# Determining resistor values for transistor with voltage divider bias

Discussion in 'Homework Help' started by Robotics Guy, Nov 23, 2011.

1. ### Robotics Guy Thread Starter New Member

Mar 11, 2011
15
1
I need to design a common emitter voltage divider biased bjt transistor circuit. I know how to solve for different values when a few of the resistor values are already given, but in this case none of the resistor values are given and so I'm confused on how to do this. The requirements are that Icsat = 25mA and Vcc = 1.2V. Beta is not given, but can be between 50 and 200.

I tried solving for the resistor values, but the values I got were very low (like 10Ω for Re and 38Ω for Rc) and R1 and R2 were negative

Any guidance on how I should do this?

2. ### samski New Member

Nov 23, 2011
20
1
There are alot of rules of thumb around these types of calcs.

Typically you want to bias Ic to around 1mA which is enough to make sure it is turned on. Bias the output to be mid way between the supply rails to give maximum possible swing. You can also get the emitter resistance using Re=25mV/Ic (with Ic in mA!!). With potential divider values, its common to start at around 10K. or i think you can do the calcs for the base current and the emitter reistance etc and get a proper value. Oh, if you need it, its useful to assume Vb is 0.65 volts (sometimes you add a little margin to this)

hope that helps. sam

3. ### hobbyist Distinguished Member

Aug 10, 2008
773
62
When you wrote 1.2v for VCC I thought you made a mistake and meant to write 12v. for VCC.

However I quickly put together a CE stage with 1.2V on my variable bench supply, and sure enough with the calculations I made, I got my bias voltages,
so I applied a small signal to it from my audio generator, and got a symetrical input and output, with a gain of 3, which was expected with the values I chose foer RC and RE.

So it is doable (don't know if it is practical) doing this with 1.2v in the real world. (NOT computer simulation)

The way I calculated values was, I assumed half the VCC at the collector terminal, and chose a value for ICQ. then I chose VE (volt at Emitter), to be around 1/3 of VC. (Volt at collector).

Assumed a min. beta of 20 and made the divider current 10 times this value.

I used a 2N3904 transistor to build it with.

So hope this helps getting you into the start of your design.

Last edited: Nov 24, 2011
4. ### Robotics Guy Thread Starter New Member

Mar 11, 2011
15
1
Thanks for both of your replies. I didn't realize that "rules of thumb" needed to be used, but that certainly makes the design easier. My question is, where did they come from?

Hobbyist -- your explanation helps a lot, but I'm wondering what you're referring to when you say the output had "a gain of 3"?

Thanks for both of your help!

5. ### samski New Member

Nov 23, 2011
20
1
so the Ic bias current makes sure that the device is 'turned on'/saturated and behaves linearly (think of it as keeping everything 'ticking over'). biasing the output to be mid way between your rails allows the maximum swing in output voltage (the output cannot go above or below the voltage rails). The 25mV thing comes from the ebers moll model for a transistor, you might not need to know about this. Vb being around 0.65v just comes from the voltage over a diode (there is a pn junction from base to emitter).

As for the gain of 3. that means that any sinusoid you place on the input has 3 times the amplitude on the output.

6. ### hobbyist Distinguished Member

Aug 10, 2008
773
62
I need to mention, biasing the transistor using a 1.2v supply made for a akward design,

because for a linear amplifier, the voltage at the collector (dc bias) needs to be higher than at its base, while the voltage at the emitter must be lower than at the base.

The amp I made had the emitter lower than the base voltage as needed, but the collector voltsge, was close to being lower than the base, this puts the transistor very close to being non linear,

usually a good CE base divider would have the (bleeder) resistor, the resistor connected to ground, much smaller than the resistor connecting to the supply rail.

The amp I made had the exact opposite, the bleeder was very high and the supply resitor was much lower, which puts the transistor close to saturation,
this is mainly due to the smallish power supplied of 1.2v. and the base emitter voltage of around 0.65v.

By making VC 1/2 of VCC @ 0.6v. and the base emitter inherent voltage of 0.65v. upsets the design, for a linear amplifier.

Other people would be able to explain this, better,

so even though I got a symetrical output with some gain, it still is a poor design for linear amplification.

A higher VCC would be more advisable when dealing with bipolar junction transistors.

Hope this helps also.