[SOLVED] Using voltage divider bias for transistor circuit

Someone asked question in forum about Transistor Hfe DC gain and I read a reply which stated that "Make Transistor circuit which is independent of Hfe". So they advice to use voltage divider bias.
If I accept that transistor circuit can be made independent of Hfe then all transistors are same despite their max ratings.
Hfe is the parameter which differentiates between different transistors.
You say that you can find value of Ic in voltage divider bias without using Hfe so it is independent of Hfe but I say can you find Ib in voltage divider bias without Hfe.
Transistor starts from base current and there is no Ic possible without Ib so only find value of IC in voltage divider bias doesn't mean that Ib is not important and if you consider Ib then you have to use Hfe

Re: Voltage Divider Bias

If you want a transistor amplifier to have a 1mA collector current and assume the hfe is 500, the base current will be 2uA. So, if you make the base bias chain consume 100uA, the 2uA will have very little effect. If another sample of the same transistor had hfe=100 then the base current would be 10uA. That would affect the base bias voltage a little, but not enough to be a problem.

The hfe is often not a critical parameter when doing design, provided it is high. If the hfe is low, such as 20, then it can be a problem and cannot be ignored. When it is >100 then transistors are chosen based on other parameters such as speed, capacitance, collector current, power dissipation.

Keith

Re: Voltage Divider Bias

The use of voltage divider bias enables you to do the opposite. Decide Ib by nailing down Ic. Ib will set itself automaticly up by drawing the current it needs from the voltage divider to maintain Ic at the value set by the emitter resistor. But because the Ib current is so small it hardly affects the voltage divider at all.

The base voltage is decided by the voltage divider is it not ?. The BE junction of a conducting transistor is around 0.7V over most of its normal operating range. Which means that the emitter voltage will be Vbase-0.7V. Since the Emiter current is approximately the Ic current we see that we can set the Ic current by simply choosing the correct value of the emitter resistor as long as we know how we set the voltage divider bias

Ie = Ve / Re = (Vbase-0.7V) / Re = (Voltage_divider_bias - 0.7V) / Re
And for Hfe >= 100 we see that Ie is within 1% or less from being as large as Ic.

See, we never used Hfe in these calculations. And we have a expression for the Ic current as a function of only the voltage divider bias and the emitter resistor since we can regard the BE junction drop as 0.7 for most applications.

If you try to design gain stages using Ib you will run into alot of trouble since you would have to measure every transistor's Hfe in your design.

Re: Voltage Divider Bias

By the way, the link Transistor Amplifiers - The Common Emitter Amplifier I posted in the other thread may be helpful as it gives a worked example.

Keith.

Re: Voltage Divider Bias

I agree with the above posts.

If you study how bipolar transistors operate, you will find that the collector current is determined by the charge in the base-emitter region. Hence they are voltage controlled current sources since the charge is proportional to the BE voltage.

The collector current is an expontential function of the base emitter voltage.

The base voltage is defined by the divider resistors and the emitter voltage will rise to a point such that Ie = (Vb - Vbe)/Re.

There is negative feedback. eg. if the transistor temperture changes, Vbe changes to compensate.

The hFE will have virtually no effect, provided that the current through the voltage divider resistors is much greater than Ib.

Consequently, changing the transistor for a different type with a different hFE will have virtually no effect.

Re: Voltage Divider Bias

In a voltage bias transistor circuit with a collector resistor and emitter resistor and r1/r2 forming a voltage divider on the base how is IB calculated ?

Re: Voltage Divider Bias

The simplest way is to use Thevenin's Theorm.

The Thevenin voltage is Vth = Vcc * R2/(R1 + R2)

Note: this is the voltage you would measure with a volt meter at the junction of R1 & R2 without the transistor connected.

The Thevenin resistance is the parallel combination of R1 & R2, ie. Rth = R1 * R2/(R1 + R2)

The base current is then Ib = (Vth - Vbe)/Rth

This is correct if there is no emitter resistor.

If there is an emitter resistor (Re), then the first step is to calculate the emitter voltage (Ve = Vth - Vbe) then the base current is

Ib = (Vth - Vbe - Ve)/Rth

Re: Voltage Divider Bias

crazyjohn said:

In a voltage bias transistor circuit with a collector resistor and emitter resistor and r1/r2 forming a voltage divider on the base how is IB calculated ?

Click to expand...

