Driving a transistor base from a voltage divider

Hi, I am trying to determine the operating specs for operating a transistor base (as a switch) from a voltage divider.

What I am trying to figure out is the appropriate values of R1 and R2, given a desired Base current and source voltage/source current available (I know the source voltage, and how much current I can pull from it).

If I eliminate the "bottom" (R2) resistor in the voltage divider and only use one resistor, I can easily determine the current that flows through that resistor (E/R), and therefore the current that flows through the base ((E-0.7)/R), which is just computing the value of a Base resistor.

Right now I am getting hung up on what effect the "bottom" resistor has on the base current. I know that without the transistor, I can figure out the voltage at the center point as a ratio of the resistances, and the current through both resistors, but what current is available from the center point to drive a transistor base?

The closest I have come is by calculating a "resistance" for the transistor Base/Emitter junction by dividing 0.7V by the desired base current, and treating the transistor as a resistance in parallel with R1, but I'm not very confident with that solution.

BTW, here is an ascii of what I have:

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Reply to
Jon Danniken
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If the current through R1 + R2 is much larger than the required base current then the junction will set the base voltage. That will allow you to set the collector current anywhere between zero and maximum.

However for a switch the transistor would be hard on or hard off. In which case a single resistor will limit base current to a safe level, at the point where its switched hard on.

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Best Regards:
                     Baron.
Reply to
Baron

Assume the base is +0.7 when on. The current through R1 is (Vin-0.7)/R1. The current through R2 is 0.7/R2. The difference current is going into the base.

John

Reply to
John Larkin

Jon,

You can't really control the base current. It draws whatever is required to support the emitter current that is flowing.

In your resistor divider setup, R1 will provide current to the base of the transistor AND to R2. So, if you measure the current through R1 (volts across R1 divided by the value of R1) and subtract the current through R2 (volts across R2 divided by the value of R2) then you'll have the base current. The only problem with this technique is that you need to know the value of the voltages and resistances very accurately. An easier approach would be to add a small resistor in series with the base of the transistor and measure the current by reading the volts across this resistor and dividing it by the value of the resistor. Or, you could simply put a current meter in series with the base of the transistor.

The current that is available at the connection of R1 and R2 depends on how many volts (from gnd) you desire. At zero volts (from gnd), the current you'll get is simply V+/R1. To better visualize what you really have, first remove the transistor from your thoughts. Now, just look at the voltage divider. It can be reduced to (thought of as) a voltage source with a series resistor coming from it. The value of the voltage source is the open circuit voltage you would get at the R1-R2 connection. That is, you would have a voltage source equal to (V+) * (R2/(R1+R2)). The effective series resistor would be equal to the parallel combination of R1 and R2, and would be equal to 1/( (1/R1)+(1/R2) ).

For example, if V+ is 12V, and R1 is 1K ohm and R2 is 400 ohms, the "equivalent" circuit would "look like" a 3.43V battery with a single series resistor of 286 ohms. So, its open circuit voltage is 3.43V. Its short circuit current is 12mA (note that this 12mA value is the same as you get from the "real" circuit -- that is 12V/1K). Now, when you hook up your transistor, it's easier to see what the base voltage would be for a given amount of base current.

Bob

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Reply to
BobW

"BobW"

Thanks, Bob, much appreciated. I think I was thinking of the voltage source as I do with a linear power supply, ie so many volts *while* delivering a certain current level.

Perhaps instead I should be thinking of voltage and current as opposites given open or closed circuit behavior, as in when plotting a load line for a tube? If that is the case, I'm guessing a minimum voltage being 0.7V and the short-circuit current as whatever I need to get sufficient base current to switch the transistor on?

Thanks again,

Jon

Reply to
Jon Danniken

Thanks Baron, that is appreciated.

Speaking of switching the transistor "hard on", how (in the absence of load charts for a particular transistor) do I determine a sufficient base current to do this?

Thanks again,

Jon

Reply to
Jon Danniken

In a linear power supply (or any type of power supply), its output voltage and current are independant quantities. For a constant voltage power supply, the output current is only determined by the load that's connected to that power supply. If the load is a resistor then the output voltage and current will be related (I=V/R), but it's not that simple for other types of loads (like the base of a transistor).

I'm not sure it's helpful to think of voltage and current as being opposites, in this case. Generally speaking, however, the output voltage delivered will decrease as the output current increases. However, it's best to try and understand each circuit on its own.

If the voltage at your emitter is fixed (e.g. at gnd) then it makes the analysis easier because the base will be (about) 0.7V away from the emitter (for any decent level of base current). So, if this is the case for your circuit, then the base current will be easy to calculate using the previous ideas that I presented.

