Common Emitter BJT Calculations Driving Me Crazy!!!

[demonstar]

Cheers Alanan! That should keep me busy for a while.

[R.G.]

But as you, RG, well know, you have to have the load be very high or distort the weighting as the loading on the emitter side will work differently to the weighting on the collector side.

I refer to that as the Univibe Memorial Mistake.

[Faber]

Okay, only 1 more question from the noob... I hope...

I understand why you voltage divide to get the voltage on the resistor to be VE+VBE (I wonder if those abbreviations were right...)  But what kind of current do you need at the base out of the voltage divider?  AHHHH!!!
Sorry for the simple questions...

[Faber]

[lightbulb over head]

never mind.  I read a college electronics textbook and found my answer.

[alanlan]

...to be VE+VBE (I wonder if those abbreviations were right...) 

Bang on.

[DougH]

This is a great thread. Not many like this one anymore. I hope a lot of people are reading this.

[gaussmarkov]

This is a great thread. Not many like this one anymore. I hope a lot of people are reading this.

agreed.  i am learning from it.

[Sir H C]

Okay, only 1 more question from the noob... I hope...

I understand why you voltage divide to get the voltage on the resistor to be VE+VBE (I wonder if those abbreviations were right...)  But what kind of current do you need at the base out of the voltage divider?  AHHHH!!!
Sorry for the simple questions...

You can first order ignore the bias current going into the base, and if you make that resistive divider small enough (using a lot more current than the base) it works pretty good.  To save current you have to work more on how much current is going to the base and how much it is going to offset your divider.

[demonstar]

This stuff seemed so complicated at first but after all of this thread (after questions, successes and mistkes) it's just clicked! Now i read the article on mosfets at geofex and i got them too. They don't seem too far apart. Total epiphany! 

What should I do next? JFETs? I'm off college for easter at moment so have more time for this stuff than usual.

[Sir H C]

Play more with the BJTs.  Get it so that you can see the ballet in all its splendor, know it cold.  jFETs would be a decent next step, or MOSFETs.  MOSFETs are more prevelent in the world, but are much easier to kill.

[JDoyle]

and not only gain but headroom.  If you load a single stage with a resistance equivalent to the output resistance of the stage then the headroom is halved i.e. it can only swing half as much even if you boost the input signal even further.[/quote]

I think this is one of those cases where explaining it as detailed as you like to is in order.

What reduces the headroom and how?

[alanlan]

and not only gain but headroom.  If you load a single stage with a resistance equivalent to the output resistance of the stage then the headroom is halved i.e. it can only swing half as much even if you boost the input signal even further.

I think this is one of those cases where explaining it as detailed as you like to is in order.

What reduces the headroom and how?
[/quote]
Well I would answer your question but I don't like your rude and abrasive tone.  Why don't you anwser the question yourself or prove what I've said to be false.  I wouldn't have a problem with that at all.

[gaussmarkov]

and not only gain but headroom.  If you load a single stage with a resistance equivalent to the output resistance of the stage then the headroom is halved i.e. it can only swing half as much even if you boost the input signal even further.

I think this is one of those cases where explaining it as detailed as you like to is in order.

What reduces the headroom and how?

i'll take a stab at this one.  i think it follows from the definition of the term "output resistance."  isn't output resistance the load resistance at which output is halved relative to a load with infinite resistance, as in when the output resistance equals the load resistance?  see, for example, Zachary Vex's post explaining this.

[alanlan]

and not only gain but headroom.  If you load a single stage with a resistance equivalent to the output resistance of the stage then the headroom is halved i.e. it can only swing half as much even if you boost the input signal even further.

I think this is one of those cases where explaining it as detailed as you like to is in order.

What reduces the headroom and how?

i'll take a stab at this one.  i think it follows from the definition of the term "output resistance."  isn't output resistance the load resistance at which output is halved relative to a load with infinite resistance, as in when the output resistance equals the load resistance?  see, for example, Zachary Vex's post explaining this.

Well it's sort of related.  True, if you consider the Thevenin equivalent of an amplifier i.e. an ideal source of gain*vin in series with a resistor of value the same as the output resistance of the amplifier, then you will as you say get half the output if you load this with a resistor of the same value.  It's a simple potential divider. 

But, there's a but. 

In the ideal situation just described, you could increase vin to compensate i.e. double vin to bring the output back up to it's unloaded level.  However, in practice, there is a limited maximum AC current supply swing available from the output and because this is shared with the load resistance, the output voltage can only swing so far.  By the way, I'm talking about the usual case where the load appears after an AC coupling cap.  i.e. usually it's the next stage which causes the loading and more often than not the stages will be AC coupled.  This effect still occurs with a DC coupled load to ground but the results are slightly different.

