[SOLVED] Need help in a Transistor Circuit ......

[kdg007]

As you can see.I built a circuit like above.I replaced 3.3Vdc instead of White LED(cant find them in my simulation software).The current for each LED suppose to get 11mA.but in practical board it is showing 4mA.Need to know what went wrong ?

 

[mister_rf]

Maybe in practice the LEDs voltage drop is higher than it was predicted in theory, somewhere up to 3.7-3.8V. Did you measure the LEDs voltage drop? :smile:

 

[Sloth]

Check DC voltages over each LED, source and resistor. That should give you some clues. Also measure resistors (first disconnecting them from the circuit).

Did you use a proper spice model for the transistor? Or is that a generic (ideal) transistor and you've just named it? I guess that's OrCad. I'm not too familiar with it.

LED voltage drop often goes up with low currents (more current -> more dissipated power -> diode threshold voltage drops with more temperature)

 

[jtrent]

View attachment 74820
As you can see.I built a circuit like above.I replaced 3.3Vdc instead of White LED(cant find them in my simulation software).The current for each LED suppose to get 11mA.but in practical board it is showing 4mA.Need to know what went wrong ?

The DC current gain in your circuit is only 11.9. That's your Ic / Ib. From the **broken link removed**, with Collector current of 10mA, you should have a minimum currrent gain of 80. Your circuit doesn't seem to have that. If this was an actual constructed circuit, it would seem there is something wrong with your transistor. If this is just through similation, then maybe the parameters are set wrong in the similator for your 2n4401.

That being said, if you want to increase your collector current, then reduce your base resistor value to increase your base current. That will result in increased collector current. One other thing I just thought of is put a jumper across the collector and emitter. See what current that gives you. That would simulate a saturated transistor.

 

[kdg007]

ok.i checked the circuit and redesigned the whole circuit again.This is the new one.
voltage across each led is 3.07v.
voltage at base = -2.15v
voltage at collector is = 1.84v
current to each LED is 4.77mA :O :O ....
-
y base is -ve ? ,i dont know whether i am keeping the probes correct to check the voltage across base of transistor.I kept the ground probe at ground and the +ve probe at the base to see the base voltage.is that right ?

 

[kdg007]

There’s nothing wrong with the transistor. At 1mA in base, the collector current of the transistor rapidly increases to a value limited only by the external circuit.
The transistors is driven into saturation. ;-)

so how do i have to built a circuit so that i can get 10mA for each LED ?

 

[mister_rf]

Normally at 1mA in base the collector current of the transistor rapidly increases to a value limited only by the external circuit. The transistors is driven into saturation…
But there’s a problem with the base voltage, you need to measure about 0.7V.
Is the emitter direct connected to the GND?

 

[kdg007]

The DC current gain in your circuit is only 11.9. That's your Ic / Ib. From the **broken link removed**, with Collector current of 10mA, you should have a minimum currrent gain of 80. Your circuit doesn't seem to have that. If this was an actual constructed circuit, it would seem there is something wrong with your transistor. If this is just through similation, then maybe the parameters are set wrong in the similator for your 2n4401.

That being said, if you want to increase your collector current, then reduce your base resistor value to increase your base current. That will result in increased collector current. One other thing I just thought of is put a jumper across the collector and emitter. See what current that gives you. That would simulate a saturated transistor.

-
Ic/Ib = 10mA/80 = 0.1mA ? i took 10mA/10 = 1mA . i kept 10 since a lot of problems took it as a standard ,i donno y .

---------- Post added at 11:26 ---------- Previous post was at 11:24 ----------

The DC current gain in your circuit is only 11.9. That's your Ic / Ib. From the **broken link removed**, with Collector current of 10mA, you should have a minimum currrent gain of 80. Your circuit doesn't seem to have that. If this was an actual constructed circuit, it would seem there is something wrong with your transistor. If this is just through similation, then maybe the parameters are set wrong in the similator for your 2n4401.

That being said, if you want to increase your collector current, then reduce your base resistor value to increase your base current. That will result in increased collector current. One other thing I just thought of is put a jumper across the collector and emitter. See what current that gives you. That would simulate a saturated transistor.

-

current from collector to emitter is 8mA

 

[mister_rf]

Check connections, maybe wiring or the emitter is faulty.

 

[kdg007]

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these are photos from the circuit i built.maybe its the transistor.i`ll try to change it ans see what happens.

 

[kdg007]

oh.i kept the transistor reverse..but i connected it correctly now.i am still facing the same problem.getting 4.7mA for each LED...

