Transistor based AC amplifier - how to build with limited details.

[AnalogKid]

To start with, the two 13v zeners will have a "dead spot" of 2v because they add up to 26v for the 24v supply.

Correct. Without that "dead spot", the circuit cannot function.

On top of this the base-emitter junctions at the top and bottom will add another 1v.

It is more like 1.2 V (0.6 V Vcesat each).

This means the input voltage will not have any effect when it is rising and falling by 3v because this is the "gap" between turning on the bottom zener then the voltage falling and turning on the top zener.

Nope. First, the input is 5 Vpp; it says so right on the drawing. The input must be greater than the dead band to prevent cross-conduction in the output stage.

Normally "biasing diodes" are just at the point of turning ON, so the rising or falling signal will pass through the diode, even when it is a few millivolts.

That is true for the circuits in posts #11 and #12, but this circuit is not a linear output stage so none of that applies.

Any voltage below 3v will not be transferred

Correct. The circuit is designed for a 5 Vpp input signal. It says so right on the drawing.

a 5v signal will have some attenuation in the 1k as well as the 2u2.

The 1K resistor is there to limit the base current in the output stage transistors. As you can see from the schematic, they are saturated switches, not linear amplifiers. Once the resistor is selected, the capacitor is calculated based on the minimum frequency or maximum 1/2 cycle pulse width of the input signal.

Most 5v signals from a digital source are less than 5v as the output of most chips is 4.5v max and 0.5v min. This gives a 4v p-p signal and the other two losses makes this circuit very unreliable.

In post #1 the TS says the signal source is a function generator, not a logic chip, and in post #3 he says he can adjust the output as needed. He mentions +/-5Vpp as an example, so once again your analysis does not apply to this thread.

What you still do not get is that this is not a linear amplifier circuit. It is not a non-inverting complimentary emitter follower. It is two complimentary inverting saturated switches. The biasing diodes are selected to guarantee that they *never* are both at the point of turning on at the same time; that would be a disaster for the output transistors.

When the input is 0 V, C1 charges up through Q1, D1, and R1, until it has approx. 10.4 V across it (24 V - 13 V - 0.6 V). When the input changes state and starts to move toward 5 V, the other end of C1 moves up. This almost immediately turns off D1 and Q1. As the input continues to rise toward 5 V it crosses 3.2 V, making the other end of the capacitor 13.6 V, and D2 and Q2 start to turn on. By the time the input is at 5 V, Q2 is fully saturated. Now the voltage across C1 is only 8.6 V. The input changes back to 0 V, pulling the capacitor down to 3.6 V, and it starts charging up through D1 and Q1 again.

ak

 

[Colin]

The OP wants a DC amplifier not an AC amplifier.

Look at the type of output he is wanting.

 

[AnalogKid]

Excellent, because this is not an AC amplifier. This is a level-shifting pulse driver with a relatively low input impedance. The input is a 5 Vpp square wave and the output is a 24 Vpp square wave. The centerpoint of the input is 2.5 V and the centerpoint of the output is 12 V. That is a 9.5 V difference. Rather than have 17 components or an opamp DC servo or whatever making that happen, one capacitor charges up to an average value of 9.5 V and it's done.

ak

 

[Colin]

"Excellent, because this is not an AC amplifier. "

You have a capacitor on the input. It is an AC amplifier.

"Frequency this thing will operate in - 1-15 Hz"

What do you think the output is going to be like at one hertz?

 

[AnalogKid]

""Frequency this thing will operate in - 1-15 Hz"
What do you think the output is going to be like at one hertz?

Yeah, I was wondering if anyone was going to catch that. Took long enough. C1 will have to increase. Also, 300 mA is on the edge for a 4401/03 but I didn't know the current back when I posted the circuit.

ak

 

[Colin]

I said, back in post 36, "will this work?" and you said yes.
You have 2u2 and 1k. The cap will only charge to 2.5v and below 2.5v the circuit turns OFF.
What do you think the time-constant for 2u2 and 1k is?
How do you say it will work?
You are just giving me fodder for my SPOT THE MISTAKE pages.

 

[throbscottle]

I am asking too much probably, but among my problems is the research of components in stores (as they are German here, and I do not know the language). So once I know the exact part name - I can find it and order - any part, any price and quantity. But going online and researching those "Verstärker"s and "Kohleschicht-Widerstand"s is painful

So thanks again for all help!

