Does C3 have to be a electrolytic cap?

  • Thread starter Planobilly
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  • #1
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4-transistor-class-ab-amplifier.png
 

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  • #2
berkeman
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Not that I can see. Sometimes folks use that same capacitor symbol for unpolarized caps, which I dislike. You should be able to use a ceramic cap there instead.
 
  • #3
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Thanks, I sort of assumed so...but better to ask...lol
I don't know much about transistor amp circuits so this is just an experiment.

Billy
 
  • #4
berkeman
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Do you understand why the polarities of the electrolytic caps in this circuit are the way they are? How would you determine what the polarity of the voltage is across those component locations where you use an electrolytic cap? The power supply cap is obvious, but what about the ones in series with the AC signal? :smile:
 
  • #5
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Give me a min. to look

The curved line on the cap symbol is positive.
All of the caps, C1 through C5 connect their positive side to the +5 Volt rail.
Also C1 is connected to the base of the 2222 which is positive and so is the collector.
C1 is preventing DC from going back to the input I assume.

Am I missing anything??

I would measure with a DMM or a scope to ground...I assume that is what you are asking.
 
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  • #6
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Perhaps if you are someone has the time we could go through each component to determine just what it does and why it is there. That would be a big help for me.

Thanks,

Billy
 
  • #7
davenn
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The curved line on the cap symbol is positive.

no, the curved side is the negative side
Note the polarised caps ( electrolytics) are marked with a +
the ones that are not marked with a + ARE NOT polarised :smile:

I also echo Berkeman's comments ... that old style cap marking is a pain and a trap for young players

Also C1 is connected to the base of the 2222 which is positive and so is the collector.

yes, but can you see what is making the base side of the transistor and C1 more positive than the other (left ) side of C1 ?

C1 is preventing DC from going back to the input I assume.

yes ... primary purposes are interstage AC (signal) coupling and DC blocking, as is C2. C4 is DC blocking as we don't want a DC offset voltage appearing across a speaker

would measure with a DMM or a scope to ground...I assume that is what you are asking.

you could, but a bit of basic circuit analysis will tell you what points/nodes are more positive than others.
the DMM or scope will give you the actual amount

Dave
 
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  • #8
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Hi Dave...sorry..yes of course I had it backwards on the caps as is easy to see on the schematic.
C1 is more positive because of R2...I think
 
  • #9
davenn
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C1 is more positive because of R2...I think

partly, as R2 does lift that node above the 0V rail, but that isn't the main reason, look higher up
 
  • #10
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There is a 1K resistor between the rail and the collector and a 150K plus the 1K between the rail and the cap.
 
  • #11
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Are the two diodes used to prevent damage from hooking up the power backwards? They also produce a voltage drop of around 1.4 v at the collector of Q4 I assume.
 
  • #12
davenn
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There is a 1K resistor between the rail and the collector and a 150K plus the 1K between the rail and the cap.

Yes .... so overall, what process are the resistor pair R2 and R3 and the single resistor R4 doing ?
 
  • #13
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Controlling current through Q1.
 
  • #14
davenn
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Are the two diodes used to prevent damage from hooking up the power backwards?

no ... but we will get to that part
 
  • #15
davenn
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Controlling current through Q1.

the single resistor R4 is doing that as well as a voltage dropper

but look at the configuration of R2 and R3 ... do you recognise that config ?
 
  • #16
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No, I don't recognise the config but I assume the following.
We need to get AC to the base of Q1 and R2 and R3 makes it go there.
We need DC to get to the base of Q1 and C1 ,R2, and R3 makes it go there.

%2F%2Fwww.circuitstoday.com%2Fwp-content%2Fuploads%2F2011%2F03%2F4-transistor-class-ab-amplifier.png
 
  • #17
davenn
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No, I don't recognise the config but I assume the following.

OK, no prob ... .it's called a voltage divider ... sound familiar ?

put 2 resistors between + and minus of a supply source and the voltage at the centre node is proportional to the value of the 2 resistors

We need to get AC to the base of Q1 and R2 and R3 makes it go there.

no, they don't help achieve that, only C1 passes the incoming AC signal to Q1

We need DC to get to the base of Q1 and C1 ,R2, and R3 makes it go there.

Yes, but for a very specific purpose
As said R2 and R3 create a voltage divider with the base of Q1 being at the centre node
This biases the transistor on and keeps it within its linear range to help avoid distortion of the signal
You also need to consider that the AC voltage at the node of C1 and Q1 base may not be high enough at
times in its cycle to allow Q1's base to emitter path to go into conduction ... providing that bias voltage overcomes that problem

D
 
  • #18
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Ok...so a NPN transistors need a certain positive voltage at it's base to conduct. Correct? What would be a nominal voltage?

As I have been only working with tubes I need to get a picture in my head of how transistors are different. The base on a transistor is like the grid on a tube and could be biased positive or negative depending on the type of transistor????

Be back in two min.

Back now
 
  • #19
davenn
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Ok...so a NPN transistors need a certain positive voltage at it's base to conduct. Correct? What would be a nominal voltage?

for a standard silicon diode or transistor it is approx. 0.7 V ( + - 0.05V )

The base on a transistor is like the grid on a tube and could be biased positive or negative depending on the type of transistor????

the principle is similar ... tubes are basically voltage operating devices where transistors are current driven devices

That sounds a contradiction in the face of my explanation of the 0.7V etc ... sometimes it's still easier to look at it that way

for a tube, varying the voltage on the grid increases or decreases the flow of current between the cathode and the plate

in a transistor varying the current level between the base and the emitter ( NPN transistor) and once the base-emitter path reaches that 0.7V conduction point
that small varying current between B and E controls the larger current in the path between Collector and Emitter



Dave
 
  • #20
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Just to be clear, is it the voltage or the current that keeps the base to collector open?
 
