Calculating the voltage at a parallel junction?

[blewisjr]

Sorry if my subject uses poor terminology.   Right now I have a preamp and I am trying to optimize the circuit.  In order for the preamp to properly amplify I need the supply voltage to be split in half so if 5V it needs to be about 2.5V at the junction of the load and bias resistors.  I am trying to figure out how to calculate this and everything I get just does not make sense.

My total resistance of the parallel resistors is about 3.8k ohms.  The voltage going in to the load resistor is 5V and since it goes through a 3.9k resistor I have 5/3900 = 1.30mA at that point.  The second half is more complicated after the mic bias resistor I have about 5/10k = 0.500mA.  This is where I get stuck because under my assumption I need to calculate from both directions.

Here is the circuit with an arrow pointing to what I am trying to figure out.

[tszaboo]

Probably it is because you did not set up a DC operating point for the transistor. Are you trying to do this (figure 3)?
http://en.wikipedia.org/wiki/Common_emitter

[Mark Hennessy]

This is slightly tricky...

To work out the voltage at the collector, you need to know the collector current. If it was 1mA, say, then you'd have 3.9 volts across R3, hence collector voltage is 1.1 volts (5-3.9).

But to determine the collector current, you need to know the base current. Ic = Hfe times Ib, hence Vc is 5 - (Hfe times Ib times 3.9)

What is Ib? It's (Vc-Vb)/120k

For our purposes, Vb is fixed at 0.6V. But Vc is variable. And before you know it, the maths gets a bit complicated. You get something like:

Vc = Vcc / (1+Hfe.Rl/Rb)

With the component values shown, and an assumed Hfe of 200, this gives around 0.7 volts. But this formula ignores the 0.6V VBE. Either way, it looks like the 120k needs to be raised somewhat - try 560k or similar, depending on the Hfe of the transistor you're using...

Rather than calculating the collector voltage for a given set of components, it's actually easier to determine the optimum base resistor for a desired collector voltage:

Vc = 2.5V

Ic = 2.5 / 3.9, which is 0.64mA

Assume Hfe of 200 (you'll need to check this).

Ib = Ic / 200, which is 3.2uA

Rb is (2.5 - 0.6) / 3.2uA, which is 594k, so try 560 or 620k

The big problem with this circuit is that it depends on Hfe, which is a decidedly shifty parameter. How much this matters to you depends - only you can decide this. For a one-of, you can tweak as necessary. For production, I'd avoid this, unless I was sure I could quantify the range of Vc that mattered, which would in turn give me an allowed range of Hfe. Operating at higher supply voltages eases the problem somewhat for a given signal level.

What is this preamp for? I mostly do audio, where this circuit would produce unacceptably high levels of distortion, but perhaps this doesn't matter for you?

Hope this helps,

Mark

[blewisjr]

First I would state everything works with the current values.  Just checking to see if I can get it slightly more responsive.  Right now it can pick up a clap at about 5 feet away which is good enough technically.  The preamp noise does not matter the only purpose of the preamp is to amplify the signal enough to trigger an interrupt on a micro controller.  There is no audio playback I just need to grab the signal so I know when the audio burst came through to set a flag.  Like I said it is probably good enough just trying to see if it is possible to get it a little better without adding a second stage or without going with the opamp.  This was working very well with a opamp based preamp but then again the gain with the opamp was around 170 where with the npn base we are looking at a max of 70 so I am trying to increase the amplification a bit through trying to get the NPN to sit at half open.  They say to do this the voltage between the base and collector needs to be about 1/2 the load voltage.

Worst case I could go back to the opamp as the math is much less complex

[Mark Hennessy]

To raise the collector voltage, you have to reduce the collector current.

This will reduce the transconductance of the transistor; probably not what you want.

Don't forget also that there is negative feedback via the base resistor, which again, is not what you want.

You'd get better results with a fully-stabilised design, using a decoupled emitter resistor - getting a gain of ~170 for a collector current of 1mA is quite feasible. But you'll need two more resistors and a cap. The maths is easy, though.

Feed the base with a potential divider, 220k and 62k might be OK, and should give just over a volt at the base.

Change the collector resistor to 2k2, and add an emitter resistor of 470 ohms. Put a cap in parallel with that.

You should find that Vc sits at roughly 2.5V.

Does your microcontroller have a Schmitt trigger input? What voltage swing does it require to detect the edge? The usual way to do this would involve a comparator between the amplifier and the interrupt input, but a Schmitt input would satisfy this. Without that, things might be a bit marginal...

[blewisjr]

Yes the Micro uses a Schmitt trigger for the inputs.  I am not sure about the swing.  I think I found the information in the datasheet but I do not quite understand what I am reading here.
So I will attach the excerpt.

I am assuming what you are saying is I should use a voltage divider going into the base and then adjust the load resistor.  I do understand the resistor on the emitter but won't that smooth out the distortion but at a cost to the overall gain?

[dr_p]

maybe if you simulate the circuit, it will help you out to properly see what's happening in there.


You can use this simple simulation program:
http://www.falstad.com/circuit/

[Mark Hennessy]

I am assuming what you are saying is I should use a voltage divider going into the base and then adjust the load resistor.  I do understand the resistor on the emitter but won't that smooth out the distortion but at a cost to the overall gain?

The circuit is here:

http://en.wikipedia.org/wiki/File:Complete_common_emitter_amplifier.png

But adding a cap in parallel with the emitter resistor will cancel the NFB at signal frequencies, so you'll get plenty of gain.

Personally, I'd recommend a 'scope over a simulator for this sort of thing...

Cheers,

Mark

[blewisjr]

Ok thanks.  I will have to tinker around with this because I am not sure how much voltage I will end up with at the base as the microphone has to be biased and if I remember correctly it is sitting around 1V already I will have to do some measurements to be sure though.  Thanks for all the input worst case scenario I will need to switch to a LM741 for the preamp which works great as well.

[blewisjr]

I have done some experimenting on this method of a true common emitter preamp with my NPN2222A and things have gained a smidge of responsiveness but still not close to the equiv of the 791 opamp.  Resistor wise on the bias divider I have a 220k and a 68k which puts me at about 1V on the base actually 0.965V ish which I think is close enough.  Then on the collector I needed to place a 3.3k which puts me at about 2.45V on the collector.  The emitter is using the 470 and the parallel cap. 

I really do not think I will be able to get the gain high enough on this circuit and am contemplating going back to the opamp.

[IanB]

It's really not reasonable to expect a similar gain from a single transistor as you can get from an op amp. But if you put two transistors in series (a darlington pair for instance), you will get much better amplification.

[blewisjr]

Yea this is what I am starting to realize through the experiment of changing to a single NPN setup.  I wanted to avoid having a second stage to the amplifier and it seems the only way to get the solid result is through an opamp which surprisingly corresponds to easier math as well on gain calculation.  There are time I wish I never started this hobby but when the aha moment comes it feels good.  Sadly I do not get too many aha moments yet.    So many times I just want to give up.