help with voltage divider biasing calculations

[Veraxis]

I have a seemingly simple problem. Perhaps the solution is obvious and I just don't see it, or perhaps I am missing something and I just have not realized it. This is not for any class or assignment, but for a small project that I am working on. I need help figuring out the general equations to solve for the resistor values in a voltage-divider-biased BJT, as shown:

**broken link removed**

Vcc, Ib, and Beta are all known, but all of the resistor values are unknown. I have been looking at guides all over the internet and have seen a number of equations listed, but I cannot find a single source that clearly lists all of the equations that I need to solve for all of the unknowns. Any time that I have seen the equations done out in an example, it was always with the resistor values already known and working backwards to find everything else from there. I have been trying to work out all of the equations on my own from what I have seen, but all I have managed to do is give myself a headache. (I looked at the Wikipedia article on transistor biasing, but, even there, I was unable to make sense of the equations listed.)

I would really appreciate it if someone could point me in the direction of a website that might give me a clearer explanation, or just list out the equations right here. If I am somehow missing some crucial piece of information to solve this, I would also appreciate advice on that as well. If it turns out that I am just an incorrigible idiot, I apologize for wasting your time.

 

[MrAl]

Hi,


This is almost like asking, "I have a handful of resistors, what value are they?". I'll explain why.

If we know Ib and the beta then we know Ic and Ie, but because we dont know what Re is we dont know what Ve is (voltage at the emitter to ground). Because we dont know what Ve is we dont know what Vb (voltage at the base) is.

Even knowing what Vb is, we dont know what R2 is so we dont know what R1 is.

So we cant really solve for any resistor values unless we know more about the circuit. If we knew what Re was it would help, but we also need to know what R2 is unless we were going to calculate a range of values for the resistors, or just an equation to relate the values to each other (as R2 below).

Here are some equations:

Ie=Ib*(Beta+1)
Ve=Ie*Re
Vb=Ve+0.7 (approximation)
Rin=Vb/Ib
R1 such that Ib=Vcc/(R1+Rin), solve for R1 max,
so R1 max=(Vcc-Ib*Rin)/Ib
If R1 is at max, then R2 is at infinity (open circuit)
If R1 is less than max, R2 is:
R2=Ib*Rin*R1/(Vcc-Ib*R1-ib*Rin)


So you can see that there are many arbitrary choices for R1 and Re without imposing more constraints, and also Rc (the collector resistor) is independent of the others within limits to keep the transistor out of saturation with the given Ib and Beta (and Re).
Also, R1 has a maximum value where R2 becomes infinite, and with R1 less than max R2 is calculated from the above.
Rc will also have a maximum value which depends on Re and Vcc to keep the transistor out of sat.

 

[Diver300]

Ie=Ib*(Beta+1)

This is correct, but needs to be approached with care. The gain, Beta, is not a constant. It varies as the collector current varies, and as temperature varies. The gain will reduce near saturation and it will not be the same for all transistors of the same type. Most transistor data sheets do not quote a maximum value of Beta.

If your design needs a certain value of Beta, it won't work in practice, or at minimum it will need adjustment if the transistor is changed.

Transistor circuits are often switching circuits or linear circuits.

For a switching circuit, allow for a Beta that is 5 - 10 times lower than the figure quoted for the transistor, so that the transistor will be properly saturated. You may be able to use a smaller margin if the Beta is quoted at or near saturation.

For linear circuits, make sure that the circuit will work with any value of Beta from about half what is quoted, up to infinity. In your circuit, you should allow for Ib to be anywhere from 0 to twice what it should be. As Ib is quite small, that isn't usually a problem, and calculating it gives you a feel for how much voltages will vary in practice.

For instance, if Ib is 0.1 mA, then see what Vb would be with Ib = 0 and Ib = 0.2 mA. If the voltage change is only 0.25 V on a 12 V circuit, that will probably work fine.

 

[Veraxis]

So you can see that there are many arbitrary choices for R1 and Re without imposing more constraints, and also Rc (the collector resistor) is independent of the others within limits to keep the transistor out of saturation with the given Ib and Beta (and Re).
Also, R1 has a maximum value where R2 becomes infinite, and with R1 less than max R2 is calculated from the above.
Rc will also have a maximum value which depends on Re and Vcc to keep the transistor out of sat.

so basically I just have to pick an arbitrary value for one of them (within limits, of course)? that makes my life a bit easier.

Geez, and I had been sitting here driving myself crazy trying to find the equations to solve for all four values!