Determining how much base current is sufficient to switch the transistor on is a little more complicated. You need to know:

1 - The maximum collector current when the transistor is on. 2 - The minimum Beta (Ic/Ib) of your transistor when the collector is close to being in saturation (e.g. when Vce = 1V)

If you know these two quantities then you can easily calculate the base current required. When I do this, I usually double or triple the base drive current to insure that the device is fully on (i.e. in full saturation).

You're welcome.

Bob

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Reply to
BobW

transistor

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transistor

Find the Thevenin equivalent of the base bias network for starters. Then you'll be working with a single equivalent resistance for setting base current. Once that's done you can look at the formula for the Thevenin resistance (how its composed of R1 and R2) to decide on what values can be chosen for R1 and R2 to set the total current draw as desired.

Reply to
Greg Neill

"Jon Danniken"

** The base current needs to be at least 10% of the collector current to ensure the transistor is biased hard on - make sure that current level is within the capabilities of the device first ( see the data sheet).

The base-emitter resistor ( R2) plays no part in biasing the transistor on - its role is helping it to switch off fast, but it will also take some drive current away from the base of the transistor.

So, the design begins with deciding on the value of R2 and the known collector current, Ic .

For non-critical applications, R2 can be between 100 ohms and 1kohms.

..... Phil

Reply to
Phil Allison

I think that's the answer the OP seemed to want.

Jon

P.S. Vth = V*R2/(R1+R2) Rth = R1*R2/(R1+R2) Assuming grounded emitter and Vth > 0.7V: Ib = (Vth-0.7V)/Rth Ic

Reply to
Jon Kirwan

"BobW"

Ah, thanks Bob, I hadn't made that leap of using the beta and Ic max to figure out the point of base current required for saturation.

Thanks again,

Jon

Reply to
Jon Danniken

Reply to
Jon Danniken

"Jon Danniken"

** Until YOU supply a few actual details of your design - NOBODY can supply any real help.

What is the transistor type ?

What Ic are you trying to switch ?

What is the switching frequency ?

You post is nothing but a STUPID DUMB TROLL without these essential facts supplied.

.... Phil

Reply to
Phil Allison

Absolutely, positively, wrong.

The base voltage has a mild dependency on current, a moderate dependency on temperature (so pay attention if you want it to work outside of room temperature!) and a moderate to strong dependency on the transistor construction.

For a silicon transistor go with John Larkin's 0.7V.

Absolutely, positively, unnecessary (but at least it's not wrong).

Yes, the base voltage will vary somewhat, and you'll have to take this into account if you want the thing to work over temperature and manufacturing variations. But you just can't trust a transistor to maintain it's B-E voltage well enough that measuring the base current would do you any good at all. Instead, you need to design your circuit so that you always have sufficient base current (I'm assuming the OP wants it to be saturated, otherwise he needs to post again and ask for clarification).

Well, at least _that_ part is correct.

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Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

** Nope - it is correct.

Cos BobW is commenting on the OP's grossly ambiguous schem ( with C and E uncommitted) which looked like an emitter follower ( rather than a saturated switch ) to him.

..... Phil

Reply to
Phil Allison

Thanks, Tim, I appreciate it. I hadn't realized about the temperature dependency, and will keep it in mind if I'll be in a temperature extreme.

And yes, this is a switch, so I'll be looking to run it with the base saturated.

BTW, am I correct in understanding that the current needed to saturate the base is independant upon the collector current?

Jon

Reply to
Jon Danniken

Uhh -- no.

Saturation happens when the B-C junction starts forward biasing, and that depends on the collector current, indeed it does. Hold the C-E potential to 5V and the transistor won't saturate, no matter how much current you pour into the base. There will be some Really Bad Things happening to the transistor for a very short while, but it won't saturate.

To do this really right you figure out the current gain when the transistor is in saturation (it'll be on the data sheet, and it's usually way less than the best value of current gain), then put in more than enough base current to provide your desired collector current with that current gain.

(By the way -- you'll sometimes see circuits for saturated switches that leave off the resistor to ground. This is bad both because it may not provide a positive "off" to the transistor, and because a saturated transistor turns off faster if there's something pulling charge out of the base. If the resistor is hooked to something that'll sink current, like a CMOS logic output, it's far less bad, but still not as fast as your circuit).

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Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

Others have answered far better than I ! The data sheet for the particular device you want to use should be the first point of reference.

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Best Regards:
                     Baron.
Reply to
Baron

I remember many years ago books of "Transistor Bias Tables" ! One for Germanium and one for Silicon transistors. I vaguely remember that they contained data for gain(hfe) and temperature along with the resistor ratios to obtain set currents.

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Best Regards:
                     Baron.
Reply to
Baron

Okay, gotcha, thanks Tim. So if I am reading you correctly, I should using the minimum Beta for when the transistor is at maximum rated Ic (indeed given by the specs, and definitely less than the best Beta), and using that Beta, figure out the Base current for my desired Ic to be switched? I hope I got that right.

Thanks again for your help,

Jon

Reply to
Jon Danniken

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