  • #21
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I guess what I am asking is does the .7 volts provide a current that actually keeps it open.
 
  • #22
davenn
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Are the two diodes used to prevent damage from hooking up the power backwards? They also produce a voltage drop of around 1.4 v at the collector of Q4 I assume.

lets now go back to that earlier Q

to save me typing an explanation, this page describes it well

http://www.sentex.ca/~mec1995/tutorial/xtor/xtor4/xtor4.html [Broken]

scroll down to Class AB variations

Dave
 
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  • #23
davenn
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Just to be clear, is it the voltage or the current that keeps the base to collector open?

once the voltage across the PN (B-E) junction reaches 0.7 V or higher, current will flow.
A side note, you don't want that voltage to go too high else the transistor goes into saturation and causes other problems
I wont go into the fine details as I'm not a semiconductor expert LOL

now you may see why in a AC - DC rectifier circuit, we loose 0.7V for each of the diodes that are in conduction mode
So the first 0.7V is lost and for a full wave rectifier dual diodes or a bridge we have a total of 1.4V lost from the final DC output voltage
 
  • #24
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I will read the tutorial and come back and ask question for what I don't understand.

One other question is what is the purpose of R1 and how would one go about changing the input impedance.
 
  • #25
davenn
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One other question is what is the purpose of R1 and how would one go about changing the input impedance.

R1 sets the input impedance so the input impedance in this case is around 1k ... pretty common value for a line level input

D
 
  • #26
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Thanks Dave for all the training.

I assumed that R1 controlled the impedance. So if I wanted to input a guitar signal is there any reason I could not use a 1Meg resistor in this transistor circuit? I normally see the 1M resistor go to ground in a guitar amp..something like this.

imgext.php?u=http%3A%2F%2Fwww.tuberadios.com%2Ftemp%2Faa4schematic.gif
 
  • #27
davenn
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I assumed that R1 controlled the impedance. So if I wanted to input a guitar signal is there any reason I could not use a 1Meg resistor in this transistor circuit? I normally see the 1M resistor go to ground in a guitar amp..something like this.

Haven't played with tubes for years ... used to restore old valve radios in another lifetime and another country

from memory the grid inputs are pretty high impedance, so some impedance matching is called for between the input and the grid
or from one stage to another within the amp ... eg also look at the high values between V1 and V2

not sure offhand what the output impedance of a guitar is ... do a google for both of us and see what you find :wink:

bi-polar transistors, on the other hand, are much lower impedance ( FETs are high impedance and are treated more like tubes)



Dave
 
  • #28
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Pickup impedance for guitars is around 6K to 17K. Normal input impedance for tube guitar amps is somewhere around 1M.

I also assume going from the guitar to the input to a 12AX7 pre amp tube for example one would be concerned about getting as much voltage as possible. I also assume that a big mismatch like 17K to 1M would have that effect. Having said that I don't know how to prove it.

I assume about 1000 ohms output impedance for a CD player
 
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  • #29
davenn
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Pickup impedance for guitars is around 6K to 17K. Normal input impedance for tube guitar amps is somewhere around 1M.

OK so the 68k and the 1M will be doing some impedance matching between the ~ 10k to the several M Ohms of the grid

here are some comments from wiki.....

Impedances[edit]
As cables between line output and line input are generally extremely short compared to the audio signal wavelength in the cable, transmission line effects can be disregarded and impedance matching need not be used. Instead, line level circuits use the impedance bridging principle, in which a low impedance output drives a high impedance input. A typical line out connection has an output impedance from 100 to 600 Ω, with lower values being more common in newer equipment. Line inputs present a much higher impedance, typically 10 kΩ or more.[4]

The two impedances form a voltage divider with a shunt element that is large relative to the size of the series element, which ensures that little of the signal is shunted to ground and that current requirements are minimized. Most of the voltage asserted by the output appears across the input impedance and almost none of the voltage is dropped across the output.[4] The line input acts similarly to a high impedance voltmeter or oscilloscope input, measuring the voltage asserted by the output while drawing minimal current (and hence minimal power) from the source. The high impedance of the line in circuit does not load down the output of the source device.

These are voltage signals (as opposed to current signals) and it is the signal information (voltage) that is desired, not power to drive a transducer, such as a speaker or antenna. The actual information that is exchanged between the devices is the variance in voltage; it is this alternating voltage signal that conveys the information, making the current irrelevant.

Line out[edit]
28px-Line_out_symbol.svg.png
42px-Line_waves03-0-out.png
32px-Line_waves03-1-out.png
35px-Line_circle_out.png
Line-out symbol. PC Guide color Lime green.

Line outputs usually present a source impedance of from 100 to 600 ohms. The voltage can reach 2 volts peak-to-peak with levels referenced to −10 dBV (300 mV) at 10 kΩ. The frequency response of most modern equipment is advertised as at least 20 Hz to 20 kHz, which corresponds to the range of human hearing. Line outputs are intended to drive a load impedance of 10,000 ohms; with only a few volts, this requires only minimal current.


Dave
 
  • #30
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If I understand the wiki comments, they are only concerned with source impedance of 100 to 600 ohms which is the normal impedance of a CD player for example.

With a guitar as the source we have 6000 to 17000 ohms impedance and a signal from say 50hz to 5000hz. I think 5000hz is about the 3rd or 4th harmonic of the highest fundamental the guitar will produce....I think....it is getting a bit late...lol

I also ASSUME the transistors in the circuit we have been looking at are concerned with voltage amplification. If we drive power transistors with a circuit the power transistors would be concerned with current amplification.

If I have this general concept wrong please let me know.

Again, thanks for all the instructions.

Good night,

Billy
 

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