Collector-Emitter current out by a factor of 20

I'm getting a current at the LED of 1.11 mA (1.1 on the 20m settings), the simulation says I should be getting about 20mA. What am I doing wrong?

My resistors are :

Br R Br = 120 Ohms

Br Gr O = 18,000 Ohms

PS. I don't see how to add an existing circuit. I can only create a new one for the post. Update - I think I worked it out, but it could be a lot clearer if you just gave us a list of existing circuits to add to posts.

PPS - How do we add images to posts? If you don't provide an image tool, how am I meant to show my actual circuit?

by Gavin.W
May 09, 2023

The simulator gives a reasonable answer given that the exact implied LED is unknown. Note that with just V1, RE and the LED, for a red LED at 1.4 volt, the current would be (5-1.4)/120 = 30mA while for a LED at 3.6 volt, we have (5-3.6)/120 = 12mA. The variation is thus large with respect of the exact LED.

Now, why you have something as low as 1.1 mA, I would first suspect a wrong pin-identification-connection for the transistor, that is, you exchanged the collector and the emitter, so that your montage end-up so that you have the reverse-active or reverse-saturation mode (for which the transistor is generally NOT certified in its behavior, but with a very low beta value in most cases), instead of the direct-mode (direct-active or direct-saturation), which is the one expected in the circuit.

It could be something else too. Is the LED emitting visible light? If so, it is possible that the multimeter that you use is either defective, either low on power, since most (but not all) LED won't provide much light under 1.1 mA and thus, the reading disagrees with a basic observable fact.

by vanderghast
May 09, 2023

The LED is giving off light. And I've just tested a range of values with this circuit. And also, with a simplified circuit without the transistor.

I found that the LED requires a voltage of 1.8 to 2.4 (I = 3 to 21 mA) so in a simple circuit, the resistor should be 1000 to 100 Ohms.

But in the transistor circuit, even with the resistor removed, the LED wouldn't exceed 1.87 V & 1.11 mA.

I flipped the transistor, and it still works. As I understand it, it shouldn't work. With a 33 Ohm resistor (RE) I get 2.64 V across the LED, giving 33.1 mA. This looks better. But I don't yet know what correct behavior looks like.

by Gavin.W
May 09, 2023

The simulation matches this much more closely. I will have to take another look at the datasheet for the transistor. I identified it as an NPN transistor. Not sure how I could make a mistake.

by Gavin.W
May 09, 2023

Though I haven't been able to find a datasheet showing a part in the same package. Every datasheet for C548 seems to describe a different part.

by Gavin.W
May 09, 2023

I have found some datasheets that may show (50/50 coz they aren't clear) the emitter on the right (with flat side facing me), so that would be the reverse of what I thought.

I will take a fresh look with the transistor flipped. And so, what, do I need to identify every single part I buy and acquire it's datasheet?

by Gavin.W
May 09, 2023

Indeed, the circuit used in the simulator is the right one for NPN: the emitter is toward the ground and the collector is toward the load.

To know which pin is the emitter, which one is the collector, and which one is the base, YOU MUST consult the datasheet for the specific component, there is NO universal convention (Brits ones and US ones are often different, as example). Alternatively, you can use these Arduino-based testers (search for "LCR-T4 Mega328 Transistor Tester Diode Triode Capacitance ESR Meter" on eBay, as example) which will identify the pinout for you in most cases).

When you exchange the collector and the emitter, the BJT STILL works, in "reverse", a little bit as normal, but with a much, much lower "beta" (hFE) and it is not a value tested neither certified by the manufacturer. But it is still working. Very rarely used in that mode, though, but not unheard of (I think some Fluke equipment used them for very specific purpose in the past).

by vanderghast
May 09, 2023

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