Newbie Question on NPN Transistors

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Hi

I have a newbie question on NPN transistors. I have built the circuit shown and have measured the current comming to the base as Ib = 0.4mA (This makes sense by doing V/R as well).
I have also measured the current comming from the emitter Ie = 2.82mA (This also makese by doin V/R).

My question is: how come I am getting such low Ie?
This gives an h = 7.

How can I boost the current comming out of the emitter say to 50mA as the NPN is rated for 100mA?

Thank you in advance.
a.

 

[Screech]

I'm not sure your maths makes sense.

transistors as like diodes, loose a bit of voltage when they pass current, so say only 4.3 volts will be going to the led.

so, 4.3volts / 83 = 0.050 Amps, or 50milliamps.
replace the 1 k resistor with an 80 ohm resistor.

Replace the base resistor with a 10k or 100k pot and adjust till you get the 50 milliamps thru the led.
Measure the resistence of the pot and replace it with a resistor.

I'm sure my maths dosen't make sense to others too.

 

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Hi

I think that you have neglected the LED V drop of 1.7V

What I find is that as I decrease the resistor from 1k to say 10ohms I get the V drop over this resistor falling substantially.
Numbers for 2 LEDs in series:

220ohms drop over R = 1V
32.7ohms drop over R = 0.65V
10ohms drop over R = 0.3V

So what is happening is that as I decrease R, the V over this R decreases, and the current while its increasing, is not increasing linearly.
Could you please tell me what I am doing wrong?

Thank you
a.

 

[Screech]

I think that you have neglected the LED V drop of 1.7V

Yeah, I did it on purpose.

So what is happening is that as I decrease R, the V over this R decreases, and the current while its increasing, is not increasing linearly.
Could you please tell me what I am doing wrong?

It's not increasing linearly because thats how your circuit and components behave.
Why you think it sould be linear?

The only thing you were doing wrong is limiting the current with the 1k resistor.

 

[Roff]

I think you have the transistor collector and emitter pins swapped. The beta (Hfe) should be much higher than 7 for almost any transistor under these conditions. A Google search for CTBC548 came up with absolutely nothing - not even a street address in Lower Slobbovia. Do you have a datasheet, or a different part number?
After we get this resolved, we can discuss what you want.

 

[Nigel Goodwin]

I think you have the transistor collector and emitter pins swapped. The beta (Hfe) should be much higher than 7 for almost any transistor under these conditions. A Google search for CTBC548 came up with absolutely nothing - not even a street address in Lower Slobbovia. Do you have a datasheet, or a different part number?
After we get this resolved, we can discuss what you want.

Presumably it's a BC548?, a standard NPN small signal transistor. I don't know where the CT has come from? - perhaps a manufacturers reference?.

 

[ljcox]

You don't need the 100 Ohm base resistor. The base can be connected directly to the +5V. The current coming out of the emitter can be calculated by:-

Ie = (5 - 0.7 - Vled)/Re approx. So for a single LED and assuming Vled = 1.7 V, R = 220

Ie = (5 - 0.7 - 1.7)/ 0.22 = 11.8 mA which should be more than enough to make a LED glow.

If you have 2 LEDs is series it will be about (5 - 0.7 -1.7 -1.7)/0.22 = 4.1 mA which is too low. A 100 Ohm would give you a current of about 9 mA.

 

[Roff]

You don't need the 100 ohm base resistor. The base can be connected directly to the +5V. The current coming out of the emitter can be calculated by:-

Ie = (5 - 0.7 - Vled)/Re approx. So for a single LED and assuming Vled = 1.7 V, R = 220

Ie = (5 - 0.7 - 1.7)/ 0.22 = 11.8 mA which should be more than enough to make a LED glow.

If you have 2 LEDs is series it will be about (5 - 0.7 -1.7 -1.7)/0.22 = 4.1 mA which is too low. A 100 ohm would give you a current of about 9 mA.

Well, I can't imagine why you would use a transistor like this, unless you were going to drive the base. But you wouldn't do that with a 10 ohm (or even 100 ohms, as Len said) load on the previous stage. You would just connect the 1k (or whatever) resistor to +5V.

 

[ljcox]

That is an interesting point.

I assumed that the base is to be driven by something.

That is how I drive LEDs from CMOS sources.

The emitter follower provides a high resistance load, you only need one resistor and there is no inversion (ie. compared with a common emitter driver).

 

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Hi

It is a BC595 low power NPN transistor. This is a dummy circuit which I am trying to figure out to be included into a LED dot matrix display. At the moment I am trying to be able to drive two LEDs in series with as much current as I specify. Next with a duty cicle I will be able to boost the current in proportion to the duty cycle

What I am thinking of is something as in the drawing below. Can someone please help me with the transistor design?

(What I have found is that if I have LED LED R NPN I can boost the current by changing the R. In the configuration of the first diagram in this post I cant boost the current above say 30mA).

Thank you in advance.
a.

 

[akg]

i think it shld be a PNP , and R be aroung 470.u also need a resistor b/w the base and 595 say 1k, only thing is the visual will be inverted , ie ic o/p low->led lit ...

 

[audioguru]

Hold down the Shift key to make straight lines in MS Paint.
Here is what you need:

 

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Hi AudioGuru

Thank you very much!!! That is exactly what I was looking for.

If you could just walk me through your calculations I would really appreciate it.

(a) How do you determine that the output current flowing from the Base to the 74HC595 is 32mA?

