Understanding of auto-illumination circuit

Using Arduino General Electronics
#1

Hi,

i got two of theese boards which automatically turn on an IR-LED if the environment is dark (and off if its bright). |500x361

I want to be able to manually overwrite the status of the LED so I identified the ICs, took out my multimeter (on continuity test) and drawed a schematic (european symbols). Most of is seems reasonable but there are a few things I dont unterstand. |500x460 |500x477

Left-hand side seems to be just a regular voltage divider between a potentiometer and a photoresistor. Right-hand side is a PNP Transistor (with current limiting resistor) to connect the LED to 3,3V. So far so good. But why is the NPN transitor in the middle connected like this? Ground at collector? Is this the so called "reverse active" mode? And when this transistor is conductive there seems to be a direct connection between 3,3V and Ground??! Why is this? I checked all the connections twice and it really is connected in such a way...

Can anybody help me unterstand this circuit?

#2

That is most probably no the circuit. Could the NPN be the other way round? The net around 3.3V and R1 also doesn't look right.

#3

Hi and thank your for your response.

you were right. One trace between the transistors was not correct. I attached the updated schematic. So this makes more sense since R1 acts as a pullup and can be pulled down via the NPN.

|500x477

But the orientation of the NPN is correct. On the second picture of the original post you can see the "HY3C" which is the NPN transistor and according to the following link the left single leg of the component is the collector which is definitely a direct connection to the Gnd-Pad.

So can you use a NPN like this? :fearful:

HY3C.

#4

qqq4000: Hi and thank your for your response.

you were right. One trace between the transistors was not correct. I attached the updated schematic. So this makes more sense since R1 acts as a pullup and can be pulled down via the NPN.

|500x477

So can you use a NPN like this?

HY3C.

Nope. I'm assuming you got that "HY3C" number off the part -- it might be a coincidence that some other part has that same number. The number on your part may be some proprietary number. If your multimeter has a Diode Tester setting, try using it across the leads on that "transistor". You can get a few clues that way--if nothing else, verification of NPN. But, if it turns out that is an NPN, then maybe you have the Collector and Emitter reversed. But, even that doesn't seem correct, because when that transistor turns on, there is the potential for a low impedance path from 3,3V to Gnd--which just doesn't make sense. If there was a resistor in series with the Base of the 2nd transistor, then it would make sense [with the Collector and Emitter swapped on the 1st transistor].

#5

ReverseEMF: [...] I'm assuming you got that "HY3C" number off the part [...] If your multimeter has a Diode Tester setting, try using it across the leads on that "transistor".[...]

Yes thats the number of the part labeling. So I tested the part with my multimeter like you said.

I am assuming this pin layout:

Resitance Mode (autorange):

Negative at Base: resistance to collector: 29k resistance to emmitter: 25,4k

Positive at Base: resistance to collector: 31k resistance to emmiter: 25k

Negative at collector: resistance to emitter: 50k

Positive at collector: resistance to emitter: 50k

Diode mode: Negative at Base: voltage to collector: 0L voltage to emitter: 0L

Positive at base: voltage to collector: 0,66V voltage to emitteR: 0,66V

positive at collector: voltage to emitter: 0,9V

negative at collector: voltage to emitter: 0L

According to this kink this doesnt make any sense because some of the resistance measurements should read open circuit.

So how do I interpret this data?

#6

qqq4000: Yes thats the number of the part labeling. So I tested the part with my multimeter like you said.

...

So how do I interpret this data?

First of all, SMD parts [especially transistors and ICs], because of the limited real estate on their surfaces, are usually marked with a code [rather that the actual part number]. It can be challenging relating this code to the actual part number--Google "SMD code lookup", or some such thing [and if there is a symbol -- Google "Semiconductor manufacturer icon lookup". Second, those resistance values [where one would expect no resistance] are probably coming from other components in the circuit. The best you can do, using the Diode Tester setting [which, I believe, is the only setting I recommended], is to get a verification of transistor polarities. But, the orientation of the Base-Emitter and Collector-Base junctions may remain ambiguous. In other words, using this method, you can distinguish between a PNP or NPN, but you may not be able to distinguish the two PN junctions.

But, based on your Diode Tester readings, it does appear to be an NPN. And, judging by the pinout of a '3904 SOT23 package [a fairly standard arrangement], the odds are good that you have the pins marked correctly -- BUT, there is still the chance this is a non-standard part, and the Base-Emitter/Collector-Base junctions may be opposite of typical. And, the only way to sort it out is probably to draw the schematic and see if it makes sense -- which is the technique you've already employed. So, did the Diode test results alter the "schematic strawman"? And, if so, and if you are still unable to evaluate its efficacy -- then, perhaps, present it to us, here, and let's have another look.[/i]

#7

MY quick research with Google shows the SOT-23, HY3C is a Richtek voltage detector, model RT9818C-12PV. There is a PDF available to explain all the details.

Paul

#8

Paul_KD7HB: MY quick research with Google shows the SOT-23, HY3C is a Richtek voltage detector, model RT9818C-12PV. There is a PDF available to explain all the details.

Paul

I checked the pinout of this part and in my oppinion it cant be right since the single leg of the SOT package is Vdd and on my pcb it is definitely connected to Gnd.

