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proper transistor circuit??

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MrDEB

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ran a simulation of this circuit in TINA and it looks like it will work??
transistor is saturated??
Was contemplating using mosfets but with only 53ma draw why reinvent the wheel (I got lost reading how tos of mosfets)
going with relay or larger transistor for powering my random cabinet light circuit (see post)
 

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ran a simulation of this circuit in TINA and it looks like it will work??
transistor is saturated??
Was contemplating using mosfets but with only 53ma draw why reinvent the wheel (I got lost reading how tos of mosfets)
going with relay or larger transistor for powering my random cabinet light circuit (see post)

it's not exactly to simulate ??

to on the led the common current is 10--> 20ma .so if you chose the current is 10ma . the Ic = 2 *10 =20 ma.
the voltage drop on the leds = 2 volt *4 = 8 volt. and if you use the transistor as the switch so if it on the Vce --0.2 volt we assume that's 0 volt. that meant the Rc we need is (12 - 8 )/20 = 200 ohm.. so if we use your circuit by simulate.. by two resistors parallel.. we need.. two resistors 400 ohm. why did you not use only a resistor 200 ohm for this .

and then. we know the Ic = 20 ma. and assume the beta = 100. so we can calculate the Ib = Ic/100 = 20/100 = 0.2 ma..

next we know the Rb = (12-0.7)/0.2 = 56 k.. we chose 47k. oke !! but you know i think you must conect Rc serial with the E pole of npn...Or you can conect all your leds and Rc seri with emiter it better..
 
i but you know i think you must conect Rc serial with the E pole of npn...Or you can conect all your leds and Rc seri with emiter it better..

hi,:)

If the LED's and series resistor were connected from emitter to 0V, how would you saturate the transistor.?
 
This is how I computed circuit

used **broken link removed** to compute led array.
I increased the resistors for less current draw.
according to what I have read (I may have misinturpted?) the collector / load connection point should be .3v or less for saturation.
 

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Hoangson said ‘the voltage drop on the LEDs = 2 volts’. LEDs commonly run 1.4 to 3 volts depending on color and type. I use one type of LEDs that use 5 volts. This voltage is very temperature dependent.

You used 4 LEDs in series. Series is a good way to use LEDs, not parallel. When using parallel LEDs do not use one resistor. “why did you not use only a resistor 200 ohm for this .” You did the right thing using two resistors!

Transistor= 2N2222. I could not find that data sheet because everyone makes 2N2222A.
For the 2N2222A (On-Semiconductor) the DC Current Gain at Ic=150mA and Vce=10 volts is 100 minimum, 300 typical and ? maximum. (probably 600) With a Vce of 10 volts this test is unusable! I just wanted to show that Hfe can vary 6:1 from part to part. Hfe at Ic=150mA Vce=1 volt has a minimum of 50 and typical about 150.

You are using the transistor at 50 mA so Hfe will be a little higher. If the transistor is cold the Hfe will be lower. 25C/-50C will (at 50mA) cause ½ the gain.

Because you want low Vce (Vsat) you need more base current. The gain will be lower with 0.2 volts of Vset.

Never use typical numbers. In this case use minimum numbers. Use numbers over a temperature range your project will be used at. Look at what condition the transistor will be used at. Example: Hfe at 0.1mA is 100, at 50mA Hfe=220, at 500mA Hfe=40. Look at the graph of Hfe and Ic and know this is typical not minimum. Look at the table of Hfe to see min/typ/max.

Look at the graph of Vce and Ib to see how well the transistor closes. Useing the 10mA curve we can see that with Ic=10mA and Ib=1mA the transistor is very ‘on’. At Ib=0.1mA the Vset is 0.1 volt but at 0.05mAIc the transistor starts to open up. Remember this is typical, at 25C.

I know your teacher will say use a Hfe of 100 and you need to pass the class. In the real world we look at min and max specs and use much more base current than the teacher said to.
 
I need more base current?

Orginaly I had a 10k resistor connected to the base.
the transistor is being controlled by a cd4029 output.
just wanted to be safe on the outputs
changing the base resistor to a 10k=base current is 1.12ma
collector voltage is 172mv, base voltage is 811mv
transistor is still saturated?
will try and locate orginal circuit
 
The 2N2222 transistor and almost every little transistor list the max saturation voltage loss on the datasheet only when the base current is 1/10th the collector current. Current gain and hFE are not used for a switching transistor but are used when the transistor has plenty of collector to emitter voltage when it is a linear amplifier.
So for a collector current of 53mA the base current must be 5.3mA for all transistors to saturate pretty well. With less base current then only some if any transistors will saturate.

The output voltage of a Cmos logic IC is 10V when it has a 12V supply and a 5.3mA load. The base voltage of the transistor is about 0.8V so the voltage across the base resistor is 10V - 0.8V= 9.2V. Its current is 5.3mA so its resistance is 9.2V/5.3mA= 1736 ohms but use 1.8k as the nearest standard value.

Your circuit has the LEDs turning on and off abruptly.
The circuit you found has the LEDs at the emitter because the NPN transistors are emitter-followers that follow the slowly rising and dropping voltage of the capacitors at the bases of the transistors for dimming and brightening. Since the base current is too low then the transistors will never saturate.
 
I hope this is what your talking about

I see the current draw to the leds is less when connecting to the emitter.
the base current is a little less (I changed the voltage on the base from 12v to 10v and base resistor to 1800 as suggested)
am assuming the transistor is going to run correctly??
 

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I see the current draw to the leds is less when connecting to the emitter.
the base current is a little less (I changed the voltage on the base from 12v to 10v and base resistor to 1800 as suggested)
am assuming the transistor is going to run correctly??

Are you wanting to switch the LED's ON and OFF? - if so that's completely the wrong way to do it, your original way was the correct one. This way means you have much less power available for the LED's, and the transistor will run much hotter.
 
Pwm ?

the leds don't need to fade but would be nice.
if the circuit I posted wlll fail what should I use?? to achieve the result.
 
The fading circuit is designed to fade a single 20mA LED. The capacitor to ground at the transistor's base does the slow fading. Then the peak power in the transistor is about 100mW.
Your circuit has a load of up to 53mA so the transistor will have a peak power of 265mW which is pretty warm but is fine since its max allowed dissipation is 625mW.
then the value of the capacitor should be 1000uf instead of 100uF.

Since the Cmos IC has a 2V loss at its output, the base-emitter of the transistor has a loss of 0.8V and the 1.8k resistor has a voltage loss then the peak current is less than if the original common-emitter switching transistor is used.
 
I need a bigger transistor?

don't like parts getting warm so I need a bigger transistor say 830mw =BF420?
or one that I can attach a heat sink to?
go with the 1000uf cap, 1800 ohm resistor,and the load leds connected to the emitter.
 
The BF420 is a 300V high voltage transistor with a max allowed continuous current of only 50mA, a low current gain and a high saturation voltage loss.
Its spec's are the opposite to what you need.
 
Use a power darlington transistor like a TIP120 to drive the dimming and brightening LEDs. It has a very high current gain so the input resistor can have a high value and the filter capacitor can have a low value. The power darlington has a max dissipation of 2W without a heatsink.
 
Here is revised circuit

Any thoughts on a replacement for relay?
this relay controls power to two 555/4029/led arrays
cap on base = 100uf?
base resistor 1800?
 

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