Help me understand a data sheet for 2N5088

[FastEdTex]

1. I need help understanding what the terms Vceo Vcbo and Vebo relate to.
    What does the "o" in Vceo signify? Output? Offset?

2. the "green" highlighted text seems to be a conflict - 1uA to 50 mA but the Ic is 100 mA?
    Is the 1 uA to 50 mA the suggested working range and 100 mA the max before the 5088 fails?

3. The yellow highlighted Vebo - Shows max at 4.5 V.
    Attempting to correctly bias for a common collector, the Vcc is 15 V  and the Vc = Vcc
    R1=900 ohm R2=1K should leave 6.85 V for Vb (7.5 V - 0.65 V =6.85 V)
    What should R3 value be calculated on? 50 mA? 10 mA?
    Or is it not the correct transistor for a Vc=15V (if the Max Vbeo is 4.5 V)



Just trying to learn something, thanks for any help.

[antonis]

1. Open..!!
i.e. Vceo means Collector to Emitter voltage with Base open

2. Quite right..!!
(suggested working range to follow transistors characteristics..)

3. Vebo means the (almost) maximum REVERSE voltage that can stand the transistor..!!
(reverse in the meaning of usually the Base voltage is greater than Emitter's one, to make transistor work properly..)

The max Vebo (and not Vbeo) is 4.5 V..!! - you have to read correctly the order of symbols..

<What should R3 value be calculated on? 50 mA? 10 mA?>
R3 value shoud be calculated on WHATEVER current we want to..!! 
(taking care of not exceed the maximum current value - 100mA in this case)

Your calculations confuse me a bit, so let's take them from blank..

Vb is 7.894 (15x1k/1k9) so Ve is 7.244 (Vb -0.65).. R3 value sets the Emitter (and actually the Collector) current which should be 7.244/R3..
i.e. for a R3 of 1k Collector current should be 7.244mA - for 10k shoud be 724.4μA and so on..

In your scheme you have a humble Emitter follower with no voltage gain but with power gain..
(not easy to explain in a few sentences..)


edit1: Maybe I was a little "rude" so let's take it more "gently"..

We bias a trasistor to work properly - the way of biasing corresponds to what we want to do the transistor (and vice - versa..)
Base biasing with voltage divider leads to firm and steady bias with (ideally) no upsets with variations on temperature, device's gain (β or hfe) e.t.c.
So, Base voltage is the product of power supply voltage & lower resistor divided by the sum of two resistors..
(Voltage divider's current should be at least 10 times greater than Base to Emitter current so we set resistors values accordingly - we may leave out impendance (mis)matching & re for the moment..)
Base voltage sets the Emitter's voltage and toghether with Emitter resistor set the Emitter current (which is almost equal to Collector current)..
Collector resistor (if any) toghether with Collector current and Power supply voltage set the Collector voltage..
Any difference between Collector & Emitter voltages is "eaten" by Vce and this "lunch" toghether with Collector current set the transistor waisting Power..

edit2: I'm pretty sure that I've confused you more so plz don't pay much attention to edit1..

[FastEdTex]

Thank you for your post antonis!

3. Vebo means the (almost) maximum REVERSE voltage that can stand the transistor..!!
(reverse in the meaning of usually the Base voltage is greater than Emitter's one, to make transistor work properly..)

From the SIM the Vb = 7.19 V and the Ve = 6.44 V is a Vebo = 0.75 V?

This is what I created in MultiSIM BLUE from Mouser.

[PRR]

> This is what I created

A unity voltage gain follower, driving a 1Meg(!) load, and presenting a 500 Ohm load to what drives it.

Voltage gain is 1 (0.99).

Current gain is therefore about 500/1Meg, or 0.000,5.

This is not even an amplifier! Output is 1/2000th of input!

There is no audio use for this. If you have a 1Meg load, and want to connect to a source that "requires" a 500 Ohm load (a situation that rarely arises), put a 500 Ohm resistor in there.

> What should R3 value be calculated on?

On the LOAD.

BTW, in a real world we never have a 1Meg load. The wiring to a 1Meg resistor tends to be <250K capacitive impedance at the top of the audio band. (100pFd is 250K at 7KHz.) In Guitar Cord World we may have 50K loads. In studio and hi-fi work we may have 10K loads.

"R3" can be 1/2 to 1/10th of the Load.

So 50K load, figure 22K to 5K for R3.

Let us use 10K for example.

The bias R1||R2 network may often be "square root of hFE" higher than R3. hFE of this part is at least 400? SqRt(400) is 20. R1||R2 should not be larger than 20*10K. Or 200K.

The fussiness of "1K or 1.07K" is not warranted. "About half" is fine. If it comes out 0.4 that is also fine. If it clips when a little off-half, making it !half! won't bring happiness, you need less signal or more supply.

So each of R1 R2 can be 400K, say 390K standard value.

Assuming "half", about 7V across the 10K R3 is 0.7mA. *NOW* look for a good 0.7mA part. Yes, the *headline* on the 2N5088 boasts that it is a good choice in a wide range including 0.7mA.

And what it is saying is that it will be a fine *amplifier* up to 50mA (which we do NOT need here). And it won't blow-up at 100mA, but they don't promise it will amplify good when worked so hard. This expands sales by suggesting the one part for your 0.1mA-10mA amplifiers +and+ your 100mA relay driver (poor "amplification" may be tolerable).