ReverseEMF: But, based on your Diode Tester readings, it does appear to be an NPN. And, judging by the pinout of a '3904 SOT23 package [a fairly standard arrangement], the odds are good that you have the pins marked correctly -- BUT, there is still the chance this is a non-standard part, and the Base-Emitter/Collector-Base junctions may be opposite of typical. And, the only way to sort it out is probably to draw the schematic and see if it makes sense -- which is the technique you've already employed. So, did the Diode test results alter the "schematic strawman"? And, if so, and if you are still unable to evaluate its efficacy -- then, perhaps, present it to us, here, and let's have another look.[/i]

If the diode test results suggest that it is an NPN then my schematic is "correct" (or at least didnt change since my last post).

Any ideas what else I can do to figure this out?

qqq4000: |500x477

#9

NPN transistors - collector always to +

#10

qqq4000: Yes thats the number of the part labeling. So I tested the part with my multimeter like you said.

I am assuming this pin layout:

Pin out is correct, most SMD NPN transistors are this way

#11

qqq4000: Hi,

i got two of theese boards which automatically turn on an IR-LED if the environment is dark (and off if its bright).

I want to be able to manually overwrite the status of the LED

What do you mean by [u]manually overwrite the status of the LED ?[/u]

Do you want to be able to turn it ON with a switch and still have automatic operation when the switch is OFF?

Thanks.. Tom.. :)

#12

Hi,
I think this is what you are trying to get drawn;
IR_Light_Detector.jpg

Tom… :slight_smile:

#13

TomGeorge:
Hi,
I think this is what you are trying to get drawn;
IR_Light_Detector.jpg

Tom… :slight_smile:

Yeah, and when Q1 turns on, it’s going to pull the Base on Q2 to ground [or to Q1 VBE(sat)]. I see three possible failure modes:

  • Q1’s Collector-Emitter junction is going to burn open

  • The BE junction of Q2 is going to burn open

  • LP2980 is going to go into over current protection, and even though Q2 is, essentially, turned on, there won’t be enough voltage to drive D1

The only other possibility is that LDR1 never causes Q1 to turn fully on – for instance if VR1 is adjusted to where Q1 turns on just enough to turn Q2 fully on, when LDR1 is at its highest value [i.e. Q1 doesn’t turn fully on], but that’s going to be temperature dependant – i.e. if this is operated at a different temperature than when VR1 was adjusted, then Q1 is either not going to turn on enough to get D1 to turn on, OR it’s going to turn on too much and smoke Q2, or pull down the 3.3V supply. There is also the issue of transistor Beta spread. Not a good design!

#14

Opps, forgot the cap;
IR_Light_Detector.jpg

Tom… :o

#15

Tom you have forgotten current limiting resistor for LED

#16

ted: Tom You forgot current limiting resistor for LED.

I thought about that but I can only see one resistor on the PCB and in the OPs circuit. The IR LED might have its own. Tom.... :)

#17

Or can be rated for 3.3V

#18

ted: Tom you have forgotten current limiting resistor for LED

Since, for IR data transmission, like in the case of a Remote Control, the IR LED is typically driven to it's MAX Pulse Current rating, perhaps the designer of this circuit is relying on the Regulator's internal current limiting to supply the proper current. As far as I can tell, it's not certain what that regulator part number is, but if it is an LP2980, then it has a max current of 50mA, which is just about right for driving an IR LED for broadcast purposes.

In fact, come to think of it, that moves in the direction of solving the Q1 shorting Q2 to ground, problem. Since the IR LED will be pulling the regulator output down to:

IR LED Forward Voltage + Q2 VCE(sat) ~= 1.6V

BUT, Q1, it seems, will, through the Q2 BE junction, pull the supply line down to more like 800mV to 1000mV, which should be too low to fire that LED! Unless, Q1's VCE(sat) is really high, like 0.8V, or something. That could be the case, if Q1 was a Darlington, but the OP's Diode tests on that part, seem to eliminate that possibility -- so, still a mystery!

#19

Infrared 3W LED - Wavelength: 850nm - Diameter: 2cm - Voltage: 1.5V~1.7V - Current: 1500mA - Replacement emitter board for flashlight

#20

ted:
Infrared 3W LED

  • Wavelength: 850nm
  • Diameter: 2cm
  • Voltage: 1.5V~1.7V
  • Current: 1500mA
  • Replacement emitter board for flashlight

Well, if that’s true [and I’m not sure how you can know it is], then there is a huge issue of power dissipation. Either that HY4D is more like a p-channel MOSFET, with a very low channel resistance [in saturation] AND that 4XY6 is somehow managing to limit current to 1500mA, without frying, or the HY4D is some mysterious marvel that can cut 3.3V down to the 1.5V~1.7V needed to properly drive that 3W LED…makes no sense what-so-ever!

To drop 3.3V down to even 1.5V [the least worst case condition] would require:

(3.3 - 1.5) * 1.5A = 2.7W!!!!

How the heck are either of those tiny little parts going to dissipate that kind of power? I mean, they gotta be linear, 'cuz there are no Switch Mode related parts–anywhere!

BUT, I’m guessing that the HY4D is a p-channel MOSFET [one with a very low Gate Threshold Voltage] – that’s the only way, so far, I can see this thing even beginning to function properly. But, that 3k3 resistor is rather odd–really doesn’t need to be that low, if all it’s doing is pulling up the Gate on a MOSFET [unless, maybe, that MOSFET needs to switch really fast–which might be the case if that LED really is 3W]! BUT, really, I doubt it’s a 3W LED [or there’s more to this circuit than has been revealed]! Another possibility is that the 3k3 resistor was chosen to set the sensitivity of the LDVR [by imposing a greater current load – but, again, the variability of transistor Beta would seem to make that a rather touchy design decision.

Image05.jpg