Why not use one of the many online translation services for this site? For example, with google translate (using the "translate this" firefox extension) I can get: https://translate.google.com/translate?ie=UTF-8&tl=en&u=https://www.conrad.de/de/elektromagnet-1000-n-24-vdc-54-w-gto-50-05000-24vdc-503669.html
It's not perfect but it's a lot easier, and it will translate into whatever language you prefer
(one thing I've found to look out for though is alternative meanings - for example it always translates the french word "fils" as it's more common meaning of "son" when it actually means "wire" in the original text. Confusing)

 

[MikeMl]

Colin AnalogKid

Since the TS wants to deflect a magnetic particle, I'm thinking that TS will be much happier with a DC-coupled linear amplifier like I showed in post #35... , the title of TS's post notwithstanding.

 

[Colin]

“When the input is 0 V, C1 charges up through Q1, D1, and R1, until it has approx. 10.4 V across it (24 V - 13 V - 0.6 V). When the input changes state and starts to move toward 5 V, the other end of C1 moves up. This almost immediately turns off D1 and Q1. As the input continues to rise toward 5 V it crosses 3.2 V, making the other end of the capacitor 13.6 V, and D2 and Q2 start to turn on. By the time the input is at 5 V, Q2 is fully saturated. Now the voltage across C1 is only 8.6 V. The input changes back to 0 V, pulling the capacitor down to 3.6 V, and it starts charging up through D1 and Q1 again.”


The last sentence is incorrect.
When the input changes to 0v, the capacitor has 8.6v across it and this means it can only charge to 10.4v and thus only 1.8v is replaced in the capacitor.

The charging will be very quick because the charging resistor is 1k.
When the input signal is 1Hz, the circuit will react in less than a few milliseconds and for the rest of the time the output will be zero.
Thus the circuit is not designed for low frequencies.

 

[AnalogKid]

When the input changes to 0v, the capacitor has 8.6v across it and this means it can only charge to 10.4v and thus only 1.8v is replaced in the capacitor.

That is correct. I'm glad you are understanding how this circuit works. 1.8 V is not much operating margin, a problem caused by the relatively small input voltage swing and the two Vbe drops. (the circuit was developed originally for a 10 Vpp input). I already agreed that the circuit as posted needs adjustments for this specific application. It was presented as an alternative circuit concept, not a finished product. A version with adjusted R and C values looks very good in simulation, but the capacitor size is approaching an impractical value.

ak

 

[Colin]

"That is correct. I'm glad you are understanding how this circuit works."

And I am glad you don't understand how the circuit works because you have given me lots of fodder for my SPOT THE MISTAKE section where I show how "technical engineers" don't know the first thing about designing a circuit.
It's already up to page 26 with hundreds of mistakes from Indian "Professors" to USA "teachers" and simple hobbyists who put things on the web without actually testing them beforehand.
Th circuit in question "ran out of steam" before it started with an input of 5v and no being able to anything with the input voltage.
You have to get off the "crutch" of relying an a simulation program to do the analysis for you.
I could see at the very beginning the circuit had flaws and a few calculations showed the 5v was "eaten up" by the gap between the zener voltages and the time-constant for the 2u2 was very short when a 1k is included.

 

[AnalogKid]

I show how "technical engineers" don't know the first thing about designing a circuit. It's already up to page 26 with hundreds of mistakes from Indian "Professors" to USA "teachers" and simple hobbyists ...

So, how does that make you feel?

 

[Colin]

"So, how does that make you feel?"

Look . . . .I get 6,000 visitors each day and over 22 million have visited my site. One extra inclusion is not going to make any difference.
I get hundreds of emails to say the reader has learnt more from my website than any of the courses he has taken.
I have closed down quite a number of Indian websites by Indian "professors" who don't know a thing about how a circuit works. They can all spout out equations and theories, but when it comes to actually designing a circuit, they make a fool of themselves.
The same with articles in magazines and text books. They are riddled with mistakes
It's obvious, that even in this Forum, the faults in the circuit under discussion have gone un-noticed and it just helps the average reader understand the finer points of designing . . . . because you will never find the answers anywhere else.

Look at the number of mistakes you made in this simple circuit and you will see why it is necessary to point them out.

I know 0.001% of electronics, but I still get 6,000 readers a day. I am only here to point out the glaring mistakes and the readers are absolutely amazed at how to go about understanding how a circuit works.
I can "see" a circuit working in my head and the Indian "Professor" said: "What absolute rubbish!" And yet he closed his website 2 weeks later.
How do you think I spotted the mistakes in this circuit if I didn't see it working and saw how the input voltage was gradually removed by the components.
And the time-constant was so short the output would never react, as well as the circuit being AC coupled.

 

[DerStrom8]

"So, how does that make you feel?"

