# Current amplifier for an IR sensor

Discussion in 'The Projects Forum' started by PsySc0rpi0n, May 17, 2016.

1. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
3
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.

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• ###### current_amp.asc
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Last edited: May 17, 2016
2. ### ErnieM AAC Fanatic!

Apr 24, 2011
7,386
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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.

3. ### ericgibbs Senior Member

Jan 29, 2010
2,499
380
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

4. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
3
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?

Yeah, that info I was aware...

Last edited: May 17, 2016
5. ### ericgibbs Senior Member

Jan 29, 2010
2,499
380
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

6. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
3
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?

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
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.

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
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Last edited: May 17, 2016
9. ### ericgibbs Senior Member

Jan 29, 2010
2,499
380
Usually a single transistor is used to drive an LED.
Look also at logic level Vgs PMOSFET's.
E

10. ### ErnieM AAC Fanatic!

Apr 24, 2011
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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.

11. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
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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?

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Vcc = sensor output voltage = 3.3V at best, 3V in data sheet.
Yes and No. Try 3 voltage + 1K resistor. Or simple measure the resistance between pin2 of a BISS001 IC and output pin.

13. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
3

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.

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
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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.

15. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
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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!

Apr 5, 2008
15,645
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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

17. ### Jony130 AAC Fanatic!

Feb 17, 2009
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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.

18. ### PsySc0rpi0n Thread Starter Well-Known Member

Mar 4, 2014
1,184
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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!

19. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
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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.

20. ### ErnieM AAC Fanatic!

Apr 24, 2011
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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?