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#1 |
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diyAudio Member
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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: ![]() Any help with this is greatly appreciated, thank you! |
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#2 |
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diyAudio Member
Join Date: Dec 2010
Location: Paris, Ontario
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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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#3 | |
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diyAudio Member
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Quote:
Thanks for the help so far |
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#4 |
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diyAudio Member
Join Date: May 2008
Location: Albany , NY (smallbany)
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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. OS
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Mongrel website , always current and updated : http://67.248.209.21/D%3A/WEBSITE/ |
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#5 |
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diyAudio Member
Join Date: Nov 2005
Location: PA
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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.
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#6 |
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diyAudio Member
Join Date: Aug 2004
Location: Cape Town and Kuwait
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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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Tony M |
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#7 | |
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diyAudio Member
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Quote:
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#8 |
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diyAudio Member
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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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#9 |
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diyAudio Member
Join Date: May 2008
Location: Albany , NY (smallbany)
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All you need is the base resistor. see below... tip120 has everything else inside.
OS
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Mongrel website , always current and updated : http://67.248.209.21/D%3A/WEBSITE/ |
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#10 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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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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regards Andrew T. Last edited by AndrewT; 24th March 2011 at 11:16 AM. |
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