(b) How do you determine the V drop between E and C? and from this the current Ic?

I am sorry to bug you like this, and
Thank you once again!

Regards
a.

 

[audioguru]

(a) How do you determine that the output current flowing from the Base to the 74HC595 is 32mA?

The BC327's typical Vbe when saturated with 560mA/32mA is about 0.9V.
Calculated from the 74HC595'd datasheet, its typical output resistance is about 30 ohms.
(5.0V-0.9V)/(30+100) ohms= 31.5mA

(b) How do you determine the V drop between E and C? and from this the current Ic?

The typical saturated Vce with 560mA collector current and 56mA base current is spec'd fairly low. With only 32mA of base current I guessed that its typical saturation voltage is 0.6V.
Therefore each string of LEDs gets only 4.4V and each 10 ohm resistor has 0.8V across it if the LEDs are 1.8V each at 80mA.
0.8V/10= 80mA.

The saturation voltage of transistors and the voltage drop of LEDs varies a lot. You might need to pick and choose transistors and LEDs, or resistors to match the current you need.

 

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Hi AudioGuru

Thank you, I think I am beginning to understand.

I don have the particular PNP you refer to, but have been fooling around with another one. I can now control the amount of current going through the LED's by changing R1 and by increasing the input V to the Emitter.

I really appreciate your help!

I have been sinking the current through the ULN2003- I am assuming that wont pose a problem?

Regards
a.

 

[Nigel Goodwin]

I don have the particular PNP you refer to, but have been fooling around with another one. I can now control the amount of current going through the LED's by changing R1 and by increasing the input V to the Emitter.

No, you control the LED current by changing the value of R2 - the transistor should be saturated when switched on.

I have been sinking the current through the ULN2003- I am assuming that wont pose a problem?

It shouldn't do!.

You might find my PIC tutorials helpful, the "Hardware Extras" section shows various ways of connecting LED's etc.

 

[audioguru]

I don't have the particular PNP you refer to, but have been fooling around with another one. I can now control the amount of current going through the LED's by changing R1 and by increasing the input V to the Emitter.

Which transistor are you using?
R2 adjusts the LED current. If R1 is less that 100 ohms, you are taking more than 32mA from the 74HC595. Its absolute max is 35mA.
The transistor's emitter voltage is supposed to be the supply voltage of the 74HC595. If they are different then you blow-up the 74HC595. The recommended max supply voltage of the 74HC595 is only 6V.

I have been sinking the current through the ULN2003- I am assuming that wont pose a problem?

Big problem. It has much more than 0.8V drop across it. Your 5V supply and even 6V is too low for two LEDs in series when you add the voltage drops.
Use Mosfets to sink the voltage down to zero.

 

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Hi AudioGuru

Thank you for your reply.

I am using the 2N4403 PNP with R1=120ohms going from the 595 to the Base and 22 ohms going from the collector to the LEDs.
I am supplying the 595 with regulated 5V.

I was wondering if I can put say 7.5V on the Emitter of the PNP? (I have tried this and this has a much more dramatic effect on current through the LED's than changing the R2 (=22ohms)).

I am using the ULN2003 right now, it "works". I drive the ULN2003 with a PIC, and it is grounded (but the COM port is left unconnected). Only sometimes does no LED light up. At this point I just switch off the power supply, give it a couple of minutes and try again and it works again. I am not sure what I am doing really, any comments would be greately appreciated.

Once again thank you for all your help.
Regards
a.

 

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Hi AudioGuru

Sorry it did not work as I reported before. What I see happen is the entire matrix lights up (quite brightly). If I put the Emitter back to 5V then the right sequence of LED's are on.

When I measure the V at the Base with the 5V supply and the 120Ohm R1 I see that V = 4.1V and 5V depending what the 595 is driving. Why is the 4.1V when 595 is showing 0 (actually its showing 0.6V or so)?

Now I am really confused.
Any comments or suggesionts would be greatley appreciated!

Regards
a.

 

[audioguru]

(With the emitter at +7.5V.) What I see happen is the entire matrix lights up (quite brightly). If I put the Emitter back to 5V then the right sequence of LED's are on.

The emitter is at +7.5V. When the 74HC595 is supposed to turn off the LEDs then its output to R1 is its supply voltage of +5V. The transistor has current through R1 so turns on the LEDs continuously. Actually, the current in the base resistor is trying to pull the output of the 74HC595 above its supply voltage, that's very bad for it.

When I measure the V at the Base with the 5V supply and the 120Ohm R1 I see that V = 4.1V and 5V depending what the 595 is driving. Why is the 4.1V when 595 is showing 0 (actually its showing 0.6V or so)?

If the display is multiplexed then each voltage is present only for a moment. A 'scope will show the transistor's base voltage at +5V when it is off, and at about +4.1V when it is on. A multimeter might average the voltage at 0.6V across the transistor's base-emitter.

The 32mA from the 74HC595 causes a voltage drop of about 0.6V across its output resistance. Its output voltage is 0V for a low without any current.
The 32mA in R1 is created by the remainder of the voltage across it.
Actually, with R1 at 120 ohms and the base at 4.1V and the output of the 74HC595 at 0.6V, the 3.5V across it limits the base current to 29.2mA.

The 2N4403 is nearly the same as a BC327, but has a different pinout.
They work very well up to about 100mA then above that their current gain drops and their saturation voltage rises.