# Differential amplifier single ended output calculations

#### jarro

Feb 19, 2015
7
Hi,
I'm trying to perform a DC analysis on a single ended output differential amplifier (I've uploaded an image of the circuit) and I'm having problems calculating the collector voltage and current of transistor 2. So far I've calculated the tail voltage and current. My calculations are as follows:

(It = tail current source)
It = V/R = 10V / 4.7KΩ = 2.13mA
Taking into account It= IE1 + IE2
Therefore IC2 = IE2 = It/2 = 1.06mA
Therefore VC2 = +Vcc - (IC2.R1) = 5.02V

I require VC2 and IC2 in order to complete the DC analysis of the rest of the circuit (not shown in the picture) as a single-ended output is taken from this point. When I add a probe on multisim and perform a simulation it is telling me that VC2 = 6.02V and IC2 is 0.846mA. I can only assume that this is due to the fact that the output is taken from the 'side' of the differential amplifier that contains a resistor whereas the other 'side' doesn't, therefore making VC1 and IC1 different to VC2 and IC2. I don't know however how to factor this into my calculations and if anyone could help me find out how to get the proper values (the ones I got from the simulation) it would be much appreciated!

(B>1 therefore the base current of each transistor can be ignored)

#### Attachments

• differential amp.png
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#### Ratch

Mar 10, 2013
1,098
Hi,
I'm trying to perform a DC analysis on a single ended output differential amplifier (I've uploaded an image of the circuit) and I'm having problems calculating the collector voltage and current of transistor 2. So far I've calculated the tail voltage and current. My calculations are as follows:

(It = tail current source)
It = V/R = 10V / 4.7KΩ = 2.13mA
Taking into account It= IE1 + IE2
Therefore IC2 = IE2 = It/2 = 1.06mA
Therefore VC2 = +Vcc - (IC2.R1) = 5.02V

I require VC2 and IC2 in order to complete the DC analysis of the rest of the circuit (not shown in the picture) as a single-ended output is taken from this point. When I add a probe on multisim and perform a simulation it is telling me that VC2 = 6.02V and IC2 is 0.846mA. I can only assume that this is due to the fact that the output is taken from the 'side' of the differential amplifier that contains a resistor whereas the other 'side' doesn't, therefore making VC1 and IC1 different to VC2 and IC2. I don't know however how to factor this into my calculations and if anyone could help me find out how to get the proper values (the ones I got from the simulation) it would be much appreciated!

(B>1 therefore the base current of each transistor can be ignored)

Your calculations look good to me. The collector circuits should act like current sources, so a little resistance in the collector should not change the current too much. I would probe and compare the emitter currents of both transistors and also vbe and Vc and see if they correlate with the calculations.

Ratch

#### jarro

Feb 19, 2015
7
Thanks for the quick reply. According to probes on the collector of each transistor: IC1 = 1.13mA and IC2 = 0.846mA, and VC1 = 10V (which is to be expected) and VC2 = 6.02V. vbe i've assumed to be 0.7. I would usually carry on regardless with the rest of the analysis but as I have to use both IC2 and VC2 in other calculations, this small difference makes a huge variation in the rest of my results. Just for an experiment I put the simulation value of VC2 into these further calculations and all of my results from then matched exactly with the simulation results! If I use the value of 5.02 instead, all of my calculations don't match up to the simulation at all! This lead me to believe that there was a problem with the calculation of VC2 and IC2. Each emitter voltage is -604mV and IE1 = 1.13mA and IE2 = -0.866mA

#### Arouse1973

Dec 18, 2013
5,178
Hello
Your tail current will be -10 Volts - about -0.635 Volts because of the Vbe drop. This gives you less than 2.13 mA and your current is not split evenly between the two. It's close as Ratch said but this might make a difference in your calculations.

#### KrisBlueNZ

##### Sadly passed away in 2015
Nov 28, 2011
8,393
Would the difference in currents be due to the Early effect? (https://en.wikipedia.org/wiki/Early_effect)

Q1 has about twice the collector-base voltage of Q2, and according to that article, this should cause a somewhat higher collector current in Q1, which is what you see.

According to the numbers in post #3, Q1's collector current is about 33% higher than Q2's collector current, which seems like quite a lot really, so I don't know whether the Early effect would account for all of it.

Both transistors have exactly the same base-emitter voltage, but the collector circuit affects the characteristics of the base-emitter junction (don't ask me why or how; I don't understand the physics of the things - that's just an observation of their behaviour) so I wouldn't be surprised if the base currents were different despite the equal VBE. What are the base currents in your simulation?

