Calculating resistor in transistor switch

[miker7940]

I've been asked to find the value of resistor 1 in the attached schematic.

What I know so far:
When the resistance of R2 is 10k the voltage at the base of the transistor is 0.6V
When the LED needs to turn on, R2 resistance will have changed to allow the voltage at the base of the transistor to become 0.7V

What will resistor 1 need to be for the LED to be almost turned on when the voltage at the base of the transistor becomes 0.6V ?
And how would I figure it out ?

Sorry if it is a silly question but I can't figure it out by myself. And won't the LED burn out with only a 330 ohm resistor between it and 12v ? I don't have any details on the led apart from it is a Red 5mm standard 80mcd.
Thanks for any replies

[uncle_bob]

Hi

Well I = E/R will let you figure out the current through the LED. You know the supply is +12 and you can look up the voltage drop on a typical LED. Best guess is you will have > 30 ma through the LED.

If indeed this is a 12V circuit (it looks more like a 5V or 3.3V design, then you need to assume the 30 ma is real.

Next you need the characteristics of your particular transistor. It will *start* to turn on based on it's gain as the base voltage approaches 0.6V. It will not be "full on" until the transistor saturates. This may be at the Ib = 3 ma point.

You have a thevinin equivalent voltage source from 12V. It has a source resistance of R1 parallel R2. There are a number of calculators on the web that will let you work out the effect of the lower resistor. One quick observation is that with the upper resistor = 10K, you have a maximum of 1.2 ma into the base.....

Since you may not be able to turn the transistor full on, the next question becomes ... how much power will the transistor handle? 30 ma at 12V is 360 mW. The max power in the transistor is about 1/4 of this. Does you part do ok at this level?

Bob

[miker7940]

Hi

Well I = E/R will let you figure out the current through the LED. You know the supply is +12 and you can look up the voltage drop on a typical LED. Best guess is you will have > 30 ma through the LED.

If indeed this is a 12V circuit (it looks more like a 5V or 3.3V design, then you need to assume the 30 ma is real.

Next you need the characteristics of your particular transistor. It will *start* to turn on based on it's gain as the base voltage approaches 0.6V. It will not be "full on" until the transistor saturates. This may be at the Ib = 3 ma point.

You have a thevinin equivalent voltage source from 12V. It has a source resistance of R1 parallel R2. There are a number of calculators on the web that will let you work out the effect of the lower resistor. One quick observation is that with the upper resistor = 10K, you have a maximum of 1.2 ma into the base.....

Since you may not be able to turn the transistor full on, the next question becomes ... how much power will the transistor handle? 30 ma at 12V is 360 mW. The max power in the transistor is about 1/4 of this. Does you part do ok at this level?

Bob

Thanks for you reply,

I don't know the spec of the transistor, infact all I know is that it is a NPN transistor. when I was copying the schematic I put in a 2n3904 as it was the first NPN transistor that came to my mind.
With no details known on the transistor, how to know when it saturates ?

Thanks

And it is definitely a 12v circuit, because it says so in the notes, but, it also says that the LED draws max 10mA, and with a 330ohm resistor doesn't that make the supply voltage 3.3V ?

[uncle_bob]

Hi

Well I = E/R will let you figure out the current through the LED. You know the supply is +12 and you can look up the voltage drop on a typical LED. Best guess is you will have > 30 ma through the LED.

If indeed this is a 12V circuit (it looks more like a 5V or 3.3V design, then you need to assume the 30 ma is real.

Next you need the characteristics of your particular transistor. It will *start* to turn on based on it's gain as the base voltage approaches 0.6V. It will not be "full on" until the transistor saturates. This may be at the Ib = 3 ma point.

You have a thevinin equivalent voltage source from 12V. It has a source resistance of R1 parallel R2. There are a number of calculators on the web that will let you work out the effect of the lower resistor. One quick observation is that with the upper resistor = 10K, you have a maximum of 1.2 ma into the base.....

Since you may not be able to turn the transistor full on, the next question becomes ... how much power will the transistor handle? 30 ma at 12V is 360 mW. The max power in the transistor is about 1/4 of this. Does you part do ok at this level?

Bob

Thanks for you reply,

I don't know the spec of the transistor, infact all I know is that it is a NPN transistor. when I was copying the schematic I put in a 2n3904 as it was the first NPN transistor that came to my mind.
With no details known on the transistor, how to know when it saturates ?

Thanks

And it is definitely a 12v circuit, because it says so in the notes, but, it also says that the LED draws max 10mA, and with a 330ohm resistor doesn't that make the supply voltage 3.3V ?

Hi

.... or the transistor never saturates and has a gain that at the max lower resistor value gives 10 ma through the LED ....

With no data on anything, it's all just a guess.

Bob

[Ian.M]

Its a very silly question about a very silly circuit - not your fault, its the academic who oversimplified the theory then asked you a question that has no usable practical answer who should be blamed.

Just for laughs lets see what happens if you are actually using a 2N3904.

See the graph "Base-Emitter ON Voltage vs Collector Current" on page 3 of the Fairchild 2N3904 datasheet.  You will note that the page is labelled "Typical Characteristics", which means that considerable device to device variation is likely.  To give you an idea of the possible range, on page 2 H_FE@Ic = 10 mA, Vce = 1.0 V is listed as 100 to 300.  That means that for a particular Ib you will get up to a 3:1 variation in Ic between different devices. Ib and to a lesser extent Ic will certainly affect Vbe.  Also note the three different lines on the graph for different temperatures, and that the voltage difference between them is about three times the difference between your supposed ON state Vbe of 0.7V and off state Vbe of 0.6V.  Furthermore the graph shows Ic is typically 0.1mA @Vb=0.6V, 25°C, so that means the LED will probably be glowing some when its supposed to be off.   

With that out of the way, to answer the question of what value R1 needs to be to give 0.6V on the base *ASS*U*ME*ing Ib is negligible, you simply solve R1,R2 as a potential divider. 

The actual voltage will be less by Ib*Rth where Rth is the Thévenin equivalent resistance of the divider ( Rth=R1*R2/(R1+R2) ). and we can estimate an upper bound on Ib from the minimum H_FE@Ic = 0.1 mA of 40, so we can be fairly certain the difference due to Ib loading the divider wont be more than a couple of mV, provided the temperature is maintained at or below 25°C.

As this looks suspiciously like a badly posed homework question, I wont solve it for you.

[uncle_bob]

As this looks suspiciously like a badly posed homework question, I wont solve it for you.

Hi

.... but by looking at a previous post and going to the appropriate pages in the text book .. it could be easily solved.

Bob

[Audioguru]

The Fairchild datasheet for the 2N3904 transistor and almost anybody's datasheet for almost any little transistor shows its guaranteed "Collector-Emitter Saturation Voltage" with a base current that is 1/10th its collector current regardless of its current gain, because its current gain number is used when it is not saturated and is a linear amplifier that has plenty of collector to emitter voltage.