Did I get Ohms Law wrong?

[angie1199]

Hi Guys,

I need to control a 2N3904 npn transistor with a 12V PS and a switch. I know the current I need at the transisotor base is between 40 and 50 mA.

Ohms law states resistance is V/I. Using this, I get 0.24R. This doesn't make sense, or does it? Did I miss something along the way?

Angie

 

[ericgibbs]

hi angie,

Base resistor is [12V - 0.7V]/0.04 = 282.5R

A base current of 40mA for a 2N3904 is high!.... whats the required collector current.?

 

[angie1199]

Hi again Eric,

This transistor is for a different part of the larger project than that of the one you helped out with the other day, but I found that the 10mA wasn't switching the transistor to saturation once I had two signals to switch, it did work with one though.

I have no idea how to work out the collector current but in Circuit 3 the pin 'Switch1' on the detector goes to the collector of T2 on the 'Switch1 Activation' circuit, the emitter goes to V0 and the base goes to the switch which has the 12v PS. This will act as a auxillary switch for the signals. The power to the base is supplied from a solonoid point motor that has a slide switch built in. (Hope you understand a little of model railways here )

The end result would activate the timer section of the detector board and automatically switch Signal1 to red and Signal2 to amber when the points are thrown and hold it in that position.

Circuit 3
**broken link removed**

I really hope this makes sense

Angie

 

[ericgibbs]

hi,
do you mean:
Circuit 3 the pin 'Switch1' on the detector goes to the collector of T11 on the 'Switch1 Activation' circuit.
I assume the 12V PS, the pulse output from the 555.?

Is it the T11 transistor thats causing the problem.?

 

[ericgibbs]

hi,
Is this how it is wired .?
You should have base resistor in both transistor base inputs.

 

[angie1199]

hi,
Is this how it is wired .?
You should have base resistor in both transistor base inputs.

Hi Eric

Yep, that's how it is eventually wired. It's those resistors I didn't know the value of. Having blown a 555 timer I realised the 'benefit' (read 'reason') of a resistor and as such was trying to work out the value of those resistors to protect the transistors. Having read about ohms law I assumed it was R = V/I but that is obviously not how you work it out given my 0.24 Ohms result.

I looked at the schematic and T11 has a 10k resistor and it does activate the timer. So I tried one on my breadboard and it works. I have no idea why though, or how to actually work out what value resistor I should use

Angie

 

[Dx3]

A rule of thumb might be to drive the base of the transistor with 10% of the current you expect to flow through the collector. There will be exceptions, but this method works most of the time.

It's OK to estimate on your first try, just double check your answer with some logical thought and then triple check the results with a volt meter. There will be times when the driving side can't quite provide that much current, so you will settle for 8% of the collector current. Sometimes you will need very good saturation and you'll drive the base with 20% of the collector current. Sometimes you will have a high gain transistor and be able to get away with 4% or 5%. High current transistors can have miserable gain and you end up driving them with 20%.

I hope I have given you something useful.

 

[DemonicSandwich]

Hi EricHaving read about ohms law I assumed it was R = V/I but that is obviously not how you work it out given my 0.24 Ohms result.

I looked at the schematic and T11 has a 10k resistor and it does activate the timer. So I tried one on my breadboard and it works. I have no idea why though, or how to actually work out what value resistor I should use

Angie

It's because you are using the wrong units in the equation. (I) is the current in Amps, while you're using milliamperes. 1 Amp = 1000 milliamperes.
So the (I) in the equation is 0.050.
The second thing is that the (V) is the amount of voltage the resistor has to remove, not the total voltage. Since the BE diode of the transistor has a voltage drop of 0.6-0.7 volts, the resistor has to drop about 11.4V so (V) is 11.4.

With those numbers, R = 11.4/0.050 = 228 Ohms.

What I don't understand is why you're trying to push 50mA into the base of a 2N3904. Such a transistor has a constant collector current of 200mA and a minimum gain is something like 50 so 10mA would be more than enough to saturate it.

