AutoBiasing Transistors

[UTMonkey]

Yes, me again.

I have worked through another of Harry Lythall's biasing examples. **broken link removed**

This time it promises to improve the control of Transistor gain using an emitter resistor and potential divider to bias the transistor.

Here is my working out:-

Assumption 1. The collector current is 10mA
Assumption 2. The transistor base is fed from a potential divider of 2V
Assumption 3. The gain (for sake of argument) is 200.

Emitter Resistor
Re = (Vb - 0.7) / Ic
Re = (2 - 0.7) / 10mA
Re = (1.3) / 0.010
Re = 130 ohms

Collector Resistor
Rc = (Vcc - Mean Level between 10V and 2V) / Ic
Rc = (10 - (10+2/2)) / 10mA
Rc = (10 - 6) / 10mA
Rc = (4) / 0.010
Rc = 400 ohm

Potential Divider
Total Resistance = Vcc / (Ic /200)
" " = 10 / (10mA / 200)
" " = 10 / (0.010 /200)
" " = 10 / (0.00005 or 50uA)
" " = 200,000 (200k)
Use values 40k\160k for 2V supply to base.

As you can see from the attachment I have plugged in the values, but as you can see the meter readings arent immediately what I expected.

For instance I was expecting to see 6V and 10mA at the collector. Also a PD of 2V at the potential divider junction feeding the base.

I have been thinking this through and I just wanted confirmation.

1. The reason the base voltage is registering 1.298v is because it has dropped 0.7v through the "be" junction of the transistor?

2. Because the simulation has done this, the Collector Voltage has dropped to 8.068v. The only way I can explain this is by Vcc - (Rc x 4.83mA) = 8.068?

Do you agree with this?

I can't explain Ic though?

Thanks in advance.

Mark

 

[kchriste]

The problem is that you've calculated the values for the potential divider wrong. As a rule of thumb the thevenin resistance of the divider should be at less than: Beta * RE * 0.1 (200 * 130 * 0.1 = 2.6K) for a "firm" divider. In other words, if your base current is going to be 50uA then you should have at least 10 times that flowing in the divider.

 

[RadioRon]

To add to what kchriste has explained, if you want 10 mA to flow in the emitter, and if the gain of the transistor is 200, then you need about 50 uA to flow into the base. This 50 uA has to flow through the 160Kohm resistor in addition to the voltage divider current. The trouble is that 50uA will cause a huge voltage drop in the voltage divider because 160K is just way too high. You want the voltage divider current to be much more than the base bias current so that the base bias current doesn't change the base voltage very much.

 

[k7elp60]

I have calculated and built many transistor circuits. I always us the minimum Beta for the specified Ic to calculate the base current. I have found what Radio Ron says about the current in the voltage divider. I calculate the value of R4(base to ground) based on the Ib x 10, then choose the next lower standard value of resistor. I calculate R3(base to Vcc the same Ib x 10 and then choose the next larger standard value.

 

[audioguru]

Your transistor amplifier has a voltage gain of only 3.
400/130= 3.08. Nearly no voltage gain.

Do you know what to add to increase the AC voltage gain to 100 (but also increase the distortion)?

 

[Sceadwian]

3x voltage gain is what, a pinch less than 4db?

 

[audioguru]

3x voltage gain is what, a pinch less than 4db?

No.
6dB is a voltage gain of 2.
10dB is a voltage gain of 3.16.
9.54dB is a voltage gain of 3.

 

[UTMonkey]

Thanks guys.

How about this then...

Based on KChriste, if I expect a current like 50uA at the base I should be working towards a divider with 10 times the amount of current going through it.

Ok:-

1. 50uA x 10 = 500uA
2. PD Resistance relative to ground = 2v / (500uA - 50uA) = 4444 ohms
3. PD Resistance relative to Vcc = (10v-2v) /500uA = 16000 ohms

But surely now that I have done that I need a resistor in series with the base of the transistor.
4. Base Resistor = (2v -0.7v) / 50uA = 26000 ohms

Is this right? I was hoping this example would do without a series resistor at the base.