Look . . . .I get 6,000 visitors each day and over 22 million have visited my site. One extra inclusion is not going to make any difference.
I get hundreds of emails to say the reader has learnt more from my website than any of the courses he has taken.
I have closed down quite a number of Indian websites by Indian "professors" who don't know a thing about how a circuit works. They can all spout out equations and theories, but when it comes to actually designing a circuit, they make a fool of themselves.
The same with articles in magazines and text books. They are riddled with mistakes
It's obvious, that even in this Forum, the faults in the circuit under discussion have gone un-noticed and it just helps the average reader understand the finer points of designing . . . . because you will never find the answers anywhere else.

Look at the number of mistakes you made in this simple circuit and you will see why it is necessary to point them out.

I know 0.001% of electronics, but I still get 6,000 readers a day. I am only here to point out the glaring mistakes and the readers are absolutely amazed at how to go about understanding how a circuit works.
I can "see" a circuit working in my head and the Indian "Professor" said: "What absolute rubbish!" And yet he closed his website 2 weeks later.
How do you think I spotted the mistakes in this circuit if I didn't see it working and saw how the input voltage was gradually removed by the components.
And the time-constant was so short the output would never react, as well as the circuit being AC coupled.

Hmm, since you seem to rely so much on numbers, and are eager to "spot others' mistakes", here's some I've found on my own.

It looks to me like you barely get 5000 visitors a day, and that's on a good month. The average visitor spends about two minutes on your page and then leaves. You have a 74% bounce rate, whereas the average bounce rate for 2015 was only 40%. Most of your viewer base seems to be students whom you force feed your content, and who are too naive and don't have the knowledge to look at your page with actual criticism. You are hypocritical--You criticize others for taking down their circuits if you spot a mistake, but you do the same thing. Furthermore, some of "your" circuits were stolen from other people, either here at ETO or on other sites, and put on your site without the explicit permission from the actual creator (there is at least one thread here at ETO discussing this very issue).

So before you start pointing your fingers at people and making yourself out to be so much smarter than everyone else, remember that everyone makes mistakes, and that includes you. Being able to point out others' mistakes does not make you any more reputable than others. In fact, it can make you even less reputable depending on how you go about it, and that seems to be the problem with you.

I suggest you knock yourself down a few pegs before you make yourself look more like a fool than you already have. Politely responding to threads with legitimate, helpful comments is fine. That's what this site is for. But once you start criticizing the posters, threatening them and bringing your lousy website into it, you're treading on thin ice.

 

[Colin]

DerStrom8

I notice you didn't come onto this discussion with any comments about the circuit, so that answers all my questions.

What does 323,000 hits per month work out to?
One page alone has received 724,581 views in 18 months.
It's not you I am aiming at. I am only aiming at those who want to learn.

 

[Colin]

It's funny, after selling over 300,000 kits, 750,000 magazines and books, you are the first to let everyone know I am a FRAUD.
And yes, I am a total fraud. Everything on my website has been copied from other sites. I cannot design anything.
And I know nothing about electronics. All the millions I stole from the customers will be sent back next week.

 

[DerStrom8]

My numbers came from a variety of sources such as Alexa, SEMrush, TrafficEstimate, and several others. They all gave very similar metrics.

My concern is not what you do with your site or your "business". My concern is the rubbish you bring to this site. Your attitude is despicable. That is why I posted to this thread -- To warn you to take it down a few notches before you get yourself into trouble.

 

[Colin]

"My numbers came from a variety of sources such as Alexa, SEMrush, TrafficEstimate, and several others. They all gave very similar metrics. "

How stupid of you to quote vague numbers when I have stats linked directly to each page on my site and see numbers updated every minute.
Get a life.

 

[DerStrom8]

Furthermore, page visits aren't everything. Considering you have a 74% bounce rate and 2 minute time average, your metrics are still pretty pathetic.

If you can post screenshots proving that you have 6000 posts a day consistently, I will take back what I said about page views, but I will not take back what I said about the other metrics.

But this thread is not about your website. That is my point. Feel free to post to the forums, but leave your website out of it.

And for the last time, watch your attitude.

 

[Colin]

I have hundreds of pages ranging from 276 per day to a few per day and average time is 5m 45s
Average visit length and 8m 49s Average visit length so I don't know where you are getting your figures from.
Show me your website !!!!!!!!!!
Show me your books ??????
So far you have just wasted time, and added nothing.
At least I have finished up with 26 pages of mistakes.
The fact that thousands are buying the CD whereas everything on the CD is already on the site shows the enthusiasm of some readers.
As I said, I am not aiming at your requirements.