 

[ericgibbs]

Hi Eric

Yep, that's how it is eventually wired. It's those resistors I didn't know the value of. Having blown a 555 timer I realised the 'benefit' (read 'reason') of a resistor and as such was trying to work out the value of those resistors to protect the transistors. Having read about ohms law I assumed it was R = V/I but that is obviously not how you work it out given my 0.24 Ohms result.

I looked at the schematic and T11 has a 10k resistor and it does activate the timer. So I tried one on my breadboard and it works. I have no idea why though, or how to actually work out what value resistor I should use

Angie

hi angie,
The collector resistor of the driven transistor is 22K, so from a 12V that would be an Ic of approx 0.5mA, which is quite low,
A 10K base resistor from 12V will give a base current of approx 1mA, so that will easily saturate the transistor.
Is it now all working.?

 

[angie1199]

A rule of thumb might be to drive the base of the transistor with 10% of the current you expect to flow through the collector. There will be exceptions, but this method works most of the time.

I hope I have given you something useful.

Dx3, you have and thank you. It's the '10% of the current you expect to flow through the collector' bit that I am stuck on. How do I know that current?

It's because you are using the wrong units in the equation. (I) is the current in Amps, while you're using milliamperes. 1 Amp = 1000 milliamperes. So the (I) in the equation is 0.050.
The second thing is that the (V) is the amount of voltage the resistor has to remove, not the total voltage. Since the BE diode of the transistor has a voltage drop of 0.6-0.7 volts, the resistor has to drop about 11.4V so (V) is 11.4.

With those numbers, R = 11.4/0.050 = 228 Ohms.

What I don't understand is why you're trying to push 50mA into the base of a 2N3904. Such a transistor has a constant collector current of 200mA and a minimum gain is something like 50 so 10mA would be more than enough to saturate it.

Aha, I knew there was something about amps and milliamps. I didn't re-check what my memory threw up and got it the wrong way round, I = Amps, not milliamps. Thanks for that

Also I wasn't aware that I had to subtract volts lost through the base/emitter. The transistor B-E is the only thing between V+ and V-. Would I be right in assuming that if there was a second component in that route I would also subtract the volts lost of that component too?

 

[angie1199]

The collector resistor of the driven transistor is 22K, so from a 12V that would be an Ic of approx 0.5mA, which is quite low,
A 10K base resistor from 12V will give a base current of approx 1mA, so that will easily saturate the transistor.
Is it now all working.?

Thanks Eric, yes it works just great now.

One thing though, looking at T7 in Circuit 3, connected to pin 3 of 555 timer (my previous forum topic you helped me with). I initially had a 1k resistor there to give me 10mA at the base of T7. When one signal control circuit was connected (switching signal 1 to red) it worked fine. But when I added the second signal control circuit (signal 1 to red, signal 2 to yellow) the greens didn't go out fully on either, and the red and yellow didn't come on fully. That's why I gradually reduced the resistor value and 470R worked, giving 20mA at T7, but 560R still wasn't quite enough.

If 1mA is sufficient to saturate the transistor then why does T7 require 20mA to switch two LEDs?

Angie

 

[angie1199]

What I don't understand is why you're trying to push 50mA into the base of a 2N3904. Such a transistor has a constant collector current of 200mA and a minimum gain is something like 50 so 10mA would be more than enough to saturate it.

This is where my logical brain fails to get physics

On Circuit 3, T7, I tried a 1k resistor (R13) which gave 10mA at T7. As I've put in my reply to Eric (Post 11) the 1k worked with one signal circuit connected but not two. I gradually lowered the R13 resistor value and got both circuits doing things right with a 470R resistor. This I took as the transistor being saturated. Thus the reason for thinking 50mA would saturated it. Obviously I'm wrong but not sure why the different resistor values on the bases of T7 and T2.

Angie