2 Stage Common Emitter Booster Impendance?

[Hallmar]

So i designed this 2 stage Common Emitter amp/booster and i'm wondering what kind of input impendance i should have?

I've heard that having it at about 1M ohm is desirable but i'm not sure.

P.s Those capacitor sizes are just guesses so don't worry about 'em, i'll adjust them later.

[Hallmar]

Also, the calculated input impendance of the first stage that i have is 261 Ohms and i've already Impedance matched the two stages.

[R.G.]

The input impedance of the first stage is the parallel combination of R2 and R8, that paralleled by the input impedance of the transistor, which can be approximated by hfe times the emitter resistor, or about 7500 ohms at a raw guess. Something like 1331 ohms.

If you're driving this with any signal with a source impedance much larger than 133 ohms, there will be heavy loading. Driving it with a guitar will give really, really bad signal loss, and this disproportionately to the treble as well.

The "1M input impedance" is a handy rule of thumb to avoid treble loss when driving the stage directly from a guitar pickup.

And then there's the supply current and bias point. Your first stage is eating about 16ma as shown. The base is pulled up to (2k/(2k+8.5K)) *9V = 1.71V. 0.5V is lost to the base-emitter, so the emitter sits at 1.21V. The emitter current is 1.21V/75 = 16.2ma. This flows in the collector resistor too, so the collector is at 9V minus 16.2ma * 375R = 6.07V, or 2.93V. The emitter is up at 1.21V, leaving 1.72V across the collector-emitter.

This is a very high current drain, and a very low imput impedance, as well as a small C-E drop for signal in the first stage.

The second stage has similar characteristics; low input impedance, high current and limited signal swing on the output.

Is there a reason for making these so low on input impedance and high current?

[Hallmar]

I'm on my 5th semester in school regarding this stuff and i just chose a lower Ic for the first stage and a bigger Ic for the second one.

With NPN CC,CE or CB connections i can't get a bigger input impendance than 500k right? Or am i bullshiting here?

And honestly i don't know why i made it low input impendance and high current.
I'm still learning about this stuff and i'm gettin' there i think.

2 weeks ago i didn't know where to start as to design an amplifier, i knew precisely what an amplifier does and what each part does but as with the math i didn't know shit.

[Hallmar]

My thought was also making a JFet buffer before to Impedance match the first stage and the guitar.

Teacher is going trough Jfets on friday.

[R.G.]

I'm on my 5th semester in school regarding this stuff and i just chose a lower Ic for the first stage and a bigger Ic for the second one.

We all start somewhere. Keep learning.

The input impedance of the CE and CC configurations of bipolars is about equal to the hfe times the unbypassed emitter resistance. There is another term, the Shockley resistance, that's an internal part of the transistor that also adds to the emitter resistance, but is generally much smaller than the emitter resistor, so it gets ignored often.

With input impedance for the transistor itself of hfe times Re, and a limit on how big Re can be, the question then becomes "how big can hfe be?"  In general, you have to guess based on the datasheet. I used 100 based on previous experience with the 3904, but it could be as high as 200 with selected 3904s, and might be as high as 1200 with selected MPSA18s.

It is often good to select Re ~ 1/10 Rc, and select the current through the stage to be on the order of 100uA to 1ma for small signal stages. There is no way you should know these bits of acquired wisdom yet, but assume they're true for now.

Try redesigning your transistor stages for Ic=100uA, Vce = 4.5V, Vbe = 0.5V, Rc = 10*Re and recalculate the input impedance for hfe = 200. Input impedance is then 200 times the Re you calculate. When that is done, calculate Vb, and then calculate the bias resistors to make that Vb be true; make the parallel combination of the two bias resistors be 10 times hfe*Re.

What do you come up with?

With NPN CC,CE or CB connections i can't get a bigger input impendance than 500k right? Or am i bullsh*ting here?

For CC and CE, it's about hfe * Re; for CB, it gets complicated.

[R.G.]

I forgot to mention: for audio uses, most often source impedances and input impedances don't need to be matched. It's often important the be MIS-matched in the right way. In general, you want loads to have an impedance that is 10x or more HIGHER than the source impedance driving them.

Matched source and load impedances are for maximum power transfer. This is almost never what you want for audio. You want maximum signal voltage transfer, not maximum power transfer.

[PRR]

Impedance is about V/I, Voltage divided by Current.

I'm guessing your first stage runs at 12mA.

You say its input is 261 ohms.

I get 22K into the transistor and 1.6K from R2 R8, so 1.5K. In-sight-of R.G.'s 1.33K estimate. why is your answer different? Show your math!

Taking 1K as a 3-person average, and knowing that impedance may scale as inverse of current, we can guess that 100K input will happen at 12mA/100 or 0.12mA, 120uA. And 1Meg input needs to go down to 12uA. However current gain will drop, so you may need to go even lower.