Normally you would design for a collector current and the base current (if required) would come from that. So, in the attached example I have chosen a 1mA collector current. Biasing the base at 6V with the simple splitter R2/R3 means the emitter will be at 6V-0.7V = 5.3V. For 1mA through R1 then R1 must be 5.3k. The collector current is now 1mA. The base current will be 1mA divided by hfe (around 127 in this example) resulting in 7.8uA of base current. The base current can then be checked against the current through R2/R3 (600uA) to make sure it is not significant, although you would have known that when you started. e.g. if you wanted 1mA collector current then you would guess at hfe=100 for example and make sure the current through R2/R3 is a lot more so the base current will have little effect.

I hope that makes sense!

Keith.

I agree, but 6 Volt is rather large. It reduces the collector voltage range.

I would make it about 2 Volt.

I was just giving an example of the calculations! If I was designing an amplifier I would have added an emitter resistor coupling capacitor as well. I thought the discussion was about how the DC conditions are calculated.

Keith

Re: Voltage Divider Bias

ljcox said:

If you study how bipolar transistors operate, you will find that the collector current is determined by the charge in the base-emitter region. Hence they are voltage controlled current sources since the charge is proportional to the BE voltage.
The collector current is an expontential function of the base emitter voltage.
.

Click to expand...

Thank you ljcox for this statement!
It's a pitty, but in a lot of papers, articles - and even textbooks - you can find the wrong formulation "BJT is current controlled".
Apparently there are some people who think the approximation Ic=hfe*Ib is identical to a physical law in the "cause-and effect" sense.

Yes, it is also in the BJT tutorial in this forum.

The expotential form is in the tutorial about the Current Mirror.

It is not possible to explain the Current Mirror using Ic = hFE * Ib

However, for most situations, the expotential form is unnecessary and is too complex.

Hence the Ic = hFE * Ib is adequate for such situations.

---------- Post added at 21:22 ---------- Previous post was at 21:21 ----------

keith1200rs said:

I was just giving an example of the calculations! If I was designing an amplifier I would have added an emitter resistor coupling capacitor as well. I thought the discussion was about how the DC conditions are calculated.

Keith

Click to expand...

Yes, I understand that, but I did not want beginners to think that Vb had to be half Vcc.

ljcox said:

Yes, I understand that, but I did not want beginners to think that Vb had to be half Vcc.

Click to expand...

I wasn't aware that I had said it should be half Vcc. Are you saying it should always be 2V?

It was an EXAMPLE.

Keith.

I did not want to start an argument.

No, you did not say it should be half, but your example showed it as half.

As I said, I did not want beginners to think that it had to be half.

No, I'm not saying that it should always be 2V.

However, I think you will find that it is usually between 1 ~ 2V in order to leave head room for the collector swing & provide reasonable thermal stabitity.

ljcox said:

I did not want to start an argument.

Click to expand...

But you did!

I design circuits to a specification. I don't have fixed rules, I have designed amplifiers for 300V supplies and 1V. I choose the design to suit the requirements. I have been trying to give a simple example which can answer the questions raised. I hope it helped the person who asked the question.

Keith

I thought a forum was where people ask a question and someone provides an answer.

Other people then respond to: -

1. Provide additional information

2. Raise a related issue

3. Provide an alternative explanation

4. Correct an error in the answer

As far as I'm concerned, my comments fall into category 1 and/or 2.

I am quite happy for anyone to comment on or correct anything that I write provided that it is not abusive.

No, what you did was say there was an error in my answer and imply my example was a poor one. I do not accept that there was anything wrong with the example I gave to illustrate the calculations involved.

Keith

ljcox said:

I agree, but 6 Volt is rather large. It reduces the collector voltage range.

I would make it about 2 Volt.

Click to expand...

This what I said, and I don't see the word "error" in it. It was simply a comment.

It is a criticism implying that the example I gave is incorrect. The example I gave was in response to this:

crazyjohn said:

In a voltage bias transistor circuit with a collector resistor and emitter resistor and r1/r2 forming a voltage divider on the base how is IB calculated ?

Click to expand...

not "how do I design a transistor amplifier".

I chose the voltages to make the maths easy. By choosing half rail you don't need a calculator to see that it should be 6V and that the effect of the base current is only to drop the voltage by 40mV. I chose 1mA to make the maths easy, and hence a 5.3k resistor.

Keith.