Transistor as a switch

Thread Starter

zoink

Joined Jul 5, 2009
5
Hiya, can anyone help me pick out one of these transistors?

I have a supply that's going to be in the range 7.5-10V.

I want to hook a microcontroller running at 5V to the collector via internal pullup of 5.6K so I can read the pin to tell when the afforementioned 7.5-10V is applied to the base of the transistor, which means I need to fully saturate the transistor with the lowest voltage in that range in order to turn it on and short the pin to ground.

So I'm using the transistor as a switch, and the emitter is tied to ground.

I want to use these ICs with built-in resistors to save board space:
http://www.onsemi.com/pub_link/Collateral/MUN2211T1-D.PDF

But I'm not seeing the stuff I normally see in transistor datasheets in this one... Can anyone help me identify which parameters I should be looking at and why?
 
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Thav

Joined Oct 13, 2009
82
Probably the most important for you is DC current gain (hFE), Vce(sat) and your resistor values. This will let you analyze the DC performance of your circuit.

Check those Output Voltage parameters. You see they rate some of the output voltages at some various values of Vb. That may give you a clue about which device to use. You may need an additional series base resistor.
 

CDRIVE

Joined Jul 1, 2008
2,219
hFE and Ib for Vce (Saturated) are listed but the data sheet doesn't make it clear if these measurements, like Ib, are measured with R1 & R2 included in the measurement or the Transistor alone.
 
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Thread Starter

zoink

Joined Jul 5, 2009
5
Hey thanks for the replies. So I guess I just pick one and make sure the Ib current with 7.5V applied to the base resistor is well above that required for Vce(sat)? The current through the pullup is 5V/5.6K = ~893uA = <1mA. So we're talking about very small Vce current.

So let's say I picked one with 35 hFE. Then I'd need 1mA/35 = ~29uA base current to push the required current of 1mA. Increase that by a factor of ten for safety margin and we come to 290uA. Am I right?

7.5V - 0.7V = 6.8V

6.8V / 0.00029A = 23448

This means I'd roughly need the equivalent of about 22k ohm resistor to the base. So the effective series resistance of the voltage divider would have to be around 22k or less.

Am I right?
 
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SgtWookie

Joined Jul 17, 2007
22,230
Choose one that has a base resistor that will allow a base current of 1/10 the desired collector current.

If Rc is 5.6k, you'd need about 1.8mA current to drop 10v across it.

So, if you put 0.2mA current through the base, you know it would be well-saturated.
 

Thread Starter

zoink

Joined Jul 5, 2009
5
Ok, cool. That's something I've heard elsewhere (about 1/10th the current to the base) so it's good to get some confirmation.

Out of curiousity, how often is this rule of thumb applied and when should I not apply it?

By the way, SgtWookie, the 7.5-10V is not attached to the collector, it's going to the base. 5V is going through the 5.6K pullup to the collector and this is used to allow the transistor to ground a pin of an MCU when the 7.5V-10V power is present at the base.
 
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mik3

Joined Feb 4, 2008
4,843
Ok, cool. That's something I've heard elsewhere (about 1/10th the current to the base) so it's good to get some confirmation.

Out of curiousity, how often is this rule of thumb applied and when should I not apply it?
This is to ensure that the transistor is in full saturation for all the changes in its gain by temperature, Ic etc. If you want to increase the switching frequency it will be better to reduce the base current as possible as to reduce the diffusion capacitance.
 

Thread Starter

zoink

Joined Jul 5, 2009
5
This is to ensure that the transistor is in full saturation for all the changes in its gain by temperature, Ic etc. If you want to increase the switching frequency it will be better to reduce the base current as possible as to reduce the diffusion capacitance.
Cool, good to know. Thanks for the information.

Frequency in this case is not important so I'll go with 1/10.
 
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