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.
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.
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?
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.
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.
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
I also echo Berkeman's comments ... that old style cap marking is a pain and a trap for young players
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 ?
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
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
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
partly, as R2 does lift that node above the 0V rail, but that isn't the main reason, look higher up
There is a 1K resistor between the rail and the collector and a 150K plus the 1K between the rail and the cap.
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.
Yes .... so overall, what process are the resistor pair R2 and R3 and the single resistor R4 doing ?
Controlling current through Q1.
no ... but we will get to that part
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 ?
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.
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
no, they don't help achieve that, only C1 passes the incoming AC signal to Q1
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
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.
for a standard silicon diode or transistor it is approx. 0.7 V ( + - 0.05V )
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
Just to be clear, is it the voltage or the current that keeps the base to collector open?
I guess what I am asking is does the .7 volts provide a current that actually keeps it open.
lets now go back to that earlier Q
to save me typing an explanation, this page describes it well
scroll down to Class AB variations
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
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.
R1 sets the input impedance so the input impedance in this case is around 1k ... pretty common value for a line level input
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