Voltage divider question

[Tadas!]

Hello all!

I have a question about voltage dividers as used for biasing transistors. Let's say i want to bias a transistor at 4,5V. So, I take a 9V battery, hook up two 100k resistors, and I have 4,5V in the middle. Also i can take two 47k resistors and have the same 4,5V in the middle. Or almost any size resistors which are not too big (so that they don't reduce +9V to less than 4,5V) or too small (so that the battery is drained very fast). My question is, what is the difference between using let's say 100k resistors and 47k resistors for that matter? They still result in 4,5V that I need. Still, a lot of the schematics show different value resistors for that purpose and I do not understand why?

P.S Please don't get angry if this is a silly question - I'm not a pro  

[R.G.]

Welcome to the forum. Now go read http://www.geofex.com . This will be useful in general, but on this one question, you'll want to read:
http://geofex.com/circuits/Biasnet.htm

You may find other things in there that help too.

[Tadas!]

Thank you! Now on to reading 

[fuzzy645]

Read your article RG...very informative.

For educational purposes, let me know if I have this correct:

Suppose we use a pair of 10K resistors to bias a transistor with a 9v supply resulting in 4.5V.   Using Ohm's law I = V/R shows the current is 450 microamps

If we then change this to pair of 47K resistors (for example), given the same parameters again using Ohm's law  I= V/R shows the current is 100 microamps

Is that correct? 

Question:  In your article, you mention it is good to determine the bias current of the opamp or transistor by inspecting the data sheet..  I just looked up the data sheet for an example transistor (2n5089) but I could not seem to find a metric for a bias current.  Is this parameter listed in some other way?  Here is the link:

http://www.mouser.com/Search/ProductDetail.aspx?R=2N5089Gvirtualkey58410000virtualkey863-2N5089G

[sault]

9v across two 10k resistors means I = V/R = 9/20k = 0.45 mA current consumption.
9v across two 47k resistors means I = V/R = 9/94k = 95.7 uA current consumption.

In these cases, with both resistors equal, the voltage will be 1/2 of the Vcc and current consumption at the tap. So 10k||10k will deliver around 0.225 mA to wherever you need it to go.

The resistors you use will depend on what you're using them for. Op-amps draw very little current (I looked up TL071's, bias current somewhere in the order of a few hundred pA's), so it's voltage you're concerned about more than current.

BJT's are different - designing a voltage divider to bias a transistor properly isn't necessarily straightforward. In a very general sense R1 is going to be several times bigger than R2.... you'll drop most of the voltage there, you only need a few volts on the base to keep it above the voltage drop. Going anything beyond that is still an exercise for me (since the currents involved will vary depending on the rest of the circuit), but at least I can bring you that for.

[R.G.]

Suppose we use a pair of 10K resistors...If we then change this to pair of 47K resistors...
Is that correct? 

In general, yes. You have the concept.

Question:  In your article, you mention it is good to determine the bias current of the opamp or transistor by inspecting the data sheet..  I just looked up the data sheet for an example transistor (2n5089) but I could not seem to find a metric for a bias current.  Is this parameter listed in some other way?

I could have been clearer.   

The datasheet for opamps will list bias current explicitly. For transistors, it's buried in the current gain specs and the rest o the circuit. As sault points out, for biasing a transistor, you use a voltage divider to place the base at a voltage; the emitter is a semi-fixed voltage different from the base, and the emitter resistor determines the emitter and collector current. The gain of the transistor at that current then determines the necessary base current. So for bipolars, the trick is to figure the emitter current some way, then divide by the (variable) current gain. The vast majority of practical circuits are designed so that the exact current gain does not matter as long as it's big enough.

If this seems roundabout and complex, that's because it is. By way of apology, it took some years of thinking on the part of EEs back in the 50s and 60s to come up with that procedure. Opamps are simpler in so many ways.

One of the bedrock principles of engineering is this: that which cannot be controlled must be made irrelevant. Opamps and digital logic do this on many levels.