Thank you, I should have come here first rather than wasting all this time trying to derive the equations myself.

 

[alec_t]

MrAl and Diver300 are right, but here are some rules of thumb that may help for a quick-and-dirty design:-
Assume beta = 10.
Make I1 at least 10 x Ib; so I2 ~= I1.
Make Rc no more than 10 x Re.

 

[Veraxis]

MrAl and Diver300 are right, but here are some rules of thumb that may help for a quick-and-dirty design:-
Assume beta = 10.
Make I1 at least 10 x Ib; so I2 ~= I1.
Make Rc no more than 10 x Re.

why would beta equal 10? perhaps I am misunderstanding, but I remember seeing something like that in an article about using transistors for switching (here), but here I am using it as an amplifier, so shouldn't beta be something closer to one or two hundred (or whatever is on the datasheet)?

 

[Veraxis]

going back to the equations listed by MrAl above, I noticed one that I have not seen before:

Rin=Vb/Ib

I assume that this relates to impedance matching? should Vb also be chosen to optimize input impedance?

 

[RCinFLA]

Beta versus DC collector current depends on physical size of device, holding other configurations of silicon the same. If you look at an ft curve, which is an AC current gain profile, the DC beta will also peak out near the ft peak.

Beta also drops as the collector to emitter voltage gets below 0.5 to 0.9 vdc. This is why a switching configuration uses a low beta number, called forced beta, to figure base drive requirements.

For silicon, the Vbe will be between 0.5v and 0.9 vdc depending on how much base current is applied for device physical size. Usually for a device biased for linear operation the base voltage will be in the 0.55 to 0.65 vdc range. For first pass approximation you can use 0.6 vdc.

The two voltage divider resistor values should be low enough so variation in beta caused base current over part to part variation and temp keeps the base voltage fairly constant. Usually you can say the input biasing resistors should have a current of 5 to 10 times the maximum base current based on worse case device dc beta.

For DC bias stability consideration you want the voltage drop across emitter resistor to be greater then the Vbe to keep the temp effect on Vbe (about -2.5 mV/deg C) from causing bias current variations. The greater the voltage drop across the emitter resistor the less the effects of Vbe temp variation on Ic current. Typical number is 1 to 3 vdc drop across emitter resistor but it depends on your overall supply voltage so you keep a minimum Vce voltage on device.

Example: 10vdc supply, desire 10 mA Ic with a device of DC beta of between 50 to 250.

At a beta of 50 the device will need 10mA/50 = 0.2 mA. Assume 0.6 v Vbe. For 1.5 vdc drop across emitter resistor you would need a 1..5v/10mA = 150 ohm emitter resistor.

Base bias voltage needs to be Vemitter resistor + Vbe = 1.5v + 0.6v = 2.1 vdc. If I consume 5 times the current for input divider resistors then current through top bias resistor would be 0.2 mA x 5 = 1 mA. With 10vdc supply and 2.1 vdc base bias voltage then top resistor divider would be (10v - 2.1v)/1mA = 7.9k. You might want to go to 8.2k for standard value resistor.

With a 8.2k top resistor then current would be (10v-2.1v)/8200 = 0.96 mA. That current is split between base current and bottom bias resistor. Bottom bias resistor would then have (0.96mA - 0.2mA)= 0.76 mA. With 2.1 base bias voltage the bottom input resistor divider would be 2.1v/0.76mA = 2.75k. You might use a 2.7k standard resistor value.

Collector resistor depends on how much gain or voltage swing on output you want but should not be any higher in value that results in a minimum Vce. For above example, keeping a minimum 3 v Vce would be 3v + Vemitter resistor = 3v + 1.5v = 4.5 vdc. 10v -4.5v = 5.5v drop across Rc maximum. At 10 mA that would be 5.5v / 10 mA = 550 ohms for maximum value.

To maximize output voltage swing the collector resistor would be based on 10v - 1.5v emitter resistor drop - min Vce. 10v - 1.5v - 0.5v = 8v. Divide this by two for equal up/down voltage swing on collector so idle bias across collector resistor would be 4v. 4v / 10 mA gives 400 ohms.

Assuming no bypass cap across emitter resistor, the AC gain will be close to 400 ohms / 150 ohms = 2.6 with maximum output voltage swing of close to +/- 4 volts.

 

[alec_t]

why would beta equal 10? perhaps I am misunderstanding, but I remember seeing something like that in an article about using transistors for switching (here), but here I am using it as an amplifier, so shouldn't beta be something closer to one or two hundred (or whatever is on the datasheet)?

