# Common Emitter / Voltage Divider issue

Discussion in 'The Projects Forum' started by ljduchez, Jul 3, 2015.

1. ### ljduchez Thread Starter New Member

Jul 3, 2015
6
0
Hello,

I'm trying to build a primitive common-emitter signal amplifier, the textbook one with collector and emitter resistors RC and RE, and then a voltage divider (resistors R1 and R2) to set the voltage for the base. The immediate mission is to set up a circuit with a Q point that keeps VCE to half the source voltage.

But I am finding that, while I believe I am doing the arithmetic correctly, the voltage drop across RE is not 1V like I want, but 0.05V. And VB is around 0.83V, not the 1.6V or 1.7V I was expecting. And VCE is maybe 0.02V.

What I think is happening is, my voltage divider is getting skewed by the transistor, and contrary to my wishes, much more current is flowing through the base than through R2. But that is to be expected, since RE is so much smaller than R2, right? Does this sound like anything anyone else has experienced, or does it sound like I've simply built it wrong? Suggestions, for example using a diode in place of R2 to force the voltage upward?

Here are specs:
Transistor: 2N3904
VCC: 9V
Desired IC: 10 mA
Desired voltage across RE: 1V
Base current that (I believe, per datasheets) will give me the desired Q point (VCE = 4.5V): 40 uA
RC: 350 Ohms (I used 330 Ohms)
RE: 100 Ohms
R1: 16.5k (I used 18k)
R2: 4.2k (I used 4.7k)

2. ### AnalogKid Distinguished Member

Aug 1, 2013
4,709
1,301
Schematic, schematic, schematic.

ak

Jul 3, 2015
6
0
4. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
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Note that V(b) is ~0.6V above V(e).
Pick V(c), that determines I(c) and everything else.
Pick I(c), that determines everything else...
V(c) and I(c) are not independent.

Last edited: Jul 4, 2015
5. ### michael8 New Member

Jan 11, 2015
18
2
But I am finding that, while I believe I am doing the arithmetic correctly, the voltage drop across RE is not 1V like I want, but 0.05V. And VB is around 0.83V, not the 1.6V or 1.7V I was expecting. And VCE is maybe 0.02V.

So the voltage at the collector from ground is VCE+V(RE) = .02 + .05 -> .07? What is the source of the 9V?

Since Rc is 330 ohm and assuming the 9V supply is still 9V then the current through Rc is (9-.07)/330 -> .027 or 27 mA.

If that 27 mA was flowing through RE then the voltage drop across RE would be .027*100 -> 2.7 volts which you don't see.
Is RC really 100 ohm? Check the 9 volt supply (top of Rc to bottom of Re) when the circuit is operating, also check that
the voltage across Rc is approx 9-.07 volts... Make sense?

6. ### ljduchez Thread Starter New Member

Jul 3, 2015
6
0
My voltage source is a 9V battery.

"Since Rc is 330 ohm and assuming the 9V supply is still 9V then the current through Rc is (9-.07)/330 -> .027 or 27 mA."

But the intention is that the transistor is only partially "open" at the Q point, so there should be a 4.5V drop across the transistor; and RE is still going to factor into the current, right?

Here are fresh readings courtesy of my multimeter:

RC: 321 Ohms (OK)
RE: 98 Ohms (OK)
R1: 17.6k (OK)
R2: 4.58k (OK)

VCC: 8.80V (OK)

V(RC): 8.53V (should be ~ 3.4 V)
V(RE): 0.16V (should be ~ 1 V)
V(CE): 0.02V (should be ~ 4.4 V)
V(R1): 7.86V (should be ~ 7.0 V)
V(R2): 0.83V (should be ~ 1.7 V)
V(BE): 0.67V (OK)
V(B): 0.83V (should be ~ 1.7 V)

I(C): 27 mA (should be ~10 uA) (calculates to 8.66 V(RC) compared to 8.53 measured value)
I(E): 1.58 mA (should be ~10 uA) (calculates to 0.15 V(RE) compared to 0.16 measured value)
I(1): 478 uA (OK) (calculates to 8.41 V(R1) compared to 7.86 measured value)
I(2): 37.8 uA (should be ~400 uA) (calculates to 0.17 V(R2) compared to 0.83 measured value)
I(B): 254 uA (should be ~40 uA)

There is something very suspicious about a couple of these current readings, though. I(E) should not be less than I(C), and I(2) and I(B) do not add up to I(1). I suspect there is something wrong with either the multimeter or its operator, but I'm posting the numbers anyway in case there's a clue in there.

