# How to prevent BJT from saturation in the case below

#### YoGMan

Joined Sep 20, 2017
76
Hello guys , How do I convert a 50mV signal to volts without any clipping. My design is as shown below.
Here is how I proceeded
1)I DC biased my amplifier so that Vce is 9.3 V
2) I assumed a 2mA collector current at Q point
3) The values of Rc and Re were then calculated
4) From the resulting voltage Ve, Vb= Ve+0.6 V and the ratio of resistor R1 and R2 was calculated.
The only problem is that the amplifier goes to saturation , I want to input signals of around 50mV and get an output around 5V pk-pk.How can I modify my circuit?

Last edited:

#### MrChips

Joined Oct 2, 2009
19,768
Reduce the gain or the level of the input signal.

Why is C2 = 2.5mF
and C3 = 110μF?

#### AlbertHall

Joined Jun 4, 2014
8,828
If your output is more than 5Vp-p then you could add a resistor in series with C3 to reduce the gain.

Joined Mar 10, 2018
3,717
Something like this, attached.

Re should be dropped to 25 ohms or so. That yields
5V and good margin on bias.

Regards, Dana.

#### Attachments

• 450.3 KB Views: 18

Joined Mar 10, 2018
3,717
Forgot you were using a 2222 transistor. Here is collector
and output with correction.

Attached.

Regards, Dana.

#### Attachments

• 178.6 KB Views: 11

#### YoGMan

Joined Sep 20, 2017
76
Reduce the gain or the level of the input signal.

Why is C2 = 2.5mF
and C3 = 110μF?
C2 should have been uF, I calculated the capacitance to get low a cutoff of 80Hz

#### Bordodynov

Joined May 20, 2015
2,455
See

#### Attachments

• 3.8 KB Views: 1

#### YoGMan

Joined Sep 20, 2017
76
seems interesting , how do we call this configuration ? I want to study the theory

#### AlbertHall

Joined Jun 4, 2014
8,828
The input impedance of my common collector is around 15k//10K = 6K ohm
Except that the 15k has a 200nF capacitor across it.

#### YoGMan

Joined Sep 20, 2017
76
Ill then have 15k//10k // 265 ohm( 1/2pi*f*C) and the gain will decrease .am I right ?

#### AlbertHall

Joined Jun 4, 2014
8,828
Yes. Remove that capacitor.

#### Bordodynov

Joined May 20, 2015
2,455
Usually in a circuit with a common emitter, emitter resistors are used to stabilize (losing voltage!) And an additional capacitor has to be used.I’m promoting a true scheme with a common emitter on this forum.Also, to stabilize the regime and gain, I use parallel negative feedback (sequential feedback is usually used in schools).
The usual version of the scheme (in our case, I cheated and upgraded the scheme):

Here is how I choose resistors:
The voltage at the collector I take about half the supply voltage.
1. For a collector, I choose a resistor of 5-10 times smaller than the load.
This is to get almost maximum output amplitude and lower voltage gain losses.
2. For a base-collector resistor, I take a resistor 10-20 times the collector (when using transistors with a higher current gain,
you can increase the value).
3. To calculate the resistor between the base and the emitter, calculate the current through the base resistor and subtract the base
current from it.To calculate the desired resistor, it is necessary to know the base-emitter voltage at the selected current.In our case, 0.71B.At lower currents less.We divide the voltage by the current and get the nominal value.With this resistor (its size) we adjust the output voltage at the collector.
4. Choose a negative feedback resistor.Determine the upper limit of the resistor.The required gain is 5V / 50mV = 100 ==>
Rbe <20kOhm / 100 = 200.Amplification of a transistor (not an amplifier) over voltage is approximately 9V * 30 = 270 (9V is the voltage across the collector resistor).There are also losses on the Rbe resistor and the input resistance of the transistor. We choose about 2 times less and check.If you need a small gain, the gain is approximately equal to the ratio of the base and the desired resistor.

#### YoGMan

Joined Sep 20, 2017
76
Yes. Remove that capacitor.
The cap is to bring the high frequency cutoff response down to 5Khz , I managed to play with the values to get both gain and frequency response I need. But a trade off has to be made as high gain will give poor freq response

#### Bordodynov

Joined May 20, 2015
2,455
You can certainly more accurately calculate the gain.But then you have to use the equivalent circuit of the transistor in conjunction with the circuit.But you can use Spice!

#### YoGMan

Joined Sep 20, 2017
76
Usually in a circuit with a common emitter, emitter resistors are used to stabilize (losing voltage!) And an additional capacitor has to be used.I’m promoting a true scheme with a common emitter on this forum.Also, to stabilize the regime and gain, I use parallel negative feedback (sequential feedback is usually used in schools).
The usual version of the scheme (in our case, I cheated and upgraded the scheme):
View attachment 162523
Here is how I choose resistors:
The voltage at the collector I take about half the supply voltage.
1. For a collector, I choose a resistor of 5-10 times smaller than the load.
This is to get almost maximum output amplitude and lower voltage gain losses.
2. For a base-collector resistor, I take a resistor 10-20 times the collector (when using transistors with a higher current gain,
you can increase the value).
3. To calculate the resistor between the base and the emitter, calculate the current through the base resistor and subtract the base
current from it.To calculate the desired resistor, it is necessary to know the base-emitter voltage at the selected current.In our case, 0.71B.At lower currents less.We divide the voltage by the current and get the nominal value.With this resistor (its size) we adjust the output voltage at the collector.
4. Choose a negative feedback resistor.Determine the upper limit of the resistor.The required gain is 5V / 50mV = 100 ==>
Rbe <20kOhm / 100 = 200.Amplification of a transistor (not an amplifier) over voltage is approximately 9V * 30 = 270 (9V is the voltage across the collector resistor).There are also losses on the Rbe resistor and the input resistance of the transistor. We choose about 2 times less and check.If you need a small gain, the gain is approximately equal to the ratio of the base and the desired resistor.
Thank you I will study the circuit in more detail. I need to add one more stage to the 2 stages i made to amplify 5mv from microphone to 50mV.I will re upload the final 3 stages in a moment Hope it works as expected