LTSpice question

Thread Starter

Guinness

Joined Dec 31, 2009
81
Hi,

I have just downloaded LTSpice and done a basic circuit to understand how it works.
I have attached a simple circuit and am getting reading of 0.18ma current going through R3 no matter what value ohms I put in, I have gone from 1 ohm up to 10k ohms and it still reads as 0.18ma on the resistor!

It is prob just me doing something wrong, but if anyone can help that would be great.

I first tried a voltage divider across the base of the transister and still got wierd reading then.
 

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Brownout

Joined Jan 10, 2012
2,390
Try it can see. Plot E, B, C voltages and see how changing R1 changes those voltages. It's a guess on my part, but I think the terminal voltages will change, thus changing the currents. Remember, there is 100's times more current in the Emitter than in the Base.
 

Thread Starter

Guinness

Joined Dec 31, 2009
81
Yes thank you, changing R1 does effect it all. Im missing an equation or some basic understanding somewhere of how it all ties together. I will go back to the tutorials and try to figure out what im missing.

Thanks again.
 

Brownout

Joined Jan 10, 2012
2,390
Well just remember that base current is tiny compared to emitter current. Changing emitter current by changing the emitter resistor has a proportionally larger effect. Here is the equation for the bias scheme ( PN voltages are left for for simplicity )

IB = VCC/(RE + (RB/beta+1))

Can you see why RE has a much bigger influence?
 

Jony130

Joined Feb 17, 2009
5,020
The base current is equal to

Ib = ( Vcc - Vbe - Vled) / ( Re + Rb/(β+1) )


Or

Ib = ( Vcc - Vbe - Vled) / ( Rb + (β+1)*Re )


And you should pick the model for the BJT and the LED
See my LTspice file
 

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Thread Starter

Guinness

Joined Dec 31, 2009
81
Yea, I think I understand now how the RE effects it looking at that equation.

I will spend a bit of time changing values and comparing results till I get get it all fully. (Hopefully)!
 

crutschow

Joined Mar 14, 2008
23,500
When R1 is small the base current is sufficient to saturate the transistor, so changes in resistor value around that point have little effect since the transistor stays fully on. If you increase R1 sufficiently so that the transistor starts to turn off, then you will see a change in the transistor current.

The R1 resistor value at which this starts to occur depends upon the Hfe (Beta) gain of the transistor. It's likely much higher than 10k ohms.
 

crutschow

Joined Mar 14, 2008
23,500
The definition of saturation in a BJT is that the base-collector junction is forward biased. This cannot happen in this circuit with any value of R3, even zero.
Wikipedia states: Saturation: base higher than emitter, but collector is not higher than base.

So the transistor is just at the edge of saturation with Vbe essentially equal to Vce instead of Vce being less than Vbe. That still doesn't change my post. ;)
 

Ron H

Joined Apr 14, 2005
7,014
When R1 is small the base current is sufficient to saturate the transistor, so changes in resistor value around that point have little effect since the transistor stays fully on. If you increase R1 sufficiently so that the transistor starts to turn off, then you will see a change in the transistor current.

The R1 resistor value at which this starts to occur depends upon the Hfe (Beta) gain of the transistor. It's likely much higher than 10k ohms.
But there is no point where you can say the NPN "starts to turn off". This is because it was never saturated. If it were a PNP, then what you say is true.
See the curves of diode current vs base resistor value. for both cases.

Wikipedia also says this:
Saturation: With both junctions forward-biased, a BJT is in saturation mode and facilitates high current conduction from the emitter to the collector (or the other direction in the case of NPN, with negatively charged carriers flowing from emitter to collector). This mode corresponds to a logical "on", or a closed switch.
 

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crutschow

Joined Mar 14, 2008
23,500
But there is no point where you can say the NPN "starts to turn off". This is because it was never saturated. If it were a PNP, then what you say is true.
Picky, picky. :rolleyes: How about I say the transistor starts to conduct less current and the collector emitter voltage starts to rise as the base becomes starved for current? That certainly is true and has nothing to do with whether the transistor is technically "saturated" or not.
 

Ron H

Joined Apr 14, 2005
7,014
Picky, picky. :rolleyes: How about I say the transistor starts to conduct less current and the collector emitter voltage starts to rise as the base becomes starved for current? That certainly is true and has nothing to do with whether the transistor is technically "saturated" or not.
OK, that works for me. I was just taking issue with your definition of saturation.
 

Potato Pudding

Joined Jun 11, 2010
688
Select any of the transistor models and LED models and that resistor will start to make a difference.

LTSpice generics can give you a really bad impression of what the program can do.

After you place a generic, change it before you forget.
 

Audioguru

Joined Dec 20, 2007
11,251
WOW!
One of the first silicon transistors I ever bought were a extremely old 2N2369's in old fashioned metal cases. I think I used as couple and still have a couple that have never been used. They must be at least 50 years old.
 

Ron H

Joined Apr 14, 2005
7,014
WOW!
One of the first silicon transistors I ever bought were a extremely old 2N2369's in old fashioned metal cases. I think I used as couple and still have a couple that have never been used. They must be at least 50 years old.
2N2369 is a GREAT high speed saturating switch. Its storage time is MUCH shorter than that of general purpose transistors like 2N3904 or 2N2222. I used it in the design of several commercially successful products.
 
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