But the measured voltages add up pretty well, even if they're not the expected values. FYI, the voltage on the battery dropped to 8.7V by the time I was done taking all my measurements, and I think I saw evidence of it dipping a bit during usage, so there's a tiny bit of play in the voltages; but overall they seem to add up.

7. ### AnalogKid Distinguished Member

Aug 1, 2013
4,709
1,301
Check your connections. An LTSpice simulation of your first post shows everything as predicted.

ak

8. ### michael8 New Member

Jan 11, 2015
18
2
I(E): 1.58 mA (should be ~10 uA) (calculates to 0.15 V(RE) compared to 0.16 measured value)

Indeed. But connecting the meter changes the circuit. Perhaps the meter has some resistance?
If the total RE (meter + 100 ohm resistor) was about 690 ohms then I'd expect 1.58 mA:
Vb of 1.7 - .6 (for Vbe) -> 1.1 V /1.58 mA -> 696 ohm implied RE.

There is something very suspicious about a couple of these current readings, though.
I(E) should not be less than I(C), and I(2) and I(B) do not add up to I(1).

Yes! And if they were all measured at the same time I'd expect them to agree. What meter are you using?

I seldom use current ranges, prefering to measure voltages across resistors and calculate the
implied current. The input impedance (resistance) of a resonable DMM is usually at least 10M ohm
which usually (but not always) has minimal effect on the circuit being measured. Current ranges seem to
me to be more intrusive.

9. ### ljduchez Thread Starter New Member

Jul 3, 2015
6
0
Craftsman 82008. Good to know, current readings can change the state of the circuit significantly. I saw something similar when verifying my resistors, sometimes they read lower while attached to the circuit because (I would assume) the current was finding multiple paths to get from one probe to the other.

10. ### ljduchez Thread Starter New Member

Jul 3, 2015
6
0
That's very good to know, thanks -- I was worried that maybe I was getting my calculation methodology from the wrong Web sites (you know how the Internet can be). I'll rebuild the circuit from scratch, and hopefully it will magically start working.

One thing that bothers me, though. In figuring out my desired base current, I looked at characteristic curves for the 2N3904, drew the diagonal line and looked at the halfway point, because to me that sounds like the only informed way to select a base current that will provide the desired result. But I've seen more guides that tell you to pick a number that you feel comfortable with, say 1/50 or 1/100 of the collector current. Isn't that, um, a hideously sloppy way to go about it? If you're doing that, how does your circuit "know" to keep V(RE) sufficiently high that current won't just blast through the base and saturate the transistor? I feel like I am missing something. (In my case, the base current is 1/250 of VC according to the datasheet, and the 1/50 or 1/100 guesses would be way off.)

11. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
The base current is (and should be) less than 10% of the current that flows in the base bias voltage divider. For the first cut, I just assume that the base current is Zero!

Another useful (first cut) assumption is that I(C) = I(E).

Last edited: Jul 4, 2015
12. ### ian field Distinguished Member

Oct 27, 2012
4,460
792

One of the radio hams has a transistor amplifier design tutorial on this that you can download as PDF.

Google his call: N1HFX and look on his site for the design tutorials.

13. ### michael8 New Member

Jan 11, 2015
18
2
The manual I found on the craftsman 82008 Autoranging Multimeter says the input impedance is 8.7M ohms (likely only on the DC volts range(s)).

It doesn't say anything about the impedance on the current ranges.

So measuring voltages changes the circuit like putting a 8.7M resistor across where you are measuring. Measuring current inserts some resistance
which isn't specified (how could you figure out how much?).

Oh, it's autoranging, I wonder if the inserted resistance on current measurement varies as it automatically switches ranges?

14. ### ian field Distinguished Member

Oct 27, 2012
4,460
792
You want current ranges resistance to be as low as possible - if the current meter's shunt drops a significant proportion of the voltage pushing the current you want to measure, it will introduce an error.

15. ### michael8 New Member

Jan 11, 2015
18
2
The beta = Ic/Ib of a transistor like the 2N3904 is a fuzzy value which depends on temperature, current (Ic) and the actual transistor (not just the type). While you might want some minumum (or sometimes maximum) beta, it's not a very stable or predictable value. So you should design a circuit using a 2N3904 so that it will still work with any 2N3904 with a beta of 100 +/- 50.

