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Gain is Vout/Vin => 695mV/7.07mV is about 100. or is it not gain? i am new in electronics, so before build this circuit on breadboard i wanted to ask here, because when i started to do DC analysis there were some troubles and it seems that it is not good circuit, but on multisim it works-_-
"That is a bit of a miracle using VCC=12V..." - i didnt understand what it means
thank you
Toffetan - are you aware that your resistance values (Rc and RE) are rather low? So you need a lot of milli-Amps as DC current through the transistor.
Why not in the lower kOhm range? Example: Ic=1mA (2mA) and Rc=6 kohms (3kOhms). The stabilizing resistor RE could be app. 5...10% of Rc.
Question: What about the load resistor (32 Ohms). Is that a given value?
Why not in the lower kOhm range? Example: Ic=1mA (2mA) and Rc=6 kohms (3kOhms). The stabilizing resistor RE could be app. 5...10% of Rc.
Question: What about the load resistor (32 Ohms). Is that a given value?
As stated above, Re should be like 1 kohm with Ic 1 mA, which is like an oldtimer rule-of-thumb.
And as BobK is stating, you cannot expect to drive such a heavy load this way. The Rc2 alone is 220 ohms and impedance of C3 at 1 kHz is about 160 ohms leaving less than 10 % of the Vcc swing to the load
And as BobK is stating, you cannot expect to drive such a heavy load this way. The Rc2 alone is 220 ohms and impedance of C3 at 1 kHz is about 160 ohms leaving less than 10 % of the Vcc swing to the load
Here's a very simple way to work out the voltage you will get across the 32R load (32R speaker).
The first thing you have to understand is this: The energy entering the speaker is not via the transistor but via the 220R load resistor.
The output transistor merely turns ON and "empties" the 1u electrolytic. If the 220R does not "fill" the electrolytic, the transistor will have nothing to remove.
So, the LOAD RESISTOR is most important.
It must be a low value to fill the electro and at the same time deliver current to the speaker.
When the output transistor turns OFF, the 220R and 32R form a voltage divider and 13% of the 12v supply will be passed to the speaker. The 1u will rapidly charge and this percentage will reduce during the first half of the cycle. So the maximum is just 13%.
When the transistor turns ON, the energy stored in the capacitor will come from the previous half-cycle, when the energy is being delivered at less than 13%.
So the overall delivery for the full cycle is less than 13%.
This is how you see a circuit "working" without any mathematics and without using any simulation software.
The first thing you have to understand is this: The energy entering the speaker is not via the transistor but via the 220R load resistor.
The output transistor merely turns ON and "empties" the 1u electrolytic. If the 220R does not "fill" the electrolytic, the transistor will have nothing to remove.
So, the LOAD RESISTOR is most important.
It must be a low value to fill the electro and at the same time deliver current to the speaker.
When the output transistor turns OFF, the 220R and 32R form a voltage divider and 13% of the 12v supply will be passed to the speaker. The 1u will rapidly charge and this percentage will reduce during the first half of the cycle. So the maximum is just 13%.
When the transistor turns ON, the energy stored in the capacitor will come from the previous half-cycle, when the energy is being delivered at less than 13%.
So the overall delivery for the full cycle is less than 13%.
This is how you see a circuit "working" without any mathematics and without using any simulation software.
As I understand the task, the amplifier should work in the quasi-linear range with a gain of 100.
Hence, the transistor is neither turned (switched) ON nor OFF.
"As I understand the task, the amplifier should work in the quasi-linear range with a gain of 100.
Hence, the transistor is neither turned (switched) ON nor OFF."
To start with, you are missing the whole point of my discussion.
The output transistor will be fully turned ON. The base bias resistors are trying to put 2v on the base and this is more than enough to turn the transistor ON.
The 10k will deliver about 1mA and if the gain of the transistor is 100, this will produce 100mA through the 220R load resistor. The voltage across the resistor will be 220 x 0.1 = 22v which is clearly more than the supply, so the output transistor will be FULLY TURNED ON.
As far as the output stage turning OFF, you will have to do some more involved calculations to determine if the 1u will turn the stage OFF.
This is not important because you can see the amplifier is so badly designed that it is not worth implementing.
As I said before, you have to look at a circuit and "see"what is happening in a general sense and do the accurate calculations after.
Hence, the transistor is neither turned (switched) ON nor OFF."
To start with, you are missing the whole point of my discussion.
The output transistor will be fully turned ON. The base bias resistors are trying to put 2v on the base and this is more than enough to turn the transistor ON.
The 10k will deliver about 1mA and if the gain of the transistor is 100, this will produce 100mA through the 220R load resistor. The voltage across the resistor will be 220 x 0.1 = 22v which is clearly more than the supply, so the output transistor will be FULLY TURNED ON.
As far as the output stage turning OFF, you will have to do some more involved calculations to determine if the 1u will turn the stage OFF.
This is not important because you can see the amplifier is so badly designed that it is not worth implementing.
As I said before, you have to look at a circuit and "see"what is happening in a general sense and do the accurate calculations after.
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