The Vce-Vcb is the maximum it will stand without blow-up. In resistor-loaded stages, this is essentially your battery. (Transformer loading can kick 2X batt voltage and more.) Power transistors will have a variety of specific cases: Base open, Base shorted, resistor on Base, Base back-biased. The little parts also have different limits depending exactly how you support them in over-voltage, but this is NOT something small-amplifier designers tend to do. (Or if they do, the foundry has other parts with higher more-complete ratings at higher prices.)

Vebo-- A happy audio amplifier never has reverse Base voltage. In real world we have too-big signals, turn-on/off transients, etc. We should design so the reverse Base voltage is never over 4.5V. Actually the majority of Si types are doped to break-down at 6V-7V, so the "4.5V" includes some margin. Reverse breakdown is not fatal if current thus power are not excessive. (It may degrade hiss and offset.) It happens but is not something you need to worry about at this stage of the path.

[R.G.]

Vebo-- A happy audio amplifier never has reverse Base voltage. In real world we have too-big signals, turn-on/off transients, etc. We should design so the reverse Base voltage is never over 4.5V. Actually the majority of Si types are doped to break-down at 6V-7V, so the "4.5V" includes some margin. Reverse breakdown is not fatal if current thus power are not excessive. (It may degrade hiss and offset.) It happens but is not something you need to worry about at this stage of the path.

Any circuit with a modestly large cap on the emitter can sometimes dump its voltage back through the emitter to a suddenly-grounded base. That usually only happens with input transistors, but the reference that gave me my fear of reverse-breaking base-emitters pointed out several circumstances that can cause similar things in not-too-unusual circuits when the inputs are shorted, the power supply shorted, or transients run amok.

The claim was that the damage to the base-emitter was tiny and proportional to the energy dumped into it, and that over time and imperceptibly it can build up.

It also pointed out that the cure for ever needing to worry about this is to put a small signal diode reverse biased across the base-emitter, so the voltage there can never exceed a reverse volt or so.

[FastEdTex]

Thanks for the feedback, and not just telling me put this value here and that value there.
Thing is I'm such a neophyte to this, I have probably mislead the gurus here.
Here is what I'm trying to understand. 2 TS (OD) schematics, one just has a volume pot to output and the other a 2N5088.
What does the 5088 add to the effect? Amplification? Output buffering?

[antonis]

3. Vebo means the (almost) maximum REVERSE voltage that can stand the transistor..!!
(reverse in the meaning of usually the Base voltage is greater than Emitter's one, to make transistor work properly..)

From the SIM the Vb = 7.19 V and the Ve = 6.44 V is a Vebo = 0.75 V?

Plz pay attention to the order of the lower case symbols..!!
Vb is 0.75V greater than Ve..
If you had a Veb = 0.75V you should have a non-working transistor..!!

P.S.
Symbols with 3 lower letters are rarely used in practice..

[antonis]

What does the 5088 add to the effect? Amplification? Output buffering?

Strictly speaking, the second..!!

"Amplification" by itself doesn't mean anything - it needs at least a parameter before it, like Voltage, Current, Power e.t.c.
(and it doesn't need to be always positive - in that case we prefer the term "Attenuation"..)

In your 2nd scheme, Q2 serves as an impedance matcher (mismatcher, actually..) between Vol pot and OUT (Input impedance of whatever is connected to..)

Considering C9 as short-circuit (which actually stands at high frequencies) you have an Emitter load of 2k5 (R13//R15 - you should ignore R14 because of it's low value) in parallel with whatever impedance comes next..

For calculating Current gain (Voltage gain is shlightly less than unity, maily because of intrinsic emitter resistance) you have to read what PRR told you above..
(and - maybe - you need a more in depth understanding of the "famous" Voltage dividing effect..)

[PRR]

> one just has a volume pot to output and the other a 2N5088.

Look at the one with output at the Vol pot wiper.

One end of the pot is ground. The other end is held very-solid by an op-amp. Neglecting C9 R9, the wiper impedance is zero at top, zero at bottom, but 50K each way for 25K at middle. The output impedance *varies*.

IMHO, 25K is not a large impedance for guitar-cord work. I hardly care if the impedance wanders zero to 25K.

Assuming the load is 50K or more, the 25K midpoint means a small change of pot taper, not very important.

Now look at the buffered job. The same 100K pot (25K worst-case) is loaded with R12 500K and Q2 which here runs near 1,000K at Base. So 330K load on 0-25K pot, pot impedance and its variation is "more insignificant" than before.

The output of the transistor is near-constant at about 60 Ohms (Shockley's Law). It will not *drive* 60r load, that is just what small signals "see". However the pot resistance "reflects-though" Q2 as about 0-25K/hFE, or about 0-200 Ohms. And there is R14 100r. So the output is 160 to 360 Ohms. Say 100X smaller than un-buffered.

Also note that I tole you to take 10K as a pencil-trial load resistor, and that's just what we find here.

[antonis]

<off topic ON>

If you ever try to precisely build the second circuit, you may find difficult to get a W taper (S curve) pot..

(off topic OFF>

[FastEdTex]

Still attempting to learn.
1st Schematic. Boost attempt.


https://postimg.org/image/pxwbkbfrr/

[duck_arse]

fasted - I/m not sure I have been paying attention closely enough, but why do you have 15k for R7 and R10? why so low a value? also, R10 is not needed; you've DC coupled (Part 1) to (Part 2).

[antonis]

I don't like single resistor biasing..

I should replace R5 & R13 with Voltage dividers of, say, 1M/1M2 and use high gain transistors..
(IMHO, halving the Emitter followers impedance shouldn't be a problem..)

Or pass to a FET datasheet..