 

[mister_rf]

There’ a 690 ohms for series collector resistance (180ohms + 510 ohms)… :-D

---------- Post added at 17:49 ---------- Previous post was at 17:47 ----------

In theory
Ic= (12 V - 3* 3.07V )/690 ohms= 4.04 mA

 

[kdg007]

its 51 ohms (green+brown+black) colour code....-> 180+51 = 231 ohms:shock:

 

[mister_rf]

Oops.. :-D

---------- Post added at 17:58 ---------- Previous post was at 17:56 ----------

Try to measure again the base and the collector voltage.

 

[RetroTechie]

voltage across each led is 3.07v.
voltage at base = -2.15v
voltage at collector is = 1.84v
current to each LED is 4.77mA :O :O ....

Those numbers don't match what's shown in the picture - what's simulated & what's measured?

Circuit looks okay. But if you put multimeter in series with the LED's, it will have a small DC voltage over it (perhaps a few tenths of a volt). Due to very non-linear behavior of LEDs + transistor, and the small voltage over R3, this might reduce the measured LED current significantly. Therefore: don't put current meter in series... measure voltage over R3, calculate current through it using known R3 value. Since in series, that same current is flowing through each LED & collector of Q1. Then measure voltages over each LED & voltage between collector & emitter of Q1.

A 10:1 ratio for collector & base current (and only 10 mA to pull), should be more than enough to drive an ordinary transistor into saturation. So: first make absolutely sure that you haven't got collector & emitter of Q1 swapped. And replace Q1 if you suspect it may have been damaged.

If you use a wire for Q1 and LED voltage is 3.07V, then you'd get (12 - 3*3.07V) / 231 Ω = 12 mA, so R3 seems decent value. So the strategy would be to basically reduce R1 value until you measure ~2.31V across R3. Current is then 2.31V / 231 Ω = 10 mA. You should then be left with about 0.48V from collector to emitter of Q1, but reading will depend on exact LED voltage @ 10 mA. With Q1 in saturation, you might have to increase R3 a little to get 10mA LED current.

 

[kdg007]

Those numbers don't match what's shown in the picture - what's simulated & what's measured?

Circuit looks okay. But if you put multimeter in series with the LED's, it will have a small DC voltage over it (perhaps a few tenths of a volt). Due to very non-linear behavior of LEDs + transistor, and the small voltage over R3, this might reduce the measured LED current significantly. Therefore: don't put current meter in series... measure voltage over R3, calculate current through it using known R3 value. Since in series, that same current is flowing through each LED & collector of Q1. Then measure voltages over each LED & voltage between collector & emitter of Q1.

A 10:1 ratio for collector & base current (and only 10 mA to pull), should be more than enough to drive an ordinary transistor into saturation. So: first make absolutely sure that you haven't got collector & emitter of Q1 swapped. And replace Q1 if you suspect it may have been damaged.

If you use a wire for Q1 and LED voltage is 3.07V, then you'd get (12 - 3*3.07V) / 231 Ω = 12 mA, so R3 seems decent value. So the strategy would be to basically reduce R1 value until you measure ~2.31V across R3. Current is then 2.31V / 231 Ω = 10 mA. You should then be left with about 0.48V from collector to emitter of Q1, but reading will depend on exact LED voltage @ 10 mA. With Q1 in saturation, you might have to increase R3 a little to get 10mA LED current.

-

the voltage across the RC is 1.29v => 1.29/231 = 5.5mA.
Voltage across C-E is 1.5v

 

[ahsan_i_h]

Negetive base voltage is the smell of oscillation. Why? because when base gets high frequency ac signal, then the positive cycle is clamped to 0.7 voltage. But the negetive cycle can not be clamped. As a result in DVM the average result will show negetive.
You have to use power supply decoupling capasitors close to the circuit. Long power supply cables of 5v and/or Long power supply cables for 12v without power supply decoupling capasitor may showing this magic.

---------- Post added at 20:09 ---------- Previous post was at 19:42 ----------

power supply decoupling capasitor means, power supply bypass capasitor. Try to add(parallel connection) 10uF electrolytic capasitor to each of your 5v and 12v power supply terminal in your circuit. If negetive base voltage is still there, then add(parallel connection) 0.1uF(104) ceremic capasitor to those 10uF capasitors.

---------- Post added at 20:12 ---------- Previous post was at 20:09 ----------

Negetive base voltage must be terminated before thinking new.

 

[mister_rf]

@ahsan_eda

Situation have changed now, initially there was a transistor mounted on reverse…

---------- Post added at 22:23 ---------- Previous post was at 22:22 ----------

@kdg007
That’s a very high Vce voltage. In order to reduce that you may need to test the circuit by reducing the series base resistor ( down to 1k up to 2.2k, just use a single 2.2k resistor or the two 2.2k resistors in parallel).
There is also probability that there is a defective transistor, try to change it for a new type…