Indeed 100K is not hard, 1Meg is difficult, with jelly-bean transistors in CE connection.

Changing current while also keeping a "decent" collector voltage (around half supply voltage) will mean a larger collector resistor. 4V drop with 120uA current is 33K collector resistor. You now need to design the second stage input to cause not-too-much loading in 33K.

> Impedance matched the two stages

Matching is rarely the best plan. (Cascade CE stages *may* be the exception.)

Make inputs 5 to 20 times outputs. OTOH, make outputs 5 to 20 times lower than the inputs they must drive. This will drive a guitar-amp. We can assume an INST jack is at least 100K. Therefore your last stage collector resistor should be 20K to 5K.

The professor has to teach you "ALL" electronics. High-power, low-power, battery audio and mains-power radio transmitters, steel welders, bio-assay instrumentation. He can't know what field you will be working in.

So he can't say "assume 1mA" because that is a lot for a cell-phone mike-amp and way too small for a cell-phone transmit stage.

When the professor is not looking, PLAGIARIZE from the field you will be working (playing) in.

Here's a good study:


I said above that 20K to 5K might be a good output collector load. Mike Matthews must have thought a similar thing, since he used 10K. Going further, we estimate the input impedance of an LPB as about 50K. This is a little low by guitar standards. R.G. says it will hurt treble. But the LPB is "supposed to distort", itself or the amp after it. Distorted treble can be painful. A low input Z on e-guitar reduces treble for a more pleasing tone, *and* a simple 1-transistor plan. (Getting this much boost plus a very high input Z probably wants 2 transistors, and when LPB was new transistors cost money!)

> capacitor sizes are just guesses so...

If you are going to listen or look at the output (even on a virtual 'scope), you need a "good guess". If your 261 ohm impedance is right, 1uFd caps suggest a bass-cut-off near 600Hz. This is _very_ close to your 1KHz test frequency. You lose several dB right there. (In real life, the mismatch between ~~5K-250K guitar and "261 Ohm" input is much-much worse.)

> NPN CC,CE or CB connections i can't get a bigger input impendance than 500k right?

Why does it matter NPN or PNP? (We can get equally-good devices in both polarities now.)

CB *can* be complicated but for most reasonable design CB input is essentially the Shockley resistance, 26 Ohms at 1mA, plus any resistance you add (which also cuts gain). The only "good" use of CB connection in audio is to block stray feedback capacitance (an advanced topic you are not ready for yet). OK, also in some voltage-translation tricks (you have logic signals out of a +supply CPU, for example, and need to switch relays hung from a negative supply).

[PRR]

> make the parallel combination of the two bias resistors be 10 times hfe*Re.

That seems high to me?

[R.G.]

> make the parallel combination of the two bias resistors be 10 times hfe*Re.
That seems high to me?

It is - but it's a good place to start.

The thing to do is to try to make it work that way, then to relax down to converting the bias divider to its Thevenin equivalent, and compute the sag of the base through the Thevenin resistance and see if you can make that work by jiggering the resistors. If there's enough room with the Thevenin equivalent conversion to hold the base up high enough, you get to a solution that's only 5* hfe*Re for an input impedance. It's about as good as you can do without going to a different biasing setup.

Sometimes you can't make it work. But a man's reach should exceed his grasp, right?   

[Hallmar]

I'm gonna have to read this through next couple of days.

Gettin there.

[Hallmar]

Impedance is about V/I, Voltage divided by Current.

I'm guessing your first stage runs at 12mA.

You say its input is 261 ohms.

I get 22K into the transistor and 1.6K from R2 R8, so 1.5K. In-sight-of R.G.'s 1.33K estimate. why is your answer different? Show your math!

According to the book i have the input impendance is    Rb1 // Rb2 // Rin.

Rin is hFE*re+ RE
re is 25m/Ie(re is the internal resistance in the transistor, it's an icelandic translation)
My hFE is 150.

Rin = 150*2.08+75 = 387 Ohms.
So

Zin = 1/( 1/8500 + 1/2000 + 1/387) = 312 Ohms(i forgot to add RE into the equation before).



>Matched source and load impedances are for maximum power transfer. This is almost never what you want for audio. You want maximum signal voltage transfer, not maximum power transfer.

Huh, i always thought that maximum signal power transfer was what is sought for.

Alright, so maximum signal voltage transfer it is.

[R.G.]

According to the book i have the input impendance is    Rb1 // Rb2 // Rin.

OK so far.

Rin is hFE*re+ RE

... but there's a problem here. Rin = hFE*(re+RE)

re is 25m/Ie(re is the internal resistance in the transistor, it's an icelandic translation)
My hFE is 150.  Rin = 150*2.08+75 = 387 Ohms.

I get Rin = 150*(2.08+75) =  11562 ohms.

Zin = 1/( 1/8500 + 1/2000 + 1/387) = 312 Ohms(i forgot to add RE into the equation before).

And Zin = 1420 ohms.

