# Differential DC amplifier circuit

Discussion in 'General Electronics Chat' started by khanh, Feb 8, 2006.

1. ### khanh Thread Starter Member

Jan 6, 2006
28
0
Does anyone have knowlege of how a differential dc amplifer circuit opertated. I try to build a kind of "single input, single output" diff amp in the multisim just like in my book "introductury semiconductor electronics", but have no idea why the output gain isn't equal to the voltage gain of the common base amp, perhap something wrong with my circuit schamatic.
I include here the picture how that circuit looks like in multisim version 8.

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
Hi,

Your differential amp is set up common emitter, not common base. There's no feedback to the Q2 base, which will not let the circuit function properly. You're driving the base of Q1 with 10 mills, but the total resistance of the collector and emitter resistors will only allow 3.3 mills current to flow. You're looking for a voltage gain, but have only applied 10 mills current @ 50 Hz to the input.

3. ### aac Active Member

Jun 13, 2005
35
0
When you asked why this circuit doesn't have the same gain as the common base amplifier, I guessing you believe the gain should be 5k/1k or 5. Is that what you meant? Anyway, this circuit will have a gain more like 300. That is because the emitter of Q2 holds node 6 at about -.7 volts. Each half of this circuit then acts more like a CE amplifier. You can control the gain of the circuit by putting a resistor between each emitter and the bias resistor R2. So if you put a resistor of 100 Ohms in between each emitter and R2, the gain would be 2*5k/100, or 100.

Does this help?

4. ### khanh Thread Starter Member

Jan 6, 2006
28
0
Sorry, I forgot to tell you that the output is he voltage read at oscillopscope 1

TO:Been There. I don't know what wrong with that circuit I posted since it is a copy from the book "introductory semiconductor electronics" by "Nigel P.Cook" published from Prentice Hall at page 553 so I suppose nothing wrong with it. Though it is weird that all the knowledge I know about the Transistor would not apply to this circuit at all. One possible mistake may be the resistors I put it, they could be not enough, but the arrangement is excact the same as in the book.

To: AAC, the circuit got the voltage of 500 mv, so its gain is 50 times, not 300 times as you reckon, I know it because I run the multisim (don't know if you could use it). Though could you tell me how you got that result since this circuit is kind of weird to predict ( normally I can calculate the output result prior to run the simulation, so I run it only for double check and make sure I understand it thoroughly).

Also the circuit you propose is correct as I got similar one and I know it is working but my question is again: whether that circuit is working or not (from the book), or the author got it wrong (very rare occassions).

5. ### khanh Thread Starter Member

Jan 6, 2006
28
0
Also, when I write the transistor Q2 in common base set up, I mean the voltage Vdd acting like a input (as the signal is transformed from Q1) and the output is between the collector node and the base so it is common base arrangement (they share the base node)

6. ### chesart1 Senior Member

Jan 23, 2006
269
1
Hi,

You have a 10 mv ac signal. Is it 10 mv peak to peak or 10 mv rms? Either way, that small variation in base voltage could be neglected in these calculations. The reason is that I am going to prove that the transistor Q1 {assuming a beta of 100) is in saturation with 0 volts at the base of Q1.

Assuming that the transistor is on, the base emitter voltage drop is about 0.6 volts. This means you have about -.6v at the emitter of Q1.

That means you have about 1.4 volts (-2.0v - -0.6v) across R2 due to Q1 base current. Using ohms law, 1.4v/1k = 1.4 ma. So Q1 base current is 1.4 ma.

Assuming a beta of 100, the collector current is 1.4 ma * 100 or 140 ma. The total Q1 collector current (about 140 ma) times the collector resistor R1 (5k) = 150 volts. However ... We know that the maximum voltage drop we can have across R1 is the supply voltage VCC (about 20 volts). Therefore the current through R1 is actually 20 volts/5k or 4 ma. And the voltage at the collector of Q1 is zero. This means that Q1 is saturated. A saturated transistor cannot follow the change in voltage at it's base.

If your transistor Q1 has a beta (base/collector current gain) of 100, the circuit would not operate correctly.

Please give me the beta (Hfe) parameter of the transistor you are using so these calculations could be made to determine why your circuit is not working as expected.

John

7. ### khanh Thread Starter Member

Jan 6, 2006
28
0
The signal source is of 10 mv p-p (not rms).
The hff (or beta) of both transistors are 100. (though when I change both of them, the result is still the same?)

8. ### khanh Thread Starter Member

Jan 6, 2006
28
0
TO: chesart1.
The transistors are both in active region. Why? because I run simulation and it tells that the emitor to collector voltage drop are around 15 V.

