Confirmation on Ib voltage and NPN Base resistor

Thread Starter

EHNeeley

Joined Nov 4, 2019
6
I am trying to use a transistor for the fist time. I have spent hours reading various post on forms which has led to more confusion than answers. If someone please point me into the right direction

I have attached my circuit I am working on. I am trying to solve the resistance for R3

The formula I have found form this site R=(V-Vbe)/Ib

If I am understanding every thing (12-.9)/.005=2220. My R3 resistor should be 2,2K ohm @ 5% or 1.26K ohm @ 1%?

Data sheet for D3, Kingbrit WP710A10SRC-E http://www.kingbrightusa.com/images/catalog/SPEC/WP710A10SRC-E.pdf

Forward voltage 1.8 Typ, If 20 mA with maximums of 30 mA and 75 mW
Using a 560 ¼ w ohm resistor I should get a load of 17.7 mA with 34 mW

Data sheet for Q1 , Fairchild BC548 https://media.digikey.com/pdf/Data Sheets/ON Semiconductor PDFs/BC546-50.pdf

Typ. Vbe (sat) 900 Typ with the conditions of Ic 100 mA , Ib 5 mA
 

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MrAl

Joined Jun 17, 2014
8,390
Hi,

Yes if your Vbe is 0.9v and the supply to the base resistor is 5v and the emitter is grounded then the formula is:
Ib=(5.0-0.9)/Rb

But since the collector load is light (20ma or less) you can assume a Beta of maybe 20 or more so we have:
Ib=Ic/20

and with Ic=0.020 we get
Ib=0.020/20=0.001

so we get an Rb:
Rb=4.1/0.001=4.1k

so 3.9k would work or probably even 5k.
 

Zeeus

Joined Apr 17, 2019
615
I am trying to use a transistor for the fist time. I have spent hours reading various post on forms which has led to more confusion than answers. If someone please point me into the right direction

I have attached my circuit I am working on. I am trying to solve the resistance for R3

The formula I have found form this site R=(V-Vbe)/Ib

If I am understanding every thing (12-.9)/.005=2220. My R3 resistor should be 2,2K ohm @ 5% or 1.26K ohm @ 1%?

Data sheet for D3, Kingbrit WP710A10SRC-E http://www.kingbrightusa.com/images/catalog/SPEC/WP710A10SRC-E.pdf

Forward voltage 1.8 Typ, If 20 mA with maximums of 30 mA and 75 mW
Using a 560 ¼ w ohm resistor I should get a load of 17.7 mA with 34 mW

Data sheet for Q1 , Fairchild BC548 https://media.digikey.com/pdf/Data Sheets/ON Semiconductor PDFs/BC546-50.pdf

Typ. Vbe (sat) 900 Typ with the conditions of Ic 100 mA , Ib 5 mA
Hi..What is your question? seems you know much already : or we both know too little

Transistor is operating as a switch
"Using a 560 ¼ w ohm resistor I should get a load of 17.7 mA with 34 mW"

Yes about 17.7mA but please why 34mW? Also have you built the circuit yet?
 

Thread Starter

EHNeeley

Joined Nov 4, 2019
6
MrAI

"Yes if your Vbe is 0.9v and the supply to the base resistor is 5v and the emitter is grounded "

I not sure on the Vbe. the specs sheet list it as 900 also the base voltage is 12 V. yes the emitter is grounded.

Zeeus,

My question was if my calculation of 2,2K ohm was correct.

The 34 mW would be the power dissipated by the LED

I read a lot and tinker with electronics. I have included as much info as i know so in the hopes that if helps those who are willing to help me fill in the blanks. No I have not built the circuit. I need to order the parts and hope with help wont order stuff that I wont/cant use.
 

BobTPH

Joined Jun 5, 2013
3,509
My R3 resistor should be 2,2K ohm @ 5% or 1.26K ohm @ 1%?
I have no idea where you got this from. The tolerance if the resistor has nothing to do with calculating the value you use. The 1% or 5% is a measure of how close the resistor will be to its stated value.

Bob
 

Thread Starter

EHNeeley

Joined Nov 4, 2019
6
BobTPH

I got those valves for a common resister listing I posted them becose I an not sure on if the vavlue I caculated is corect and how close the value I must be for the circuit to work.
 

MrAl

Joined Jun 17, 2014
8,390
MrAI

"Yes if your Vbe is 0.9v and the supply to the base resistor is 5v and the emitter is grounded "

I not sure on the Vbe. the specs sheet list it as 900 also the base voltage is 12 V. yes the emitter is grounded.

Zeeus,

My question was if my calculation of 2,2K ohm was correct.

The 34 mW would be the power dissipated by the LED

I read a lot and tinker with electronics. I have included as much info as i know so in the hopes that if helps those who are willing to help me fill in the blanks. No I have not built the circuit. I need to order the parts and hope with help wont order stuff that I wont/cant use.
Hi,

I just quoted 20ma because that is the max for a small LED and it looks like you have an LED there, but if it is an LED then the current is much less because of the voltage drop caused by the LED itself. So 5k would work, 2.2k would work with a bit more current drive. Probably even 10k because the collector load is so very light.

