# Troubleshooting circuit

1. Nov 26, 2009

### StealthRay

Hi,

Can anyone help me check the circuit and if possible tell me what could be the problems?

At 12V supply,the circuit works fine and I am getting approx 60mA with resistance 100 ohms.
Now I wanted to increase the current by reducing the resistance to 50 ohms.And I got 115mA.

Since there isnt much resistance to reduce,I thought of increasing the current by raising the voltage to 24V.

Strangely the current only increase by 1 to 2 mA and the npn transitor gets hot with some burning smell.

The npn transistor 2SD1857A has a high reverse voltage Vcbo,vceo at 160V.I have no idea why the transistor heat up but no significant increase of current to my LED.

Furthermore,at 12V supply,the vce is about 5V.But as I increase the Vs,the Vce keeps increasing.

So my question is,

1)Why the transistor heated up but there was no increase in current when the supply raised from 12V to 24V.

2)Some of my transistor has very low VCE when current is applied at the base but this transistor has a high VCE.What is the parameter that affects the VCE?I think the VCE suppose to be low when current applied at base.

Thank you.

2. Nov 26, 2009

### raithrovers1

Why do you want more current?
The LED has certain requirements. It will have a maximum current rating and voltage rating. Most Led's will only tolerate 2V accross them and a current of approximately 20mA. They also require a current limiting resistor in series with to drop the voltage required at the required current.
You have Vs= 24V. Say your LED is a 2V device with max current of 20mA. R=(Vs-Vled)/I Therefore your resistance should be approx 1.1k.
Increasing the current to the LED above this is only going to damage it.
Also, you say the current you are measuring is 115mA. How are you measuring this. If we negate the transistor CE resistance and the LED resistance you would have approxiamately 1A flowing in the circuit. This is of course only when the transistor is on. The timers duty will reduce the rms value. It still means that the peak current is quite high. Therefore your transtor will get hot and therefore should be connected to a heat sink.

http://www.calibrepower.co.uk

3. Nov 26, 2009

### StealthRay

I am sure I am measuring the current correctly.Yes,there was no significant increase in current when i raise my voltage from 12V to 24V.

The problem was not with the IR emitter.The reason to increase the current was to increase the brightness of the led and thus improves the distance of detection.

At 12V supple I am getting 115mA and at 24V supply I am getting 117mA.Most of the voltages are dropped at the CE junction of the transistor.

So I wish to find out why is the current not increasing and why the transistor getting hot.I would be happy if I get 1A and burn the transistor because that would be making sense.Anyhow this is a high power transistor and 1A is not a problem at all.

4. Nov 26, 2009

### dlgoff

Your transistor is getting hot because it is dissipating a voltage drop of almost 20 volts. This drop has to go somewhere; hence heat. You see, the LED will have a forward voltage at about 2 volts (see http://en.wikipedia.org/wiki/Light-emitting_diode#Colors_and_materials") and the more voltage you apply, only the current increases until its junction fails.

Last edited by a moderator: Apr 24, 2017
5. Nov 26, 2009

### StealthRay

I am sorry but I still dont understand:shy:

Yes,I can understand that diode characteristic curve.

The confusion was why the current didnt increase.If I am not mistaken,the VCE is suppose to goes low once the transistor on.I dont think the transistor is faulty because it was working fine at 12V supply.

Now,let see on the equation,

Vcc=Vce+IR+Vled

If Vce is the same as it was in 12V.

24=5+I(50)+0.9
I=(24-5-0.9)/50
= 362mA

But the experiment showed that the Vce increased to 17.2V.
Hence,

24=17.2+I(50)+0.9
I=(24-17.2-0.9)/50
= 118 mA <-------------------I=117mA when the supply =12V

As you can see from the second calculation;

The current didnt increase but the VCE was the shooting up.If the current didnt increase,why is the heat?

I think something about the transistor is limiting the current.

Could it be hfe has reaches its maximum value? Ic=hfeIb ??

The datasheet of the transistor states:
hfe= min 82 and max 270 (condition Vce=5V and Ic=1A)

The ib= 0.28mA and ic=118mA
hfe=ic/ib=421 gain (way above the parameter stated)

I dont really understand how the ic can shoot until 118mA in the first place if gain is max at 270.

Last edited by a moderator: Apr 24, 2017
6. Nov 26, 2009

### vk6kro

Power in the transistor is voltage across it times the current. So, if the current stayed the same and the voltage went up, then the power increased. So, it got hot.

The voltage at the emitter cannot exceed the voltage from the 555. In fact it must be less than the 555 voltage by the base emitter voltage plus whatever voltage is across the series base resistor.
So, this sets a limit on the emitter current and it doesn't matter what you do with the collector voltage (except reduce it), the emitter current will not change, so the collector current will not change.

When you measure these currents are you allowing for the duty cycle of the 555? A normal meter will give you an average, but the "on" current will be higher.

7. Nov 27, 2009

### StealthRay

Hi Mike,

So the limitation is actually on the timer output?

Vout timer = Vb+Vbe+IeRe+Vled
Ie=(Vout-Vb-Vbe-Vled)/Re

So if this Ie cannot increase,Ic will not increase too.
So if I raised Vcc to 24V,the extra voltage will dropped at the CE junction of the transistor and hence the heat by P=VI.

Is it something like this?

8. Nov 27, 2009

### vk6kro

Yes, exactly right.

If the emitter voltage exceeded the base voltage, the base emitter junction would be reverse biassed and the base current would drop to zero. So, then there couldn't be any collector current.

I think you should try common emitter and get that working properly. With a frequency counter and an oscilloscope, it should be easy to get it working better than this emitter follower design.
Did you manage to get a loan of a frequency counter?

9. Nov 27, 2009

### StealthRay

Nope,the lab doesnt have a frequency counter.I think the oscilloscope will work just fine.

By the way,I did try the common emitter configuration but the signal was only recognized at a very close distance(probably of different frequency).The current to the IR emitter was good.I didnt measure the frequency or investigate at that time because I was rushing to get my model work.

Also my subsequent experiment showed that the frequency does not change with load current.

So something must have gone wrong during that common-emitter experiment.I will find out thoroughly in my next experiment.

Thank you.

10. Nov 27, 2009

### vk6kro

The 555 may not have been turning the transistor off properly, because it doesn't really give zero output when you would expect it to.

You can put a diode in the base line to help with this and putting in a bigger base resistor helps too.

Sometime when you have access to an oscilloscope it would be worth trying this again.

Cmos 555s work much better than the normal ones and give an output that is pretty close to the supply rails. They can't give quite as much current, though.

11. Nov 27, 2009

### StealthRay

Ok,thank you everyone for the valuable advices and suggestions.

I will look into it in my next lab session.

12. Nov 30, 2009

### rolerbe

Another way to think about what's going on: Assume the transistor is 100% on. As others have stated, the voltage Base to Ground is limited by the 555 to be < 5V. Assume it's 5 V. Take away 1.2 to 2 Volts for the LED drop. Call it 2 volts and ignore Vbe for a moment (about .6 volts). That means that 3 volts are dropping across the 25ohm resistor. V=IR, so the current is limited to 120ma, just what you are seeing.

Doesn't matter what Vs is. Except that at 24 Volts, you have to then have 19 volts Vce. 19V X 120 ma = 2.28Watts dissapated by the transistor. That's a lot, therefore heat buildup if no heatsink.

13. Nov 30, 2009

### Bob S

You can always put a series resistor from +24 volts to the collector, so that the IR voltage drop is ~15 volts (R = 15/0.12 = 125 ohms, I2R = 1.8 watts), and reducing the transistor power loss correspondingly.
Bob S