are you requiring to amplify the current from a fixed 3V voltage level.?
Add a 1K resistor in series between the PIR output and transistor Base, the Collector current will be sufficient to drive a small relay or LED circuit.
E

the led was just an example! I would like to have about a 40mA up to 200mA, for instance to drive a fan, or a DC motor or so!

You guys mean something like the attached image?

 

Why I should move the load to the Q2 collector?

Write a KVL for this circuit
Vcc = Vbe1 + Vbe2 + Ie*R3+ Vled
2Vbe = 1.2V and LED needs 2V this give us already 3.2V.
And I think that your module already have a built-in resistor to limit the base current. And yes , you now can build a circuit from post #6.
And remember, you should never short the output pin of a any device.

 

And from data sheet we can tell that your PIP module is using this IC
http://img.ozdisan.com/ETicaret_Dosya/354460_2643812.pdf as a 3.3V voltage regulator.
And the main circuit is built around this IC
http://www.sc-tech.cn/en/BISS0001.pdf
http://www.elecfreaks.com/wiki/index.php?title=PIR_Motion_Sensor_Module:DYP-ME003

 

You guys mean something like the attached image?

Your circuit will work OK,
Usually a single transistor is used to drive an LED.
Look also at logic level Vgs PMOSFET's.
E

 

the led was just an example! I would like to have about a 40mA up to 200mA, for instance to drive a fan, or a DC motor or so!

You guys mean something like the attached image?
View attachment 106276

Yep, that's what i mean. Reason your first attempt will not work is your 3v input looses around 1.4 v going thru the darlington, that leaves just 1.6 volts to drive the led and resistor... May get a red led to light, not a white one, and not a motor either.

Do it the way you redrew it and you get close to that 12 volt supply as useful voltage on your load, all you loose is the sat voltage of the darlington that's maybe a volt.

11 volts is much more useful than 1.6 volts.

 

Write a KVL for this circuit
Vcc = Vbe1 + Vbe2 + Ie*R3+ Vled
2Vbe = 1.2V and LED needs 2V this give us already 3.2V.
And I think that your module already have a built-in resistor to limit the base current. And yes , you now can build a circuit from post #6.
And remember, you should never short the output pin of a any device.

What is the Vcc of your equation? The one of 5V or the one of 12V?

Another question:
Can I replace the IR sensor by a 3.3V supply in the circuit?

 

What is the Vcc of your equation? The one of 5V or the one of 12V?

Vcc = sensor output voltage = 3.3V at best, 3V in data sheet.

Can I replace the IR sensor by a 3.3V supply in the circuit?

Yes and No. Try 3 voltage + 1K resistor. Or simple measure the resistance between pin2 of a BISS001 IC and output pin.

 

Ok, I already built the circuit on the breadboard...

I can measure 2.90V from the IR sensor output pin to GND when movement is detected! Let's call this voltage Vo_IR.

So:
Vo_IR = 2.90V
Vce1 = 2.03V
Vce2 = 0.6V (when no motion is detected is about 8.69V)

As I'm using 2 BC548 transistors, I'm affraid that too much current is in the Q2 collector branch and therefore I placed a 100kΩ resistor there. But I can't make the LED to turn ON.

 

Always show us the circuit diagram. What is the IR sensor output pin voltage when no motion?
Try add a 10K resistor from Q1 base to GND.

 

This is my circuit at the moment:

The ouput voltage of the IR sensor is 0V when no motion! The LED drops 2V, and not 2.7 as the one in the circuit! It's a green one!

There is something weird... I can't measure the current through the LED. It just says 0A, no mater the scale I choose! And the LED doesn't turns ON when I'm using the multmeter to close the circuit and measure the current!

 

Hello,

Looking at the given datasheet, they do not use a resistor in series with the base of the transistor:



I think they rely on the current limiting capability of the module.

For your schematic, I would make Rb1 1K and an 10K resistor from the base to ground, to assure the transitor is switching off.

Bertus

 

And the problem is ?? The LED does not want to turn ON?
First connect a LED + 100R resistor directly into the IR output pin. And see if it will work . And the longer leg in the LED is anode pin.

 

And the problem is ?? The LED does not want to turn ON?
First connect a LED + 100R resistor directly into the IR output pin. And see if it will work . And the longer leg in the LED is anode pin.

I'm sorry... I forgot to mention that I managed to get the LED on.

This is working OK, not... Or at least is turning th eLED ON...

But somehow I can't measure the current that is flowing through the LED as I described above!

 

But somehow I can't measure the current that is flowing through the LED as I described above!

Maybe you broke the fuse in your multimeter. You can always measure the voltage across the series resistor with the LED when the led is ON. And use the Ohm's law to find the current.

 

Just to make sure we are all talking the same thing... you break the circuit with the led and put the meter in amps scale into the break, yes?