# npn emitter follower confirmation

Discussion in 'The Projects Forum' started by jetbat, Oct 23, 2009.

1. ### jetbat Thread Starter Member

Oct 21, 2009
14
0
Hi. I am new to transistors and I think I have figured out what I am doing, but I would like to confirm that I have the right resistor values for a emitter follower circuit. I wanted 100mA output, but the closest 5 watt resistor I have is 68R. (If I figured correctly, the resistor will need to be over 1 watt) So I am working around that with what I think will be 117mA output. The NPN transistor is a MPS2222A.

Attached is a schematic of the circuit.

With the voltage divider bias resistors, the closest values I have are 680R and 880R.

So I have two questions. 1. Have I done the math correctly for figuring out the resistor values? 2. Are the 680R and 880R resistors close enough to work?

Thanks
Scott

2. ### SgtWookie Expert

Jul 17, 2007
22,199
1,801
Scott,
It's best to post images in a .png format; as those type images are both compact and not "lossy" like .jpg images are.

1) In your schematic capture program, do a screen print by pressing Ctrl+PrintScreen
2) Start MSPaint.
3) Paste the image into MSPaint using Ctrl+V or Edit/Paste.
4) Save the file as type .png
5) Upload to the forum using the "Go Advanced", "Manage Attachments" buttons.

3. ### jetbat Thread Starter Member

Oct 21, 2009
14
0
Oops. I must have not clicked on the upload button. Here it is.

File size:
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4. ### Wendy Moderator

Mar 24, 2008
21,500
2,961
Actually the emitter resistors isn't that important. It sets the current through the transistor, and controls the current from +16V to ground when the transistor is off. The current from the transistor to ground is only limited by the load and the gain of the transistor.

The base to ground resistance is determined by the gain of the transistor and the emitter resistor. If the transistor has a hfe (also known as beta, β) of 150 then the base to ground resistance is β X Re (68Ω X 150), or 10.2KΩ. This resistance will interact with the bias resistors.

5. ### spacewrench Member

Oct 5, 2009
58
1
Your numbers look pretty good to me. Since you're using resistors to bias the transistor base, the voltage where you have marked 8.6V will actually depend on how much current goes into the base (as opposed to down through R1). With your numbers, you're apparently assuming that the transistor beta is (exactly!) 112. If it is, then you'll get 11mA going down through R2, just less than 10mA through R1, and just more than 1mA into the base. The base current times beta gives you the 117mA you expect through Re.

Both your 8.6V and 8V nodes will vary depending on the actual transistor beta, which will itself change with temperature and age.

Depending on what you're trying to accomplish, you may want to adjust your biasing circuit to reduce the dependence on beta.

6. ### Ron H AAC Fanatic!

Apr 14, 2005
7,018
682
If the load (not shown) is AC coupled, the emitter pulldown resistor determines the available sink current of the emitter follower.

7. ### SgtWookie Expert

Jul 17, 2007
22,199
1,801
Scott,
Your 117mA current is flowing through the emitter resistor.

Is that what you intended?

Also, with that much of a load, your transistor will be dissipating nearly 940mW, which exceeds the package capability of TO-92 transistors; usually 625mW.

8. ### jetbat Thread Starter Member

Oct 21, 2009
14
0
I have attached a more detailed schematic. This is an audio circuit and the emitter follower is driving a tone stack. I want to experiment with transistors at different drive/current levels to see how it affects the sound. Also transistors are a bit cheaper then IC buffer chips.

As you can see, I am still trying to understand how the transistor circuit works. I am assuming that since the transistor is a current amplifier, increasing the current through it will increase the signal current into the tone stack. Is this correct?

I know that the emitter follower has a gain of less then 1 but it has a low output impedance which is good for driving the tone stack. If the way I am thinking is not the correct, then maybe I could use another transistor to raise the current before the emitter follower.

SgtWookie- Yes the datasheet does say a 625mW limit. But it also says 600mA collector current. This confuses me. How could this be a limit if my 117mA is already over the 625mW limit?

Thanks for the help
Scott

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9. ### SgtWookie Expert

Jul 17, 2007
22,199
1,801
OK, you have 117mA current flowing through the 68 Ohm resistor Re.
117mA x 68 Ohms = 7.956 Volts dropped across Re.
16v - 7.956v = 8.044v dropped across the transistor.
Since power = voltage x current, 8.044 x 117mA = 941mW.

You must stay within both the current AND power limitations of the transistor. For better performance and longer life, use transistors that have double the current and power ratings that you need.

If the voltage dropped across the transistor were only 1v, then you could pass up to 600mA, since then the power dissipation would be 600mW.

Your TL072 opamp will not perform well trying to drive a load with such a low impedance.

Your tone controls should be in a lower current area of the schematic.

10. ### jetbat Thread Starter Member

Oct 21, 2009
14
0
Thanks SgtWookie. Another piece of the puzzle falls into place in my head. And I got so wrapped up in trying to figure out the math involved in the transistor, I forgot about impedance matching/bridging.

Scott

11. ### Ron H AAC Fanatic!

Apr 14, 2005
7,018
682
You don't need the emitter follower. The TL072 should drive your tone stack directly.

12. ### Bernard Expert

Aug 7, 2008
5,019
567
Why not increase Re to about 680 ohms as you are driving into a fairly high impedance?