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help bulding BJT amp

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hello everyone
am trying to design a BJT amp ...
i came out with 2 circuits and i used multisim to simulate it
this are my results

my input is 50mv.

when i tried this cicuit on the bread board it didn't work ...
can any one point to me why is that??

hi,
The resistor values are very high, how did you calculate the biasing resistors and collector loads.?
 
using this example
**broken link removed**

i chose the Rc randomly then continue the equations...

hi,
If you relook at the final circuit in your link, you have chosen resistor values almost ten times higher than the circuit suggests.
Your version may work in simulation but I would say the design would be problematic in a real circuit.

I would suggest you re-sim your circuit using lower value resistors as shown in your link.:)
 
Like I said on the other website the value of the emitter bypass capacitors are very low and will cause low gain at low audio frequencies and high distortion. The gain is maximum at about 200kHz.
Use 4.7uF emitter bypass capacitors for max gain at all audio frequencies.
 
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Like I said on the other website the value of the emitter bypass resistors are very low and will cause low gain at low audio frequencies and high distortion. The gain is maximum at about 200kHz.
Use 4.7uF emitter bypass capacitors for max gain at all audio frequencies.

hi agu,
Is this a typo.?
 
Like I said on the other website the value of the emitter bypass capacitors are very low and will cause low gain at low audio frequencies and high distortion. The gain is maximum at about 200kHz.
Use 4.7uF emitter bypass capacitors for max gain at all audio frequencies.

so do i have to change the capacitors only?or do i hv to recalculate the whole circuit?
 
hi,
If you relook at the final circuit in your link, you have chosen resistor values almost ten times higher than the circuit suggests.
Your version may work in simulation but I would say the design would be problematic in a real circuit.

I would suggest you re-sim your circuit using lower value resistors as shown in your link.:)

but if use small value resistors the gain is very low and i need at least a gain of 100
 
A gain of 100 is quite a lot to ask of a single common emitter amplifier.
 
I reduced the values of the base bias resistors and increased the value of the capacitors.
Before your second circuit had a voltage gain of only 35.4 at 3kHz.
Mine has a voltage gain of 6667.
 

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I reduced the values of the base bias resistors and increased the value of the capacitors.
Before your second circuit had a voltage gain of only 35.4 at 3kHz.
Mine has a voltage gain of 6667.
thnx alot for helpin me.
i simulated your circuit and it worked but i got few questions
1) your input is very low 0.6mv ,while i use input of 50-150mv and when i apply this amount your upgraded circuit i get clipped output wave.
2) i want to understand how you got the new resistors values ,becouse folowing those equations (**broken link removed**) i cant get those values .
 
i meant 100 from the sum of the 2 stages

Sorry, I didn't know how I missed that.:eek:

You only need a gain of 100?

Why not get rid of the bypass capacitors and select resistor values that will give a gain of 10 per stage?
 

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Two transistors with a total voltage gain of 100 each have a voltage gain of only 10.
My circuit showed that its gain is very high and the low frequencies gain in your circuit is very low.
The circuit posted by Hero has its base bias resistor values too high.
 
The circuit posted by Hero has its base bias resistor values too high.
Why do you say that?

The simulator said it was fine, although I accept that simulators can't predict everything.

Reducing the resistor values won't be good because it will reduce the input impedance which might not be acceptable.
 
Why do you say that?

The article says that the divider current should be 10 times the typical base current. The total resistance of your divider is 253k.
Their current is only 59.3uA
They create an unloaded voltage of 1.96V for the base of the transistor.
The transistor loads down the voltage divider so its voltage will be about 1.67V and the emitter of the transistor will be about 1.02V and its current will be 3.78mA.
The typical curent gain of the transistor is about 200 so its base current is about 18.9uA. The values of the divider should have a total resistance of 79.4k ohms, not 253k ohms.
 
The article says that the divider current should be 10 times the typical base current. The total resistance of your divider is 253k.
Their current is only 59.3uA
They create an unloaded voltage of 1.96V for the base of the transistor.
The transistor loads down the voltage divider so its voltage will be about 1.67V and the emitter of the transistor will be about 1.02V and its current will be 3.78mA.
The typical curent gain of the transistor is about 200 so its base current is about 18.9uA. The values of the divider should have a total resistance of 79.4k ohms, not 253k ohms.

Sorry but that doesn't explain why it isn't acceptable for the resistance to be that high.

Is it due to the stability of the transistor?

I simulated the circuit with an Hfe of 100, the Vc = 7.3V, I reduced the gain to 50 and it was 9.87, I increased it to 400 and it was 2.5V.

Reducing the potential divider resistance will make the bias voltages more stable. Using the values you say will mean Vc will vary between 9.75V and 6.9V when the Hfe is varied from 50 to 400.

So having a lower resistance potential divider reduces the effect of loading by the transistor so the bias voltage will change less.

That makes sense but what if the voltage swing is just 1V?

Fair enough I can see that being a problem for the second stage where the voltage swing will be quite large but who cares if the bias varies widely on the first stage?

The voltage swing will be under 1V so it doesn't matter.

See the revised schematic.

The first stage uses higher value resistors to provide a higher input impedance. The voltage swing is tiny because the gain is low.

The second stage has lower value resistors in the potential divider to make it more stable.

It works just as well when the Hfe is 50 or 500.
 

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Now I think your circuit will have a very wide operating point for its transistors since its emitter voltages are so low.
One transistor has a Vbe of 0.7V and it will be almost cutoff. Another transistor has a Vbe 0f 0.6V and it will be almost saturated.
Temperature will also affect the operating points. The transistors will be thermometers.
 
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