Using common emitter configuration for single ended to differential signal converter

[kevinraphael]

Greetings!

I am a new member and I really need guidance in my university project. I am required to create a single ended to differential signal converter by using transistors and passive components. I was told that with a minor changes to the common emitter configuration, this is possible. I hereby attached an image of my circuit. It should work theoretically but I am unclear about the values of resistors I should use in the circuit. Can anyone be kind enough to teach me about the calculation that I should do to get the correct values of those resistors? Thanks in advance

 

[crutschow]

For the differential circuit to have equal plus and minus output gains the collector resistor must equal the emitter resistor value. The reason for this is:
The emitter output voltage gain is ≈1 so the current through R4 is ≈ Vin/R4. This same emitter current (minus the small base current) flows through R3. Thus to get the same gain at the collector as the emitter requires the R3 to be the same value as R4. And the voltage at the collector is decreasing while the emitter voltage is increasing (and vice versa) giving the differential signal you want.

If you need to compensate for the small difference in collector and emitter current due to the base current, then you could make R3 slightly larger than R4 by ratio of the collector/emitter currents. This problem is eliminated if you use a FET since they have negligible gate current.

 

[kevinraphael]

Thanks for your kind reply crutschow however, is there any specific values of R3/R4 that i should use? I mean the basic requirement is to have two similar values of resistors. But is 1k a good value?

Another question is about the values for R1 and R2. I understand that they act as a potential divider bias to the circuit. In your opinion, what should be the value of Vb? The npn transistor that i used is 2N3904 and seems like I have to determine the value of Ic I want(or is there any fixed value of Ic?), and then proceed with calculations to get the suitable resistors value for the circuit. I hope that you can enlighten me regarding this matter. Thank you so much and do have a nice day!

 

[crutschow]

The value of R1 and R2 determines the quiescent current through the transistor based upon the base bias voltage. They also determine the frequency response of the circuit as determined by the stray capacitances in the circuit. So it's generally a trade-off between power (current) and frequency response.

The bias value of Vb is determined by the maximum peak-to-peak voltage of the signal you have. You need to select a voltage that does not clip either the upper peak or lower peak of the signal. So you calculate the output voltages at both the collector and emitter for both the positive and negative peak signal to insure that neither output is clipped.

 

[kevinraphael]

the input AC signal that I currently feed to the circuit is 1 Vpp. I did some tinkering with the circuit and I realized that when I increased the input signal, the output signal is indeed clipped. But it is resolved when I increased the Vcc value. Correct me if I am wrong, but are you saying that Vb can be of any value as long as the output signal is not clipped? =) If it correct if I start the calculation by setting the value of Vcc and Ic I want? And then assuming that Vbe = 0.7v and the voltage drop across R4 is about 1v. This is the information I obtained from other websites. I do not know whether they are accurate or not.

 

[crutschow]

Normally the value of Vb is selected so that the positive and negative clipping points are about equal with the given supply voltage. Then the circuit is biased for the maximum dynamic range.

Your description of the calculations should give a proper bias point.

 

[kevinraphael]

Hi crutschow, thanks for your valuable feedback! I hope you do not mind my countless questions. There are two output in the circuits. Hence normally I see that Vb is biased for only one output. In the event where the circuit has two outputs, just like the one I showed you. What will be the suitable biased point? I know how to calculate the DC value of Ve. But I am unclear about the value of Vc. Again I see that from examples, people tend to set the value of Vc at about 1/2 of Vcc.
Hence if my Vcc is 10V, can I set Vc as 5V and Ve to be any value that is 0.7 lower than Vb?

 

[crutschow]

No. As I stated, you usually bias it to meet the criteria of the maximum dynamic range for a given supply voltage.

Say the supply voltage is 10V. Then the maximum dynamic range would be for the emitter to swing 5Vpp and the collector to swing 5Vpp (since the signals are out-of-phase).

Thus the emitter voltage bias should be 2.5V.

This gives a collector bias of 10V - 2.5V = 7.5V (for equal resistor values).

A +2.5V signal then generates 2.5V +2.5V =5V at the emitter and 7.5V -2.5V =5V at the collector (the transistor is saturated).

A -2.5V signal would generate -2.5V +2.5V = 0V at the emitter and 7.5V +2.5V =10V at the collector (the transistor is in cut-off).

Make sense?

 

[kevinraphael]

Make awesome sense! Thank you so much! =D Since there are 3 currents involved, Ib, Ic and Ie. With the supply voltage of 10V, can I set Ic to a value that I desire, lets say 10mA and Ib to, lets say 1mA? If I set Ib and Ic to these two values, then I believe that I will be able to calculate the value of resistors I need =)

 

[jjw]

You can't set Ib and Ic arbitrary, because they are related by hfe.
Ic can be set simply by Vb and Re and is about ( Vb-0.7V )/Re

 

[kevinraphael]

Hi jjw, thanks for commenting! Hmm... Okay... How do I set value of Re? Can I set it arbitrarily? =D Please assume that I do not know the values of R1, R2, R3, R4, Ic, Ib and Ie initially =D [I know the function of those resistors but I do not know the exact value to use in the circuit=( ]

 

[crutschow]

The transistor bias current is equal to V(R4) / R4. So, for example, if you want 1mA emitter current with a 10V supply, then you want the emitter biased at 2.5V, This gives R4 = 2.5V/1mA = 2.5kΩ. The collector resistor is then selected to be the same value.

Since the emitter current is determined by the emitter voltage across R4, you want to set Vb to equal V(R4) + 0.7V = 3.2V in this example. To minimize the effect of the base current in setting Vb, you make the current through the series value of R1 and R2 to be at least 1/10 of the emitter current. The ratio of R1 and R2 is selected to give the desired Vb, of course. That math for that is left to the reader.

 

[Bob Scott]

Crutschow, good advice! Someone should mention output impedance.

The output impedance of the line connected tothe emitter is near zero.
The output impedance of the line connected to the collector is equal to R3.
(..unless someone adds more negative feedback circuitry, but the emitter resistor should be enough.)

 

[kevinraphael]

Haha! Thank you so much crutschow! Your help is very much appreciated =D

 

[kevinraphael]

Hello Bob! Do you mind to explain a lil more in depth? So sorry, I am a newbie in electronics. Still have a lot to learn... >.<