# BJT dynamics

#### roxi60

Joined Sep 2, 2018
73
Hello.
In enclosed file there's a BJT with a sine signal on base, through simply a capacitor.
I don't understand the reason why if I put no resistor from base to ground, I can't saturate the BJT, whereas I get it saturated if I put the resistor.
Why the BJT conduct better with the resistor from base to ground?
It seems that the base current is higher, but I don't understand physically why.

Thank you for any kind explanation.
Regards
Mod: extracted files from PDF/docx.E

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#### ericgibbs

Joined Jan 29, 2010
18,212
hi roxi,
I guess you know that a transistor Base to Emitter junction is a forward biassed diode.?

What exactly are you trying to do.?

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#### Wolframore

Joined Jan 21, 2019
2,593
Try another simulation with R2 between base and power.

#### roxi60

Joined Sep 2, 2018
73
Thank you Eric,Wolframore.

What Eric shows is quite the same signal on collector, but in Proteus or real circuit on breadboard is not so: in real circuit I have constant 8.1V on collector, when no resistance is between base and ground.
I would only know why, what does the added resistor do on the base current and voltage, that make the BJT conduct in one case and no conduction in the other?

Thank you.

#### Wolframore

Joined Jan 21, 2019
2,593
Answer this: what does the capacitor do?

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#### Ian0

Joined Aug 7, 2020
8,935
Shouldn't the capacitor charge up and stop it switching? It can charge through the B-E diode, but where's the discharge path?
It should do the transistor equivalent of blocking distortion.

#### Ian0

Joined Aug 7, 2020
8,935

#### crutschow

Joined Mar 14, 2008
33,323
what does the added resistor do on the base current and voltage, that make the BJT conduct in one case and no conduction in the other?
It provides a path for the capacitor to recover the charge it used when the base-emitter junction is forward biased at the peak of the positive input.
It should do the transistor equivalent of blocking distortion.
I agree that makes no sense.
What "distortion"?

#### roxi60

Joined Sep 2, 2018
73
Thank you all for your kind replies.
I just want to explain that the simple example I asked for is part of circuit you see enclosed.
Here a hand clapping make a sound that is amplified by a two stage (or not really a two stage) amplifier, the signal operates a monostable and jk flip flop and endly the power stage of the motor.
Here the sound amplified by the first stage Q1, is an impulsive signal that saturates the second BJT.
I know that the second stage is not biased, so can not be treated by a two stage amplifier, but I studied it alone and I was curious to understand why with only the capacitor C2 on the base, a sine signal can not saturate the Q2 transistor.
So the answer I'm looking for is:
With only the capacitor on the base (so not biased), the Vbe (or the Ib) is not enough to saturate the Q2 BJT because ... (fill in the blank).
Adding the resistor between the base and ground (or putting a probe of oscilloscope) the Vbe (or the Ib) is enough to saturate the Q2 BJT because ... (fill in the blank).
Sorry for the silly question, but I don't understand what happens in one case that lower the Ib.
Thank you again.

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#### Ian0

Joined Aug 7, 2020
8,935
I agree that makes no sense.
What "distortion"?
Blocking distortion - obviously something you've not heard of. Perhaps it's called something different in your part of the world.
Anyway - it's a thermionic valve phenomenon. (I can post the explanation from Merlin Blencowe's book if you want).
If there is a signal that overload a capacitively coupled amplification stage and forward biases the grid-cathode diode then the capacitor charges up rapidly, and cannot discharge through the grid-cathode resistor which is generally in the order of megohms. The charged-up capacitor stops the stage from working until it recovers, giving a few tens of milliseconds of missing signal.
I would expect the same thing to occur here especially in the stage without the base-emitter resistor. The capacitor will charge through the base-emitter diode to Vpeak-Vdiode. (about 0.3V), but cannot discharge. On the next cycle, the capacitor voltage will be subtracted from the signal voltage and there may then be insufficient voltage to turn the transistor on.

#### crutschow

Joined Mar 14, 2008
33,323
Blocking distortion - obviously something you've not heard of. Perhaps it's called something different in your part of the world.
No, I've never heard of that in my part of the world.
Why is that phenomenon called blocking distortion?
To me that means in stops distortion.

#### Ian0

Joined Aug 7, 2020
8,935
No, I've never heard of that in my part of the world.
Why is that phenomenon called blocking distortion?
To me that means in stops distortion.
I didn't name it!
Presumably because the voltage on the capacitor blocks the signal.
According to Mr. Blencowe it goes unnoticed until the amplifier input is overloaded - so it must be a phenomenon from the guitarist domain.

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#### crutschow

Joined Mar 14, 2008
33,323
I didn't name it!
Okay, I understand the meaning now.
It means it a type of distortion cause by the blocking of the signal by the capacitor and grid overdrive.

#### Audioguru again

Joined Oct 21, 2019
6,432
When the input to C2 goes positive then the base-emitter current of Q2 charges C2 to the peak voltage (minus the base-emitter voltage) of the signal. Then when the signal goes negative the base-emitter of Q2 stops conducting which allows C2 to hold its charge. When the signal goes positive again then since C2 is charged then Q2 does not get enough base voltage to conduct again.

When you add a resistor to ground at the base of Q2 then it discharges C2 each time the signal goes negative.

I could not read the parts numbers because the parts on the schematic are so far apart that the text became tiny. So I cropped the schematic and enlarged it but the text still cannot be seen properly.

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#### Ian0

Joined Aug 7, 2020
8,935
Pretty much what I said in post #6 (before I realised that "blocking distortion" wasn't a widely-known term) and post #11; so why do BOTH waveforms back in post #2 show the transistor turning on and off?

#### roxi60

Joined Sep 2, 2018
73
But if you open the previous with no enlarging it should also be readable.

To fully understand I'm simulating the first post simple circuit as a clamper circuit, since B_E junction is a diode.
When there's the resistor, as you said, there's the discharge of the capacitor and the average signal of the clamper increases, probably enough to make the transistor conduct.
I'm trying to demostrate this with formula, but it's not easy also to find in literature examples of clamper with and without load resistance.
Thanks.

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#### Wolframore

Joined Jan 21, 2019
2,593

#### ericgibbs

Joined Jan 29, 2010
18,212
hi roxi,
This is the front end of your project, uses an actual clap sound wav to test simulation.

E
Clap wav as a zip

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#### Wolframore

Joined Jan 21, 2019
2,593
uses an actual clap sound wav to test simulation.
how cool, always learning from you guys