So guess a minimum beta, looking at the data sheet say 50 for the 2N3904. Pick an Ic value, assume Ie is about the same (since beta is at least 50, Ie can't be more than
(1+1/50.)*Ic).

Then pick RE and the Vb to get that value of Ie. Then roughly estimate an Ib as Ic divided by your minimum beta.

Pick R1 & R2 to get 10 times the assume Ib to flow through them (ignoring the Ib current through R1) and get the Vb you need.

There will be some pull down effect on Vb from the base current but with 10 times Ib through R1 & R2 (or more if the real beta is higher than the estimated/minimum beta) the error in Vb will be small for normal cases. Remember that slightly higher Vb will cause a slightly higher Ie but less than you'd expect since the higher Ie increases the voltage across RE.

16. ### ian field Distinguished Member

Oct 27, 2012
4,460
792

The data sheet usually specifies a minimum and maximum figure for gain - you base your calculations on the minimum figure, if the actual gain turns out to be nearer the maximum figure, DC nfb from the emitter resistor compensates for it.

17. ### MikeML AAC Fanatic!

Oct 2, 2009
5,451
1,066
I'll modify my design procedure a bit:

Ic = Ie, Ib=Ic/100 and make current in bias resistors ~10*Ib

Lets do it for Rc=330Ω, Re=100Ω, Vcc=9V, Vc=Vcc/2=4.5V

Ic = (Vcc-Vc)/Rc = (9-4.5)V/330Ω = 13.6mA
Ve = Ie*Re = Ic*Re = 13.6mA*100Ω = 1.36V
Vb= Ve+0.65V = 1.36V+0.65B = 2.01V
Make Rb1 (the upper resistor) carry 10*Ib; so Rb1= (Vcc-Vb)/(10*Ic/100) = 7V/1.39mA = 5KΩ.
This requires that Rb2 carry 9*Ib; so Rb2=Vb/(9*Ic/100) = 2.01V/(9*Ic/100) = 1.6KΩ

Let's see how we did:

Last edited: Jul 4, 2015
18. ### ian field Distinguished Member

Oct 27, 2012
4,460
792
If the transistor has high enough gain, Ib can be too small to influence the result much.

But higher power devices tend to sacrifice gain a bit and Ib can make a pretty big dent in the calculations.

Around the late 80s/early 90s - a typical horizontal output transistor usually had a gain spread something like 2 - 8!

Years ago I wanted to use one in a transistor assisted contact on a small motorcycle - the alternator had nothing in reserve and the PTC LT winding could take up to 8A when cold, as it was theoretically possible for the transistor I obtained to have a gain as low as 2 - I would have had to design for as much as 4A to be available for base drive.

19. ### michael8 New Member

Jan 11, 2015
18
2
Going back to the original problem, the strange Ic value. Looking at the measurements in #6 I'm ignoring the Ie measurement as I suspect it's vastly altered by the meter being in series with RE (and thus increasing the RE value, but demonstrating that higher RE values do lower Ic & Ie).

So there's the V(RE) of .16 V and RE is 98 ohms. This implies a RE current of 1.6 mA, however V(RC) of 8.53 V and RC of 321 ohms implies an Ic of 25.5 mA. The provided schematic doesn't show any low resistance paths which would allow for an Ic of 25+ mA to flow with the specified conditions so either the measurements are wrong (sounds very unlikely) or else the conditions are different. A much lower RE value (wiring short?) seems the simplest possiblity, something like .16V/27mA -> about 6 ohms.

How is this circuit assembled?

20. ### ljduchez Thread Starter New Member

Jul 3, 2015
6
0
Update: I think I found the culprit, a busted-off piece of wire stuck in the breadboard on the row where the collector was. I rebuilt the circuit elsewhere on the breadboard and it's behaving properly now (at least in terms of getting me the Q point I was looking for). Next step is to try to find a signal to amplify; I'll have to root around in the basement for a microphone or something.

I don't know what the wire was connecting to what, but it's a ratty old second-hand breadboard and I should probably invest in a new one.

Thanks for all the help, and I'm sorry to waste people's time on what turned out to be a short or at least an unintended connection. I guess that explains where the missing current was going.