That's the problem; we don't know what the beta actually is. So when designing a circuit we have to assume the absolute minimum value it possibly could be. We're talking rule of thumb here. Suppose you decide you want 10mA collector current. If we take beta = 10 that gives design base current = 1mA. That will comfortably exceed the actual base current needed and is the basis for then deciding on the current through the base bias resistors.

 

[audioguru]

For a low current amplifying transistor, beta is much higher than only 10. It is typically 230 for a 2N3904 but could be as low as 100 or as high as 300. At very low or very high currents the beta is as low as 30.

A BC54x transistor has a very wide range of beta (110 to 800 at 2mA) but you can buy a low beta "A", a medium beta "B" a high beta "C" version where they test, sort and mark them.

A general rule is that the current in the input voltage divider resistors should be 10 times the typical base current. Then the circuit will work if the beta is half or is double the typical value.

 

[KeepItSimpleStupid]

Could this be moved to tutorials? It's quite good.

 

[MrAl]

Hi again,

Yes the operating point can change if the Beta changes but also if Vd (the base emitter diode voltage) changes with temperature.

To investigate these effects, you can use the following equations...

To start, you'll want to select an output operating point like 1/2 of Vcc. Vc is the collector voltage and we'll call that the output.
With Rc the collector resistor and Re the emitter resistor selected by external circuit requirements, we have:
Ic=(Vcc-Vc)/Rc

and this means that the base current is:
Ib=Ic/Beta

and so the emitter current is:
Ie=Ib+Ic

and so the emitter voltage is:
Ve=Ie*Re

and so the base voltage is:
Vb=Ve+Vd

and so what we are calling Rin is:
Rin=Vb/Ib

This means the maximum value of R1 with R2 open is:
R1max=Vcc/Ib-Rin

That's the max value of R1 with no R2 present, but we get better results with R2 present and R1 lower than that, so with a value of R1 lower than that we have a value for R2:
R2=Ib*Rin*R1/(Vcc-Ib*R1-Ib*Rin)

where Ib was calculated above.

So Rc and Re are selected based on external circuit requirements and of course gain, and R1max is the max value of R1 allowed, and R2 is calculated as R1 is brought down lower than R1max in order to get a more stable circuit with temperature.
The drawback to lowering R1 is increased sensitivity to Vd change in temperature because the bias point is more stable with larger R1, but the sensitivity to change in Beta is worse so it's usually going to be better to lower R1 to some acceptable value and then calculate R2.

To investigate the effects of what happens with different R1, you can calculate R1max and then calculate the output bias point (should come out to exactly Vc that we started with) and then calculate Vc again with a change in Beta of 100 percent (2 times original Beta) and then again with a change in Beta of 50 percent (one half the original value) and see what effect this has on Vc with different R1.
As R1 is lowered more and more, the value of R2 has to be recalculated of course to maintain the original bias point with the original Beta.
As the Beta is changed, Vc will change, but Vc will change by a different amount for each R1 and R2 pair.

The secondary effect as mentioned is the change in Vd, which can also be investigated. The usually approximation is -2mv/degreeC so changing Vd by 20mv will show the approximate effect as the temperature decreases by 10 degrees C.

The effect of change of Beta reduces with decreasing R1 R2 pair impedance, but some limit has to be placed on this too because we dont want our R1 R2 divider drawing too much current just to bias the transistor.

 

[Veraxis]

wow. that's a lot of information coming at me!

**broken link removed**

But I really appreciate that you're taking the time to really explain this all to me. You have no idea how helpful this is.

 

[MrAl]

wow. that's a lot of information coming at me!

**broken link removed**

But I really appreciate that you're taking the time to really explain this all to me. You have no idea how helpful this is.

Hi,

You can take it one step at a time, using a hand calculator or something like that. Do the first equation, then the second, etc.
Once you get a feel for it, vary Beta and see how that changes the output Vc.
Once you do these a few times you start to get very used to it and it becomes much easier.

What's that video about?

 

[Veraxis]

What's that video about?

never mind. I was trying to make an analogy. I thought it would be funny...

**broken link removed**

 

[ljcox]

I don't bother considering Beta.

The way I do it is roughly as follows:-

Decide on the emitter voltage, say 1 Volt for example.
Add 0.7 Volt for Vbe

Select R1 & R2 to give this voltage.

Then calculate Rc for about half Vcc - Ve

The current thrpough R1 needs to be significantly greater than Ib, so you can use Beta at this point to calculate the worst case Ib, or simply make the current through R1 about 1/10th of Ic.