The easiest way to investigate this is to simulate. 

[gaussmarkov]

and not only gain but headroom.  If you load a single stage with a resistance equivalent to the output resistance of the stage then the headroom is halved i.e. it can only swing half as much even if you boost the input signal even further.

I think this is one of those cases where explaining it as detailed as you like to is in order.

What reduces the headroom and how?

i'll take a stab at this one.  i think it follows from the definition of the term "output resistance."  isn't output resistance the load resistance at which output is halved relative to a load with infinite resistance, as in when the output resistance equals the load resistance?  see, for example, Zachary Vex's post explaining this.

Well it's sort of related.  True, if you consider the Thevenin equivalent of an amplifier i.e. an ideal source of gain*vin in series with a resistor of value the same as the output resistance of the amplifier, then you will as you say get half the output if you load this with a resistor of the same value.  It's a simple potential divider. 

But, there's a but. 

In the ideal situation just described, you could increase vin to compensate i.e. double vin to bring the output back up to it's unloaded level.  However, in practice, there is a limited maximum AC current supply swing available from the output and because this is shared with the load resistance, the output voltage can only swing so far.  By the way, I'm talking about the usual case where the load appears after an AC coupling cap.  i.e. usually it's the next stage which causes the loading and more often than not the stages will be AC coupled.  This effect still occurs with a DC coupled load to ground but the results are slightly different.

The easiest way to investigate this is to simulate. 

o.k.  thanks! 

i was thinking of a signal swinging at the maximum amplitude and no AC coupling.  if you put a load on equal to the collector resistor (so its equal to the output resistance -- my guess at output resistance) aren't you are going to run into a problem with forward biasing the base-collector junction when the output voltage swings down?  if so, then you cannot just increase the amplitude of the input signal to make up for loading down the output.  that will just make the problem worse. (i am picking this up from Perro Grande's comment back on page 1 where he mentions watching out for both extremes of the signal.)

cheers, paul

[alanlan]

...aren't you are going to run into a problem with forward biasing the base-collector junction when the output voltage swings down?

Not exactly because if you think about it, the emitter follows the base minus a diode drop.  If the transistor is saturated then the collector will be a tad higher than the emitter, but no lower than the emitter.  So the collector-base junction can't become severely reversed biased.  It will all limit out and the output will clip.

[gaussmarkov]

cheers!

i worked through some maths to see if i was getting some of this.  if i take the bounds of operation as  VB>0.6 (for the diode drop) and VC>VB (which could be loosened slightly?) then, for maximum swing, i get a VB operating point equal to

(6g + 3  + 5 Vcc)/(10(g+1)) =   0.6    +    (1/2 Vcc - 0.3)/(g+1)

(edit in green) and a VC operating point equal to

0.3 g /(1 + g)    +     Vcc/2 x (g + 2)/(g + 1)

where Vcc is the supply voltage and g = Rc/Re is the voltage gain of the amplifier.  so for unity gain (g = 1), that's VC = 0.15 + 3/4 Vcc, which is essentially what JDoyle observed.  for high g, VC is just above 1/2 Vcc.  and for a moderate value like  g=4, it's VC=5.6V.

given that  VE = VB - 0.6V, i get a VCE = VC - VE of

0.3 + 1/2 Vcc

which does not depend on g at all and is slightly above 1/2 Vcc, just like alanlan said.

[gaussmarkov]

how do you think about VBE when there is a bypass capacitor across Re?  i see that the constant diode drop doesn't apply anymore.

any help is appreciated, paul

[alanlan]

how do you think about VBE when there is a bypass capacitor across Re?  i see that the constant diode drop doesn't apply anymore.

any help is appreciated, paul

When you add a bypass cap, the signal on the emitter will be very much reduced providing less negative feedback which in turn means higher gain but also lower impedance looking into the base.  The base emitter junction is still a diode so you can't really make Vbe rise very much above normal diode forward drop voltages.  In practice what will happen is that a large signal going into the base will be clamped by the base emitter diode junction and the base current will rise causing more collector current to flow.  Hence higher gain and distortion (due to reduced negative feedback).

[gaussmarkov]

thanks, alanlan.

as i understand it, when current gain is low base current is no longer negligible and these approximations don't apply.  how low would you go with current gain (hfe) before considering these approximations inaccurate for biasing and other calcs?  i am thinking of the popularity of low gain transistors in fuzz face circuits, where below 50 is common.