#### LvW

Apr 12, 2014
604
Yes - I agree to Adams (Vbe drop) as well as KrisBlues comments (Early effect).
Moreover, we can expect a corresponding (small) difference in base currents because Vbe is equal for both transistoirs.

#### jarro

Feb 19, 2015
7
Cheers for all the replies. As beta > 1, all the base currents can be ignored. Which is why I assumed that IE2 = IC2 (approx) etc.

Now the early effect has been mentioned I realised that in my original question the early voltage is given. Looking through my notes I found the equation (which has been simplified):

IC=(aprox)=IS.e^(VBE/VT)

After looking at the datasheet for the BC547, apparrently the reverse leakage current is 0.1uA. Which gives me:

0.1uA . e^(0.7/26mV) = 49.3kA which is obviously too high. I believe that maybe this early voltage is what's causing my calculations to be off (especially since the early voltage values are given!) although I don't fully understand how to factor it in.

#### jarro

Feb 19, 2015
7
I also have the equation rce = (VA + VCE) / IC

(early voltage VA is given as 100V (NPN) and 75V (PNP))

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#### LvW

Apr 12, 2014
604
IC=(aprox)=IS.e^(VBE/VT)

No - thats not an equation for the Early voltage. Note this is the equation for the collector current dependence on the B-E voltage. Hence - the input-output transfer characteristic for the BJT.
The Early voltage is a measure for the slope (g,ce=1/r,ce) of the output curves Ic=f(Vce) with Ib=const. It is fictive voltage that cannot be measured.
It can be found graphically using the asymptotic lines for set of output characteristics, which meet in a common point on the negative voltage axis. This point is the Early voltage.
And - yes - the following formula is correct: rce = (VA + VCE) / IC.

#### jarro

Feb 19, 2015
7
No - thats not an equation for the Early voltage. Note this is the equation for the collector current dependence on the B-E voltage. Hence - the input-output transfer characteristic for the BJT.
The Early voltage is a measure for the slope (g,ce=1/r,ce) of the output curves Ic=f(Vce) with Ib=const. It is fictive voltage that cannot be measured.
It can be found graphically using the asymptotic lines for set of output characteristics, which meet in a common point on the negative voltage axis. This point is the Early voltage.
And - yes - the following formula is correct: rce = (VA + VCE) / IC.
Sorry I should have been more clear, I tried calculating the collector current with that equation but still got an unlikely result. How would I factor in rce to try and get the correct IC2 and VC2 values? I am only realising now that I am being expected to use the early voltage value in some kind of equations to work out the collector voltages and currents, if not I see no reason why it was stated in my assignment. I'm guessing this is why all my calculations are slightly out although I still don't fully understand how and where to use this early voltage.

#### LvW

Apr 12, 2014
604
I am not quite sure what do you intend to do.
You did a simulation - and you want to find the simulation results by hand calculations? Is that true?
In this case, I must tell you that this seems to be impossible because you have used during simulations a real BJT model with many non-idealities which all cannot considered in your hand calculations.
As an example: Because both collectors are loaded differently (unsymmetric) you must - from the beginning - assume that the tail current is NOT equally divided between both transistors (due to the Early effect). But the division ratio again depends on the currents (on the voltage drop across the coll. resistance) - thus, the calculation is rather involved.

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#### jarro

Feb 19, 2015
7
Yes that's correct. Aah I understand now, I wasn't even considering that any other factors would play a part in the analysis. However I am now confused as to why the early voltage value was specifically stated if such calculation is so complex.

#### LvW

Apr 12, 2014
604
However I am now confused as to why the early voltage value was specifically stated if such calculation is so complex.

What do you mean with this part of the sentence?

#### jarro

Feb 19, 2015
7
Sorry I mean that I was given certain values at the start to perform the analysis on this circuit. For example I was given resistor values, the beta value and also the early voltage value.

#### Arouse1973

Dec 18, 2013
5,178
I am not quite sure what do you intend to do.
You did a simulation - and you want to find the simulation results by hand calculations? Is that true?
In this case, I must tell you that this seems to be impossible because you have used during simulations a real BJT model with many non-idealities which all cannot considered in your hand calculations.
As an example: Because both collectors are loaded differently (unsymmetric) you must - from the beginning - assume that the tail current is NOT equally divided between both transistors (due to the Early effect). But the division ratio again depends on the currents (on the voltage drop across the coll. resistance) - thus, the calculation is rather involved.

Rather complex, yeah I would go along with that This is some of the stuff you will need to work out.

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