BJT amp - special case

Hello,

I want to understand how to calc the values of resistors to make a schematic about a simple bc547 preamp work

I know how to do it with a regular common emiter schematic that one learn at school and wich is really theoric, but in practice, the diagrams changes a bit

there is no Re, no R2 and R1 is connected between collector and base, after Rc



how do I calc the two resistors on the left ?

thanks

With a battery voltage of X volts you want about .5X on the collector. A good current for the transistor is in the range of .1-> 2mA, lets use 1mA. This give the collector resistor of .5X/ 1 K ohms. To find the base current, you divide the Ic by the Hfe (beta), this is a high gain transistor so lets guess Hfe = 200, so Ib = 1/200 mA, 5 micro amps. The base will be about .7V above the emitter, so the voltage drop across the base resistor will be .5 X -.7 and the current through it will be 5 micro amps so you can find its value.
Frank

What's the value of the capacitor on the right ?

0.1ùf->1ùf

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wow thanks dude I'll check this out

I do this differently.

The transistor config is a common emitter with negative feedback for bias and gain.

Since the Re=0 , it can only be used for small signals like <<100 mVpp

The output impedance of the electret is a open drain buffered affects gain and Q operating point which is not critical for small signals, so 10k is a common value.

If you recognize that negative feedback BJT's lower both the input and output impedance by the amount of negative feedback or forward gain and that the voltage gain is limited by the current gain of the device, it is quite easy to achieve approx voltage gain of 100 with an hFE of 200 +/-50%

I would choose a feedback resistor ratio of Rbc/Rc= 100 for this configuration (with high hFE transistors at low current.)

To be close to mic output impedance for maximum power transfer, Since the feedback R also controls AC gain lets make Rbc= 100 * 10k
where the gain is Av ~= Rbc/Rc (as long as Av < hFE with sufficient margin.

The cap value determines the high pass filter ( HPF ) breakpoint where f =1/(2pi*RC) is -20 dB per decade below cutoff.

Conclusion
Rbc=1M
Rc=10k
Choose C for 100Hz
Gain = Rbc/Rc=100
BJT Zin= Rbc/Av = Rc for Av=100
BJT Zout = Rc/Av = 10k/100 = 100

Since base current is approx 0.5%, OHm's Law determines Q operating point from resistor ratio and using Vbe = 0.6 (i) for low current. or slightly less for determining since Ic=hFE*Ib (ii) and (Vce-Vbe)/Rbc=Ib ...(iii) and (Vbat-Vce)/Rc=Ic ...(iv) you have 4 equations and 4 unknowns so you can solve for to verify what Vc is and that it is close to Vbat/2 for max undistorted switch. Normally you alllow Vce>2V min for linear operation, but with negative feedback, this increases the range to Vce>0.2V.

Let's see if you can figure this out!

thanks alot people for these awesome infos
have a great day

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thanks to spend time explaining in details

I think you should select a low noise transistor that has a narrow range of hFE like a BC549B.
We draw a schematic with the input on the left, the output on the right, the positive power supply voltage at the top and the 0V ground at the bottom like this:

Audioguru said:

I think you should select a low noise transistor that has a narrow range of hFE like a BC549B.
We draw a schematic with the input on the left, the output on the right, the positive power supply voltage at the top and the 0V ground at the bottom like this:

Click to expand...

no schematic...

Hi Frank

just had time to check all this

for the resistor between C and B
4.5V0.7V=>3.8V , with Ic= 5µA => 3.8V/5µA = 760K right ?

but what I dont understand is why would the current flowing through C->B resistor be the same as Ic ?

thanks

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Hi SunnySkyguy,
been a while, I just had some time to get back to this one,

SunnySkyguy said:

I do this differently.
Since the Re=0 , it can only be used for small signals like <<100 mVpp

Click to expand...

why ? what about big signals and Re value, why is it related ?

SunnySkyguy said:

The output impedance of the electret is a open drain buffered affects gain and Q operating point which is not critical for small signals, so 10k is a common value.

Click to expand...

mmmmh ok, seems everyone uses 10k for electrets anyway

SunnySkyguy said:

If you recognize that negative feedback BJT's lower both the input and output impedance by the amount of negative feedback or forward gain and that the voltage gain is limited by the current gain of the device, it is quite easy to achieve approx voltage gain of 100 with an hFE of 200 +/-50%

Click to expand...

I am completely lost

SunnySkyguy said:

I would choose a feedback resistor ratio of Rbc/Rc= 100 for this configuration (with high hFE transistors at low current.)

Click to expand...

why ?

SunnySkyguy said:

The cap value determines the high pass filter ( HPF ) breakpoint where f =1/(2pi*RC) is -20 dB per decade below cutoff.

Click to expand...

AC considerations are way out my leage, I saw these a long time ago when I was at school, ,I got no idea what you refer to
I just remember Ce was the capacitor making things more stable or having a bigger AC gain or something

SunnySkyguy said:

Since base current is approx 0.5%

Click to expand...

why 0.5% ?

The resistor from C to B provides Ib which is base current (5uA), not Ic which is the much higher collector current through the collector resistor which connects to the positive power supply voltage.
The transistor does not have an emitter resistor so its voltage gain will be very high (if its signal source has a low input resistance) then 0.1V of input will try to produce an output of maybe 20V. Re will provide some negative feedback which reduces the gain then the input signal level can be higher.
The datasheet for an electret mic shows that it is tested with a current of 0.5mA. Then with a 9V supply, the resistance to produce 5V across it is 10k.
There are different ways to calculate voltage gain. Hfe is not used when calculating voltage gain.
The ratio of the base biasing resistor and the collector resistor should be determined and calculated from the hFE value on the datasheet, not just a wild guess.
The simple calculation for a coupling capacitor to pass low frequencies into a resistance was shown.
The datasheet of the BC549B shows that its typical hFE is 520 when its collector current is 2mA so its base current is 0.2% of its collector current, not 0.5%, because of the wild guess of the hFE of 200.

actualy I made calculations with a typical hfe of 300 for a bc547 (datasheet)

and the "wild guess" was actually an example answering my question about how to calc resistor values

and it is the same infos I found in "Chapman - Transistor Circuit Techniques" and I tested it, it works like a charm

A BC547 has noise and has a very wide range of hFE from 110 to 800 when its collector current is 2mA. Therfore it is a general purpose switch that can turn on an LED or something.
I selected a BC549B because it is low noise and has its hFE selected into a narrow range from 420 to 800 and is typically 520.