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Old 24th March 2011, 03:16 AM   #1
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Default Transistor Base Voltage Question

I am looking at my MPS2222A transistor and it says the absolute maximum Vebo=6v. (others iv read all say 5v). I am wondering, does this mean the voltage connected to the resistor on the base must not be more than 6 volts? Or does it mean something else? I had someone say all of the extra voltage (if 12v is put to the base resistor) will be dropped across the resistor (minus about .7v for the drop across the base) so it doesnt matter if you put more than the Vebo to the base resistor.

The purpose of this question is that I am wanting to use this as a switch for a computer fan which draws 140mA, the transistor has a max of 600. Will this schematic work for this intended purpose? This is going inside an xbox so it will be powered by a computer type of board. When plugged in the 12voltage rail is held at about .4volts; I am wanting this transistor to switch on the extra fan so that .4 wont be flowing through the fan:

Click the image to open in full size.

Any help with this is greatly appreciated, thank you!
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Old 24th March 2011, 03:41 AM   #2
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This would be the base-emitter reverse breakdown voltage, the max that can be applied before breaking down the base-emitter junction. This comes into play if there is any way for the circuit to reverse bias the junction. Typically it's about 5v, but will vary.
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Old 24th March 2011, 04:02 AM   #3
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Quote:
Originally Posted by jkuetemann View Post
This would be the base-emitter reverse breakdown voltage, the max that can be applied before breaking down the base-emitter junction. This comes into play if there is any way for the circuit to reverse bias the junction. Typically it's about 5v, but will vary.
Alright, so if I put 12V at the base with the proper resistor (lets say a 200 ohm, 240ohms at the base with a 12v signal gets a 500mA emitter-collector current according to the data sheet) to limit the base current, it will work as intended? If 6V is the breakdown voltage, i dont think this comes into play since this is just DC.

Thanks for the help so far
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Old 24th March 2011, 04:14 AM   #4
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I did this with a 200mA pc fan. Use a tip 121 darlington/w small HS , fan will still have about 11V across it (most PC 12v supplies are really 12.25-12.40V - 1.2V for the tip). 2n2222 is kinda" on the edge" for this app.

I think radio shack has the TIP.

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Old 24th March 2011, 04:19 AM   #5
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I'd also recommend a bigger transistor. Even though the signal transistor is rated for compatible steady state current the internal filter capacitor and startup current of the fan might eventually make your 2222 switch stay on all the time. With the darlington you wont need more than a milliamp into the base.
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Old 24th March 2011, 04:21 AM   #6
polsol is offline polsol  South Africa
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Basically the voltage at 'B' (the base) will depend on the current flowing through R1.
if the current is 0 then the vlotage at "B" will be 12V.
As the current through R1 increases so does the voltage drop (V=IR).
The current through R1 will depend on the gain of the tranny.
Probably easiest way for you to control the current through the fan itself is to put a pot across the rails with the 'wiper' connected to the input of R1 then adjust it from zero (i.e. earth as far as the wiper is concerned) until the fan starts to turn.
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Old 24th March 2011, 05:05 AM   #7
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Quote:
Originally Posted by ostripper View Post
I did this with a 200mA pc fan. Use a tip 121 darlington/w small HS , fan will still have about 11V across it (most PC 12v supplies are really 12.25-12.40V - 1.2V for the tip). 2n2222 is kinda" on the edge" for this app.

I think radio shack has the TIP.

OS
Thanks for the tip (bad pun, i know!). I will pick up a TIP120 at radio shack. this seems like this is the new schematic, a small transistor with max Ic=200mA, and the second transistor connected to the fan is the TIP120:
Click the image to open in full size.
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Old 24th March 2011, 05:27 AM   #8
agdr is offline agdr  United States
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Almost. The TIP120 is both the first transistor and second transistor in one package.

http://www.fairchildsemi.com/ds/TI/TIP120.pdf

You just need the one TIP120 and feed the base from 12V with something like a 4.7k resistor.
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Old 24th March 2011, 05:42 AM   #9
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All you need is the base resistor. see below... tip120 has everything else inside.

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File Type: gif tip121.gif (50.6 KB, 122 views)
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Old 24th March 2011, 11:05 AM   #10
AndrewT is offline AndrewT  Scotland
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the transistor is being ask to operate as a switch.
This operation is different from standard amplifier operations.

When the transistor switch is open, no current flows and so no power is dissipated in the transistor junction.
When the transistor closes (passes motor current) the Power dissipated in the junction is Vce * Ic
.
To keep the transistor switch cool and reliable the Vce when closed must be low.
This is ensured by running the transistor "saturated".
Saturated is when the base current : collector current ratio moves towards 1:1
hFE necessarily becomes very low. Saturated hFE are commonly in the range 5 to 20 and most would use 10 to obtain a low Vce when switched on. The datasheet will show Vce sat for various Ic and Ib.

The switching circuit must supply sufficient base current to obtain the low hFE demanded for saturated operation.
A base resistor fed from a stable voltage source can ensure a known base current.
A second transistor can supply this base current, but this transistor must also be set up to provide that large base current.

Using an integrated Darlington does not guarantee a saturated switch. It depends on the resistors fitted inside the integrated Darlington.

I recommend a single transistor with "controlled" high base current
or
two transistors where the first passes sufficiently high "controlled" current to the base of the second.

Feeding the collector of the first transistor from the low side of the motor fails to meet the saturated condition required. There is an exception. If both transistors in the Darlington are saturated then the first collector can be fed from the low side of the motor. But then the "Darlington gain" has gone from 5000 to 50 or so. The switching control circuit must pass sufficient current to saturate both transistors in the integrated Darlington.
If the motor has a running current of 30mA then the base must have a supply of 3mA. This must be provided by a resistor or a an extra transistor. The starting current for a 30mA motor could be ~100mA. A 200mA transistor will meet this current requirement. A 200mA transistor will have a peak pulse current rating of 300mA to 400mA. Easily capable of starting a 30mA motor.
Note that using a high hFE transistor has no advantage in this switching mode.
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Last edited by AndrewT; 24th March 2011 at 11:16 AM.
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