Amplifier with direct coupling dont work

Thread Starter

Виктор Борщёв

Joined Nov 2, 2016
11
Is this design correct(I mean places with zener diodes)? And how can be calculated right values for that area resistors? I need to get Ku(amplification factor) = 150. But on first stage i get only 1 normal part of signal, other part is distorsioned. Can some1 help with that? Ty

 
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AlbertHall

Joined Jun 4, 2014
12,346
D1 is 7.5V and D2 is 6.2V and that definitely will not work with an 8V supply. Both Q1 and Q3 will be switched off. The positive cycle of large input signals will manage to get through which is why you see that distorted output.

Was the original design of this circuit for a higher supply voltage?
In any case this is not a good way of biasing the transistors for linear operation. It will be very dependent on the supply voltage.
 

Thread Starter

Виктор Борщёв

Joined Nov 2, 2016
11
D1 is 7.5V and D2 is 6.2V and that definitely will not work with an 8V supply. Both Q1 and Q3 will be switched off. The positive cycle of large input signals will manage to get through which is why you see that distorted output.

Was the original design of this circuit for a higher supply voltage?
In any case this is not a good way of biasing the transistors for linear operation. It will be very dependent on the supply voltage.

original design was for 6.3 V both, but there was used D1 for 6,8 and D2 for 4.8 V
 

AlbertHall

Joined Jun 4, 2014
12,346
This circuit shows positive and negative supplies. If the are both 6.3V then the total supply voltage would be 12.6V and the circuit would have a better chance of working. Your circuit uses a PNP transistor for VT2 (Q3 in your circuit) and this means the biasing would need to be changed.
What do the dotted lines in the circuit represent?
 

Thread Starter

Виктор Борщёв

Joined Nov 2, 2016
11
This circuit shows positive and negative supplies. If the are both 6.3V then the total supply voltage would be 12.6V and the circuit would have a better chance of working. Your circuit uses a PNP transistor for VT2 (Q3 in your circuit) and this means the biasing would need to be changed.
What do the dotted lines in the circuit represent?
In the guide said that with dotted line(with zener diode) i need to replace resistance what comes to emitters
 
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Thread Starter

Виктор Борщёв

Joined Nov 2, 2016
11
so there really a lot of text,i will introduce only some of it, something i guess more important

Replacing of Emitter resistor with a zener diode( in figure 1.2 it shown as dotted line) give us a possibility to increase factor om amplification of tension(voltage). For stabilization of thie circuit we use resistors Rg and Rrn what forming a negative reaction in this circuit. This negative reaction exist for initial signal too what comes from Eg. That way amplification factor will be reduced and will be stable.
This example we will do a amplifier with direct coupling what will give us (Out Voltage) = +-3V, on Rs = 5kOhm. We need to got amplification factor F=100, in range 0-10 Khz.
Calculations

Is=Uout/Rs=3v/5kohm=0.6mA
Ec1 and Ec2 = 6.3
Ic3min =0,2mA (for germanium transistors)
Rc3=Ec1-Uout/Is+Ic3min=6.3-3/0.6+0.2=4,12kOhm
Ic30=Irc30=Ec1/Rc3
Irc3max=Ec1-(-Vout)/Rc3=2.2 mA
Ic3max=Is+Irc3max=0.6+2.2=2.8mA
Vce3min=Vce3saturation=0.8V
Re3=|Ec2|-Vout-Vce3min/Ic3max=6.3-3-0.8/2,8=0.89kOhm
Vce30=|Ec2|- Ic30*Re3=6.3-0.89*1.5=5V
Ib30=Ic30/Hfe(min)=1.5/20=0.075mA
Ube30=0,3V
Urc20=Ec2-Uce30+Ube30=6.3-5+0.3=1.6V
Ic20=0,5mA
Rc2=Urc20=1.6/0,5=3.2kOhm
Ib10=Ib20=0,25mA
Ure10=Ec2-Ube10=6.3-0.3=6V
Re1=Ure10/Ic10=6/0.5=12kOhm

We choose 6V zener
R1=Ec1+|Ec2|-Ustabil.1/Ist1(min)=6.3+6.3-6/3=2.2kOhm

Uce10=1.5V
Urc10=Ec1+|Ec2|-Vre10-Vce10=6.3+6.3-6-1.5=5,1V
Rc1=Vrc10/Ic10=5.1/0.5=10.2
Ure20=Urc10-Ueb20=5.1-0.3=4.8V
Ust2=4.8V
R2=Ec1+|Ec2|-Ust2/Ist2min=2.6kOhm
 

AlbertHall

Joined Jun 4, 2014
12,346
This is designed for germanium transistors not silicon :rolleyes:
It isn't a very practical circuit. Do you actually want to build it or are you just investigating it?
To achieve DC - 10kHz, gain = 100, it would be much easier, more stable, fewer parts to use an op-amp
 

Thread Starter

Виктор Борщёв

Joined Nov 2, 2016
11
i do it as additional task for my university :) anyway i assembled this circuit but with gain = 10 and now just investigating it. If you can give some advices what to correct(or change but still with using direct coupling) will be good
 
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