 

[audioguru]

You don't need to have a resistor in series with the base. The 16k resistor feeds the 4.4k resistor and also feeds the small current to the base. Then the base will be biased at 2.0V. The emitter will be about 1.3V.

Now re-calculate the input resistors using standard resistor values (4444 ohms????)

 

[UTMonkey]

hokey cokey, let me see.

Using the E12 Series
The closest value to 16k is 15k
8v / 15000 = 533uA
and the closest to 4.4k is 4.7k
2v / 4700 = 426uA

Does this mean using these values there is (533uA - 426uA) 107uA feeding the base?

Am I doing this right?

 

[audioguru]

You want the base current to be 50uA and 2V.
The 4.7k resistor will draw 426uA at 2V and the 15k resistor will supply up to 533uA to 2V. But the base current is typically only 50uA so the base voltage will be a little higher than 2.0V then the collector voltage will be lower than 6V.

Readjust the value of the emitter resistor to get the operating voltage at the collector correct.

 

[UTMonkey]

I like this!

ok, if I want the 4.7k resistor to supply 450uA then:-
4700 / 0.000450 = 2.115v

So to reconsider Re:

Re = (Vb - 0.7) / Ic
Re = (2.115 - 0.7) /0.010
Re = 141.5 ohms
Re = 150 ohm (e12 standard)

Close?

 

[audioguru]

You are very close, if the hFE of the transistor is 200, not 100 and not 300.

 

[UTMonkey]

Thanks AudioGuru, what do you mean by (200, not 100 and not 300)?

 

[audioguru]

Thanks AudioGuru, what do you mean by (200, not 100 and not 300)?

The hFE (current gain) of the 2N2222 is from 100 to 300, but is typically 200.
You calculated only when the hFE is 200.
Calculate what happens when the hFE is 100 and when it is 300.

 

[UTMonkey]

Will do, at work at the moment so will look at it at lunch time.

Thanks for this.

 

[UTMonkey]

Just looking at this now, one question though. looking at your diagram you notes currents 520uA,472uA and voltages 6.16v, 2.2v.

How did you come by these?

 

[audioguru]

Just looking at this now, one question though. looking at your diagram you notes currents 520uA,472uA and voltages 6.16v, 2.2v.

How did you come by these?

I just used Ohm's Law. My numbers this time are slightly different.

 

[UTMonkey]

I think I am just a little confused that you calculated "from" the emitter to get everything else. But when you say "your figures are slightly different" be honest was I wrong?

Anyway, that little exercise.

What if Hfe is 100?

Ic = 10mA
Ib = Ic / 100 = 100uA
Desired PD resistance = 100uA x 10 = 1mA

Resistance relative to gnd = 2v / (1mA - 100uA) = 2222 (e12 = 2.2k)
Resistance relative to Vcc = 8v / 1mA = 8000 (e2= 8.2k)

Adjusting the Emitter resistor
Vb = 2.2k / 900uA = 1.98v
Re = 1.98v - 0.7 / 10mA = 128 (e12 = 120)


If this is ok I will let you have the 300 gain results. (bet you cant wait)

Mark

 

[Nigel Goodwin]

You can simplify things VERY easily - by making a number of assumptions:

1) The emitter is 0.7V lower than the base - a reasonable average value.

2) The emitter current is the same as the collector current - you can ignore the much smaller base current.

3) There is no base current - as long as the base bias resistors are fairly low values, you can safely assume this (at least five times the current down the potential divider as the base draws).

Notice I made no mention at all of the Hfe of the transistor - it doesn't matter - as long as it's high enough, which it should be for reasonable emitter and collector resistor values.

These simple assumptions make life MUCH easier, yet the calculations are still going to be far more accurate than the tolerance of the resistors you use in the circuit.