# Resistor values for BJT Differential Amp

Discussion in 'The Projects Forum' started by circuitBoy818, Dec 20, 2013.

1. ### circuitBoy818 Thread Starter New Member

Dec 13, 2013
8
1
Hey guys,
for a project im trying to build a differential amplifier. I've done alot of searching and based on the classes i've taken so far, im a bit lost.

Specs
works up to 2Mhz
Gain: 5 -10
Transistor: 2N2222 BJT
DC: +- 10 V

I'm having alot of trouble getting started and finding the resistor values. Can someone please help me get started. What should i do first and what equations i should look at, how do i break it down to smaller chunks? what should i look for in the data sheet and how can i apply it?
I am attaching a diagram i have so far.

2. ### #12 Expert

Nov 30, 2010
18,076
9,686
You can't do current mirrors like that with discrete transistors. They will not track properly. The usual method is to add resistance in the emitter leg to get them to track each other better. Why would you even try that method? You don't need fabulous impedance on the collectors to get a gain of 10.

And why use up half your power supply with a grounded base on Q3?

If you want to play, "constant current", you can do it all with Q3. Just use 2 resistors to set the base at about 2 above -10V. Another resistor to set the total current around 1ma to 10 ma. 140 ohms to 1.4k. Then pick a collector resistor to add a current follower.

If you want a grounded input, you must be doing an inverted gain. This requires a resistor in series with the input and therefore a current load on the input signal. Otherwise, you would be building a non-inverting amplifier with a resistor network to the inverting input. Is that what has you crossed up? Pick one.

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3. ### circuitBoy818 Thread Starter New Member

Dec 13, 2013
8
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ok, lets say i do go with the design you introduced in the picture. I still don't know how to get the resistor values. What do i need from the 2n222a datasheet and how can it be used?

4. ### circuitBoy818 Thread Starter New Member

Dec 13, 2013
8
1
Another problem is that, it needs to work for up 3Mhz, will the design above work that?

5. ### #12 Expert

Nov 30, 2010
18,076
9,686
Are you saying you don't know how to design an op-amp with a gain of 10?

Non-inverting Gain = 1 + (Rf/Rc) where Rc is the resistance from the inverting pin to ground and Rf is the resistance from the output to the inverting input.
Inverting gain is -Rf/Rin where Rf is the resistance from the output to the inverting input and Rin is the resistance in series from the source to the inverting input.

I don't know if this will work at 3Mhz. I don't work in frequencies that high. I just assumed that you chose a transistor that can work at 3 MHz.

ps, the emitter resistors are the wrong size. Wait until I correct that.

Page 1, circuit 1 and circuit 2.

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• ###### differential pair.png
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Last edited: Dec 20, 2013
6. ### circuitBoy818 Thread Starter New Member

Dec 13, 2013
8
1
what about the beta and rpi and all that stuff. I don't think its a simple as your saying. This is going to be completely discrete

here look at this

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7. ### #12 Expert

Nov 30, 2010
18,076
9,686
Apparently I'm doing figure 6.
I did not intend to hand you the completed circuit. I'm trying to find out what has you stopped, and, as for me, all that math in the thing you posted has me stopped. One of the first things it says is that we assume the transistors are identical. That just can't be true for a discrete design. I'm trying to get you away from that assumption because it can't work unless this design will be built on one silicon substrate.

Do you realize you are building an operational amplifier?
Do you understand that the gain is almost entirely controlled by the feedback resistors?

8. ### #12 Expert

Nov 30, 2010
18,076
9,686
You could get a gain of ten without a feedback path, but you could do that with just one transistor. So, why use a differential pair if you're not going to use it to get an externally controlled gain?

9. ### #12 Expert

Nov 30, 2010
18,076
9,686
Let's try this. If the input voltage changes by .1 volt, you want the collector voltage to change by 1 volt. If the emitter voltage changes by .1 volt, the current through the emitter resistor will change by .1/2k = 50 ua
To get 50 ua to show up as 1 volt at the collector, that resistor must be 1/50ua = 20k.
20k in a 1/2 ma leg will use up 10 volts.
That will make your collector 0 volts when the base is 0 volts. Not good.
Must decrease the emitter resistor.
Try 1k
.1 volt difference over 1 k is 100ua.
to get the collector resistor to change its voltage drop by 1 volt, it must be 1v/.0001A
That makes it 10k. It will use up 5 volts at the 1/2 ma idle current.

Is this what you're looking for?

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10. ### #12 Expert

Nov 30, 2010
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Still, you can see that the same values will work if you do not use a differential pair.

11. ### circuitBoy818 Thread Starter New Member

Dec 13, 2013
8
1
thank you very much for taking all this time to explain to me in details, i will try to use all the advice you gave me.

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12. ### #12 Expert

Nov 30, 2010
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The way I see it, you didn't understand how to start and I didn't understand how lost you were. I went completely over your head, and I apologize.

13. ### circuitBoy818 Thread Starter New Member

Dec 13, 2013
8
1
Is there anyway i can send you a private message? so i can explain my projects in full details, that way if possible maybe you can help me a little bit. I just need to know how to get started, then i will use my knowledge from my other classes to continue working on it.

14. ### #12 Expert

Nov 30, 2010
18,076
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You can send me a private message by clicking on my name, but this is a public forum. The intent is to discuss these things in public for the benefit of all. Other people catch my mistakes, for one thing.

I think you have a lot of building blocks but they aren't organized. You have to go through the process of making mistakes like this one until you get a feel for which building block goes where. Other learners do, too. That's why this is a public forum.