Your proof got a lof of mistake as I can see: the base current in Q1 should be 1.4mA /100 not 1.4 mA since the resistor you are taking into calculation is emitor resistor (not base resistor), hence the voltage drop across collector : 1.4 mA *5*1000 = 6V. This explained why the multisim got that result.

9. ### pebe AAC Fanatic!

Oct 11, 2004
628
3
Try a 10K base resistor in the feed to Q1.

10. ### chesart1 Senior Member

Jan 23, 2006
269
1
Hi,

Thank you for the correction to my calculations.

John

11. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
The gain is determined by the collector resistors and the dynamic emitter resistances. Since the tail current is about 1.4ma, each emitter current is about 0.7ma. From the diode equation (look it up if you don't know it), each emitter resistance is about (26/0.7) 37 ohms. The simplified differential gain is approximately (2*5k)/(2*37)=135. The actual gain will be somewhat less, due to the attenuation caused by the 1k resistor, and by non-infinite dynamic collector resistances, which wind up in parallel with the 5k resistors. I ran a sim on "generic" transistors and got a gain of about 119. I ran the same sim with 2N3904s and got a gain of about 127.

12. ### khanh Thread Starter Member

Jan 6, 2006
28
0
Ron H, my test on sim only gives 60 in gain for 2n3904 transistor type. Also I do look for the diode equation you'r talking about. does it look like this:

Id = Is[e^(Vd/0.026) -1]
I dont know how u use it to calculate the diode resistor??????????

13. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The results I obtained from simulation agreed closely with RonH's values. In the absense of emitter resistors, the dynamic emitter resistance is dominant as RonH has indicated. The equation used to estimate the value of this dynamic resistance is 26mV/Ie again as RonH has stated. RonH's value of 37 ohms based on an emitter current of 0.7 mA is a good estimate for the value of the dynamic emitter resistance.

The only thing I can add to RonH's comments is that the quiescent DC at the two collectors will be somewhere near 20V-(5K*0.7mA) or approximately +16.5 volts. I say approximate since Ic is a function of the assumption made for the value of Vbe. Depending on the individual designer, assumed Vbe values range from 0.6V to 0.7V. Of course when designing with real components it is prudent to consult the device manufacturer's datasheet for the critical device parameters including the nominal value for Vbe.

The assumptions that underlie this analysis are possible because this problem is based on the small-signal behaviour of the transistors.

As I recall, the derivation of the equation for estimating the dynamic emitter resistance is based on the slope of the diode I-V curve. The slope is obtained by taking the derivative of the diode equation in that portion of the curve that occurs once the diode is forward biased.

Diode I-V Curve explanation

Scroll down to the section on Semiconductor Diodes for an explanation on how the value of 26 millivolts is obtained.

hgmjr

14. ### khanh Thread Starter Member

Jan 6, 2006
28
0
You're right, the gain is about 120. And thank for your explaination in Diode equation. I've finally got it.

15. ### aac Active Member

Jun 13, 2005
35
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About half the 1.4ma goes through each collector not the base. The base current of each transistor is .7ma/beta. The transistors are not in saturation.

16. ### aac Active Member

Jun 13, 2005
35
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Are you taking the gain to be the voltage to be one of the collector voltages with respect to ground divided by 10mV? Ususally in a differential pair, we take the gain as the difference between the collectors divided by 10mV. If this what you did, the second method would double your gain giving you something closer to the 127 value. For my part, the 300 was just some calculations in my head. So I believe your 50 if you are going from either collector to ground.

17. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Greetings, Khanh.

I am glad I could help out. In all honesty, I simply built on the excellent response from RonH.

Based on chesart1's previous informative and enlightening replies on other topics on this forum, I believe he just got his base current and emitter current momentarily swapped. I myself have had those moments here in the forum when my braincells misfired. It can happen to anyone.

Thanks for posting your topic. It is an interesting one and I am sure that others have benefited from following the lively exchange it has engendered.

I hope your experience has been a positive one and that you will return to the forum in the future with new topics as well as replies to the topics of other members.

Good Luck,
hgmjr

18. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
The derivation of emitter resistance is as follows: (Google "diode equation" for definitions of k, q, and T)

I=Is(e^(qv/kT)-1)

where kT/q=26mv at T~300 degrees Kelvin

since Is<<I at normal bias currents,

I/Is~e^(V/.026)

V/.026~ln(I/Is)

V~.026(ln(I)-ln(Is))

Recall that R=dV/dI

dV/dI=.026/I QED

19. ### CoulombMagician Active Member

Jan 10, 2006
37
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If you're interested in the design of diff amps start with this app note

I_AMPS

There are other pertinent notes there also.

The AD622 has an interesting closed loop scheme for improving CMRR and it is only one of many such amplifiers.

Good luck!