Notice:
- Zin is dominated by the 2k bias resistor
- Zin can never be higher than 11.56K
- RE has a huge effect on the input impedance of the transistor base.

By the way, the trick of adding conductances (1/R) then taking the reciprocal to get a value for parallel resistors is something I use all the time. It's really, really handy.



>Matched source and load impedances are for maximum power transfer. This is almost never what you want for audio. You want maximum signal voltage transfer, not maximum power transfer.

Huh, i always thought that maximum signal power transfer was what is sought for.

Alright, so maximum signal voltage transfer it is.
[/quote]

[Hallmar]

Rin is hFE*re+ RE

... but there's a problem here. Rin = hFE*(re+RE)

So that's what i did wrong.

I'll keep that in mind.

Thanks R.G, always a big help.
(You too PRR)

[PRR]

Food for thought. Don't eat it all at one sitting.

> i always thought that maximum signal power transfer was what is sought for.

Rarely.

The power-line into my house is long, has (say{*}) 200V and 0.5 Ohm resistance. A "matched load" would be 0.5 Ohms. Total of Line and Load is 1 Ohm. At 200V, that is 200 Amps. I get 100V dropped in Load *and* 100V dropped in Line. I get 100V*200A= 20,000 Watts in my load (house) AND 20,000 Watts of _waste_heat_ in the power line. If I switch-on/off my 20KW load and leave-on one small 10W lamp, it gets 100V and 200V... either too-much or too-dim.

In fact I never pull more than (say) 30 Amps. This causes 0.5r*30A= 15V drop. 200V no-load becomes 185V with 30A load. 185V/30A= 6 Ohm load.

0.5 Ohm line, 6 Ohm load, is >10:1 "mis-match". However the power IN my house is "good", the heat in the power line is "bad".

The ideal would be 0.0000... Ohm in the power line. Or even 0.1 Ohm (2% drop). However this means wire 5 times bigger than what I have. My wire is 500 feet (160m) long. Replacing it with bigger wire costs more than a new car.

Transistor loudspeaker amplifier outputs are about 1 Ohm (before negative feedback). We load them in 8 Ohms. If we used 1 Ohm loads (or 8 Ohm transistors) there would be a LOT of heat in the transistors, and less sound in the loudspeaker. (Yes, car-radio amplifiers drive 1-Ohm loads, but use massive parallel arrays of fat transistors to get sub-Ohm internal impedance.)

When vacuum tubes were very costly, 1920s, we "matched". (Actually we matched plates to transformers versus desired treble response.) When transistors were very costly, 1950s, we "matched". Indeed some 1950s transistor-amp designs are hard to follow with modern understanding.

But by 1972, transistors were cheap. And very-very often we find that one or two transistors is not enough for a job, yet 2 or 3 transistors is more than we need.

And circuit calculations are *easier* when mis-match is assumed. Go back to my house. With load matched to line, 0.5 Ohms each, I get 100V 100A at the outlets. But if the big load is disconnected I get 200V. what do I get for a 10 ohm load ? For a 10 Amp load? For a 0.2 Ohm load? Where do I get light-bulbs which burn the same at 100V, 200V, or 57V, so I can see to turn on/off my big loads? With 6 Ohm load on 0.5 Ohm line, I can rely on having 200V to 184.6V, a "200V" lamp will burn well-enough, whether solo or with electric cooker and dryer and heater and welder running.

Know your source and load impedances. (Or the best-practice interface impedances, ~~1Meg for electric guitar.) If 1Meg in and 50K load, same voltage (unity voltage gain), you need a Current Gain of 1meg/50K= 20. Remember that bias networks may use-up much of a transistor's Beta; I start by assuming Beta=400 means the whole stage can have current-gain nearer 40. So this may be a 1-transistor job.

If the stage has voltage gain, you must multiply by that. So Gv=10 between 1Meg and 50K needs current gain of 200. You can not easily get that from one transistor. Two transistors offers MUCH more than Gi=200. Two Beta=400 may be Gi=40 each and 40*40= 1600 overall. Which is much-more than the required Gi=200. We can be "sloppy" about "matching" and still do the job. And the math can be more approximate, thus easier/faster.

{*} In fact I have 250V, split 125V+125V, nominal 100 Amp service, and 0.4 Ohms of line resistance. With 20 Ohm load I have 2% drop (5V) at 12 Amps. That's about where the house runs early evening (lamps and TV but no big loads). However when the well-pump, burner-blower, or microwave kick-in, the lights flicker a bit. At 75 Amps total load my voltage would be 220V total, 110V on "120V" lamps, which is pretty low.

[tubegeek]

Great thread! This is very helpful for me - I have a lot to learn about transistors, having sort of skipped over them in the evolution of gain from tube->transistor->op amp.

I have little to add except:

A 1M grid resistor is looking pretty good to me right now, and,

My dog is cuter than your dog, Hallmar!