2N4401 simple circuit design ..............................

I have built a simple circuit .I needed Ic = 23ma.. i have calculated and kept the resistor values for the requirement.Please give me suggestions if i have done any thing wrong ??? thank you
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Calculation :
1)Rc = (Vcc-Vce-Vled)/Ib = (5-0.06-3.5) / 23mA = 62 (kept 75 ohms as nearest value )

2) Ic = 19mA => Ib = 23mA/10 = 2.3mA

3)Rb=(Vin-Vbe) / Ib = (5-0.8) / 2.3mA = 1.8kOhms (took 1.5kohms as nearest value)

4)Iceo = 0.1microamp => so, i have to choose R3 such that it should drop 0.1microamp at the worst case. after trying few values i got 4.7k as good choice.... 4.7k drops 0.16microamps.

Hi kdg007,

What Ic did you get? Can you provide this information?

Microwave123

Re Point 4...

Do you mean microamp or milliamp? Because your schematic shows R3 drops 160 uA. My simulator agrees within a reason margin, depending on what is the volt level of the central node.

Microwave123
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i wanted to get Ic around 23mA range

---------- Post added at 11:48 ---------- Previous post was at 11:47 ----------

BradtheRad
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if i am not wrong... i datasheet showed microamps !!!

---------- Post added at 11:55 ---------- Previous post was at 11:48 ----------

0.1microamp or 100 nano amp for icbo --- sorry i kept Iceo but its Icbo

kdg007 said:

View attachment 67274

I have built a simple circuit .I needed Ic = 23ma.. i have calculated and kept the resistor values for the requirement.Please give me suggestions if i have done any thing wrong ??? thank you
-
Calculation :
1)Rc = (Vcc-Vce-Vled)/Ib = (5-0.06-3.5) / 23mA = 62 (kept 75 ohms as nearest value )

Click to expand...

62 ohms is one of the standard values in the E24 series. If you want to choose from the more commonb E12 series, the nearest value to 62 ohms is 56 or 68 ohms.

2) Ic = 19mA => Ib = 23mA/10 = 2.3mA

Click to expand...

If you've selected 75 ohms as Rc and calculated 19mA as Ic, you should use 19mA/10 as Ib, not the original 23mA. If you choose 68 ohms as Rc, Ic = 21mA, and you should calculate Ib as 21mA/10.

3)Rb=(Vin-Vbe) / Ib = (5-0.8) / 2.3mA = 1.8kOhms (took 1.5kohms as nearest value)

Click to expand...

1.8k is also a common standard value. But you should first choose the proper value of Ib as above.

4)Iceo = 0.1microamp => so, i have to choose R3 such that it should drop 0.1microamp at the worst case. after trying few values i got 4.7k as good choice.... 4.7k drops 0.16microamps.

Click to expand...

Don't mix up your units. The unit of drop is Volt or mV. In the other thread, the worst-case drop was 100mV which is 0.1V. 4.7k will work, but I think you're still a little mixed up about the reasoning.

It's important that you understand the reasoning behind the calculations: Picture the collector-base leakage current Icbo as flowing from the collector to the base. If this current is not diverted, it will flow into the base-emitter junction and act just like a base current coming from outside, and that base current will be amplified by the transistor's beta. If we place a resistor of proper value between base and emitter, most of the leakage current will flow into the resistor instead of going into the b-e junction.

There is no exact value for the external b-e resistor. There's a fairly wide range of values that we can use. This is how we get the usable range:

If the resistance is too high, it will not divert enough of the leakage current. If it's too low, it will divert too much of the normal base drive current and not enough will enter the base. We choose a resistor value somewhere in between those two extremes.

To look at it another way, the resistance should be low enough that the voltage drop across it due to Icbo is well below the normal Vbe. It should be high enough that the current diverted by it in the 'on' condition is much lower than the normal base drive current. That is,
R*Icbo << 0.5V
0.8V/R << Ib

The voltages 0.5V and 0.8V are not exact figures. 0.5V is roughly the base-emitter voltage that starts to cause current to flow into the base-emitter junction.
0.8V is the approximate value of Vbe when the transistor is switched on. Thus, choose R so that it satisfies these two conditions:
R << 0.5V/Icbo
R >> 0.8V/Ib

thank you for the great explanation
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yes as you said i still need more understanding into these things,,,
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you have a great weekend

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