The Vbe voltage spec is not that important for a circuit that is as non critical as this. you can assume 0.6 to 1.0 volts with almost the same results. 2.2k to 10k i would say, with 5k or 5.1k best.
 

Thread Starter

EHNeeley

Joined Nov 4, 2019
6
MrAI

Thank your for your second reply.

I need to do the same calculations for a 5 volt circuit. based on your answer I have a few questions so i can understand.

how did you calculate the 5K ?

Also was my understanding of the data sheet information as it inputs into the formula correct? I ask this become most of the data sheet is way over my current undstanding.
 

MrAl

Joined Jun 17, 2014
8,390
MrAI

Thank your for your second reply.

I need to do the same calculations for a 5 volt circuit. based on your answer I have a few questions so i can understand.

how did you calculate the 5K ?

Also was my understanding of the data sheet information as it inputs into the formula correct? I ask this become most of the data sheet is way over my current undstanding.
Hi again,

It all starts from the collector load and the Beta of the transistor. Because the collector load is very light we can assume at least a gain of 20. It's probably more like 50 but if we assume 20 we are sure to get it right.
Since the collector load would be a max of 20ma which is 0.020 amps and the gain is 20, we can divide the 0.020 by 20 and we get 0.001 which is 1ma base current.
With 1ma base current and a drive voltage of 5v we have a resistor value:
R=5/0.001=5000 Ohms.

Now if we want to get more particular, we can subtract the max of 1.0v Vbe from the 5v and get 4v so the base resistor would be:
R=(5-1)/0.001=4/0.001=4000 Ohms.

So we have a resistor value of 4k, but we can go lower than that if we like. Half that is 2k so 2.2k would be good.
So we really have a range of 2k to 5k with around 4k more optimal.

The base resistor is not that critical as long as we use min and max values to obtain the value. We used a min gain of 20 and max collector current of 20ma. Ir reality, the LED will drop some voltage so the max current could be even lower like 5ma so 4k will definitely work.

Feel free to ask more questions if you still dont understand yet. That's what this forum is for :)
 

Thread Starter

EHNeeley

Joined Nov 4, 2019
6
MrAI

I can not tell you how thankful I am for your help. I am at least starting to put some of the puzzle together. Its baby steps but once I get it I will own it. It may take a few good wacks in the head to make it stick tho ;)

After watching a few more videos it appears that the NPN is the wrong part for my application. My idea was if the base was saturated it would would turn off the emitter. If I am understating correctly I need a PNP. This way any voltage after the fuse feeding the base to saturation would shut off the emitter. If the fuse was blown then the base would have no voltage and the power would pass from the collector to the emitter.

The goal of this part of the circuit was to turn on the LED when the fuse was blow regardless if there was a load after the fuse. This board is part of a larger project and I want a sure fire way to quickly trouble shoot.

Even tho I need to rethink and redesign this part of the circuit I still want to learn the NPN. So for my education and any who may find this thread useful I going modify the design where this transistor is monitoring the switch. I will then add an PNP to monitor the fuse.

I will re-post when I have done my math for some one to conferm I got it right or giggle some and then point me into the correct direction.

I did stumble onto a nice web sites on this and thought I share.

https://www.nutsvolts.com/magazine/article/may2015_Secura

Respectfully

Ed

Do not trust on line calculators. Learn and do the math yourself. This way you own the math.
 

DickCappels

Joined Aug 21, 2008
7,570
One thing to keep in mind is that whatever you learn about using a NPN also applies to a PNP, except the polarities of the voltages are reversed.

In the 1950's and most of the 1960's it seemed every circuit used PNP transistors with a negative power supply. Then, for reasons I still do not understand, high performance silicon NPN transistors began to appear and NPNs dominated. It took me a year to adjust to the change. I hope you can avoid this kind of problem.
 

Thread Starter

EHNeeley

Joined Nov 4, 2019
6
MrAI

Outstanding and thank you. Ill do the math and post my results.

It may be a day or two. My granddaughter is coming over to tomorrow. Oh the adventures a 5 year and her granddad can have. Last time we were left to our self's I was grounded for a month afterwords. Seams my wife and daughter were not fond of the home made catapult.

This week were going to try our hands at an ballista. :)

DickCapples

Thank you for your insight. Its nice to know that a lot of the math etc will be the same.
 

MrAl

Joined Jun 17, 2014
8,390
MrAI

Outstanding and thank you. Ill do the math and post my results.

It may be a day or two. My granddaughter is coming over to tomorrow. Oh the adventures a 5 year and her granddad can have. Last time we were left to our self's I was grounded for a month afterwords. Seams my wife and daughter were not fond of the home made catapult.

This week were going to try our hands at an ballista. :)

DickCapples

Thank you for your insight. Its nice to know that a lot of the math etc will be the same.
:)cheers :) :)
::)
 
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