# Common-Emitter Amplifier (determining base voltage and current)

#### BarryBozeman

Joined Apr 1, 2016
41
Hi folks -
I'm currently trying to design a Common-Emitter amplifier based on the "Comprehensive AC Coupled Design" on this page: http://www.radio-electronics.com/info/circuits/transistor/common-emitter-amplifier-design.php

I'm a little bit uncertain about a few things and had the following questions:

1. I'm calculating this circuit based on a speculation about what sort of current I might need at the output stage. I'm mainly using this circuit for voltage gain, though. How does one design this circuit using the intended voltage gain as the starting point?

2. On Step 4, the instructions are to "determine the base current by dividing the collector current by hFE." I'm a little unsure about where the decimal point goes here. Let's say I've got a 1mA current divded by an hFE of 160. Is the solution here going to be 0.00625mA?

3. Step 5 explains the voltage needed at the base. It's going to be 1.6v, in my particular application. If I know this, can I just use the voltage divider provided by R1 and R2? Or do I need to factor the current into this decision somehow? I'm a little unclear on what is being recommended here. Do I add R1+R2 together and use the V+ supply to get the range?

Thanks all...

#### #12

Joined Nov 30, 2010
18,224
Here's what you do:
Subtract 1.5V from the power supply voltage.
Set R3 to use up half the remaining volts when the necessary current is flowing through it.
Set R4 to use up a volt at that current.
Calculate R1 and R2 to flow 1/10th of the collector current.
Adjust the values of R1 and R2 to hold the base 1.6 volts above ground.

Did you notice I never said anything about the gain of the transistor?
The goal is to design circuits that do not depend on any particular gain number.

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#### Jony130

Joined Feb 17, 2009
5,457
1 - You need to know how much voltage gain do you want and what is the load resistance.
On Step 4, the instructions are to "determine the base current by dividing the collector current by hFE." I'm a little unsure about where the decimal point goes here. Let's say I've got a 1mA current divded by an hFE of 160. Is the solution here going to be 0.00625mA?
Very good. But notice that 0.00625mA if we move the decimal point three places to the right we get 6.25μA (microampers)

#### BarryBozeman

Joined Apr 1, 2016
41
Calculate R1 and R2 to flow 1/10th of the collector current.
Adjust the values of R1 and R2 to hold the base 1.6 volts above ground.
Did you notice I never said anything about the gain of the transistor?
The goal is to design circuits that do not depend on any particular gain number.
Not sure why that's the goal. I need a particular range of gain, that's why I'm building the circuit..

Also, I'm a little uncertain about how to calculate R1 and R2, which is what question 3 was about..

Thanks..

#### BarryBozeman

Joined Apr 1, 2016
41
Hoping to get a voltage gain of somewhere between 6 and 8. Not sure how to calculate the load resistance, though.

Thank you for the link re: current/voltage for the transistor base. That makes it a lot clearer.

#### #12

Joined Nov 30, 2010
18,224
Give me the power supply voltage so I can do the math. Without that, I can only answer in formulas.

#### BarryBozeman

Joined Apr 1, 2016
41
Just to clarify...the current for R1 would be 11 the base current, and the current for the R2 would be 10 times the base current?

#### BarryBozeman

Joined Apr 1, 2016
41

#### Jony130

Joined Feb 17, 2009
5,457
Not sure how to calculate the load resistance, though.
You do not know the load resistance ?? What you are gonna to connect to your amplifier output?
As for R1 and R2 simply try use the Ohm's law and rule of thumb
R1 = (Vcc - Vb)/(11*Ib)
R2 = Vb/(10*Ib)
And for for good thermal stability we usually choose Re = (0.1 ... 0.4)*Vcc/IC or Ve large then 1V (Ve>>Vbe)
Vb - Voltage at base
Ve- Voltage at emitter.

#### #12

Joined Nov 30, 2010
18,224
So, you're using up a volt in the emitter resistor and you have to allow for the transistor to use up some voltage because the gain of the transistor approaches zero when the Vce approaches zero. I choose 1/2 volt for the transistor. You have 3.5 volts left, so the collector resistor uses up 1.75 volts at the required current. Now we look at the ratio of the collector resistor to the emitter resistor. It is 1.75:1 That's your DC gain. You want an AC gain of 7? You are going to have to use a capacitor in parallel with the emitter resistor to boost the AC gain. Pretending the collector current is 1 ma, the collector resistor would be 1.75K and the emitter impedance should be one seventh of that, 250 ohms. Add 330 ohms in parallel with the emitter resistor but place a capacitor in series with the 330 ohm resistor. Calculate the capacitor to have an impedance of 330 ohms at the lowest frequency you want.

Now, back to the base bias. At a certain current and a certain Vce, the transistor has a current gain of 160. You can see that the Vce will become 1/2 volt during some parts of the AC signal. That is NOT the 5 volts Vce at which the gain was specified. That is why you throw in a bias circuit carrying a lot more current than what is required when the transistor has 5 volts across it. You are thinking, "10 times what the advertised gain would require." OK. Try that and watch the distortion increase as the Vce becomes 1/2 volt.

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#### dannyf

Joined Sep 13, 2015
2,197
How does one design this circuit using the intended voltage gain as the starting point?
The voltage gain of this circuit is unknowable: because you don't know the input impedance of the next stage, and more importantly, the Re resistor has a capacitor shunt so its ac impedance is "zero". In most those cases, you have to experiment with it, or go by your experience.

The key in many of such designs is to pick Ic. Typically, they range from 1ma to 5ma. Sometimes higher and sometimes lower.

From there, pick R3 to maximize power transfer to the next stage. Pick R4 to maintain bias / thermal stability (typically < 110ohm). Pick C4 for bandwidth / RF concerns. 11n is a fairly typical figure.

Once Ic/R4 is picked, you know Vb. Pick R1/R2 (sufficiently small) so Ib is not a big concern.

Done.

#### BarryBozeman

Joined Apr 1, 2016
41
I'm sorry.. I feel a little confused again by all of this new information..I'm going to trace my steps and maybe if one of you is feeling patient, you can tell me if I'm on the right track.

1. I have a +/-0.25v signal. I need it to be closer to 0-5v, but I know that with a 5v power supply this isn't going to happen. My goal is to feed an audio signal into an Arduino. I'm not sure what current is expected there.

2. Using Ohms Law, I take half of the power supply (2.5v) and the arbitrarily-selected 1mA of current to determine a resistor value of 5k for R3 (which I think we're calling Rc now, due to it being attached to the collector of the transistor.)

3. 10% of the supply is 0.5v, which I use to calculate Re (resistor attached to the emitter) with the same current value. I get 500 ohms...but I'm probably going to round down to 470 ohms, since the world is unfair.

4. Base current is the collector current divided by the HFE. The HFE is 200, the base current is 1mA. Base current = 0.005mA.

5. Resistor 1 is calculated for the following:
Current = 11 * Base Current (0.055mA).
Voltage = 5v(power supply) minus Re(0.5v) minus diode drop (0.6v). That means 3.9v.
The resistor value: +/- 65k...though 68k will be substituted.

6. Resistor 2
Current = 10 * Base Current (0.05ma)
Voltage = 1.1v (Re + Diode Drop)
Resistor value: 22k.

Is this right so far?

#### dannyf

Joined Sep 13, 2015
2,197
4. Base current is the collector current divided by the HFE. The HFE is 200, the base current is 1mA. Base current = 0.005mA.
As long as hFE is sufficiently large (and bias network sufficiently small, resistively), you don't really need this. Instead, pick R1/R2 based on your desired input impedance - if you care. Typically for audio, anything less than 47K will work.

since Ic = 1ma, Re = 500ohm, the voltage drop on Re is 0.5v. So Vb sits at 0.5v+0.7v = 1.2v. So pick R1/R2 so that

1) R2 / (R1+R2)*5v = 1.2v, and
2) R1//R2 = 22k (for example).

The 10x of base current isn't a bad rule of thumb as well - but it may result large resistors for transistors with high hFE. I usually assume hFE=100 for small signal applications.

#### dannyf

Joined Sep 13, 2015
2,197
BTW, don't shun Re with that capacitor. Or your gain will go through the roof.

#### ian field

Joined Oct 27, 2012
6,536
Here's what you do:
Subtract 1.5V from the power supply voltage.
Set R3 to use up half the remaining volts when the necessary current is flowing through it.
Set R4 to use up a volt at that current.
Calculate R1 and R2 to flow 1/10th of the collector current.
Adjust the values of R1 and R2 to hold the base 1.6 volts above ground.

Did you notice I never said anything about the gain of the transistor?
The goal is to design circuits that do not depend on any particular gain number.
The design sheet I read suggests: Decide how much voltage you can afford to drop across R4, then calculate R1 & R2 to set the bias 0.7V more than that. Decide what collector current you want and calculate R4 to set that value. Finally; calculate the value of the collector resistor to drop Vcc/2 at the collector current you've set for.

#### eetech00

Joined Jun 8, 2013
3,429
Hello

If this an audio signal being read by the ADC of a Microcontroller (OP mentions "input to arduino"), It seems like an op amp buffer/level shifter would be easier.