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Need Help With Tip-127 Transistor

RatShack

Aug 11, 2009
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hello everyone this is my first time being here at EP and am glad i found this place. I am having some problems getting my TIP-127 PNP to work properly. here is some background info on what i am trying to do.

Pic16f88 programmed with 3 outputs each of these 3 outputs go to the Base pin on the TIP-127. One of these outputs Will be a PWM output.
My logic for switching the TIP127 on is correct in the software (Logic 0 turns Tip on)
if i measure the output pin of the PIC outputs are acting proper to what i have programmed.

I plan on driving 3 solenoids at 22ohms@12v here is my calculations
using ohms law i calculate .545 amps draw at solenoid.
Minimum base current=545ma/1000hfe=.545ma+a lil to ensure full saturation so .700 sounds good?
Rbase=12/(.545/1000*.7)
Rbase=32K ohms
Power dissappated= .3(.545)=163mw

my problem is when i have a high condition on the output pin,my TIP is at 12v on the collector(no load-just measured with a volt meter) When the output pin is brought low the TIP is still at 12v.

This is the first time i have worked with a TIP transistor and am clearly doing somthing wrong. can anyone help me understand whats going on here. ill check back often if anyone has further questions.

Thanks
Jason
 
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Resqueline

Jul 31, 2009
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If you hooked the solenoid in it would work as intended.
A darlington has a very high hFE at low currents. The voltmeter draws next to nothing (1.2uA). Divide this by a hFE of 10000 and you find that only 0.12nA base current is needed to drive the transistor. The outputs of the PIC surely has much higher leakage currents.
The solution? Use a pull-up resistor B-E to absorb a portion of the (calculated) base current. (10%=30K.)
Even then the transistor might have enough leakage current to spoil such unloaded measurements.
 

neon

Oct 21, 2006
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If you hooked the solenoid in it would work as intended.
A darlington has a very high hFE at low currents. The voltmeter draws next to nothing (1.2uA). Divide this by a hFE of 10000 and you find that only 0.12nA base current is needed to drive the transistor. The outputs of the PIC surely has much higher leakage currents.
The solution? Use a pull-up resistor B-E to absorb a portion of the (calculated) base current. (10%=30K.)
Even then the transistor might have enough leakage current to spoil such unloaded measurements.

Darlingtons are not for switching but for current driving capbablities. if you use a switch to drive slenoid better think of a force be beta of 10 not 1000. for .5 amps that becomes 50ma of base current now you know
 

ElectronWorks

Aug 20, 2009
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First, let's get some things right. I am assuming you have 3 outputs going to 3 individual transistors. Obvious, I know, but I want to assume nothing here.

Secondly, if you have a pnp transistor, your emitter will be at 12V and your base has to be driven from 0V to 12V. If you are driving a PIC that is powered from 5V, your base will be driven from 0V to 5V, so your pnp will never switch off. If this is the case, you need to have an npn (driven from the output of the PIC), in between the PIC and the pnp. You then connect a collector resistor to the npn and the base of the pnp to the collector of the npn. The npn will then level shift the 0-5V signal to 0-12V (inverting it at the same time). You will then be able to fully switch off and on your TIP tranny.

You will also need a diode across the coil of your relay (pointing opposite to normal current flow) to stop the back emf when the coil switches off

Base resistor for npn - rough guess - 10k. Collector resistor 10k. diode for coil 1n4148.

Hope this helps :D

Bill
 
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RatShack

Aug 11, 2009
6
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Aug 11, 2009
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First, let's get some things right. I am assuming you have 3 outputs going to 3 individual transistors. Obvious, I know, but I want to assume nothing here.

Secondly, if you have a pnp transistor, your emitter will be at 12V and your base has to be driven from 0V to 12V. If you are driving a PIC that is powered from 5V, your base will be driven from 0V to 5V, so your pnp will never switch off. If this is the case, you need to have an npn (driven from the output of the PIC), in between the PIC and the pnp. You then connect a collector resistor to the npn and the base of the pnp to the collector of the npn. The npn will then level shift the 0-5V signal to 0-12V (inverting it at the same time). You will then be able to fully switch off and on your TIP tranny.

You will also need a diode across the coil of your relay (pointing opposite to normal current flow) to stop the back emf when the coil switches off

Base resistor for npn - rough guess - 10k. Collector resistor 10k. diode for coil 1n4148.

Hope this helps :D

Bill

Bill,
i have got the circuit to work on the breadboard. If i attach a schematic is it possible for you to check it over for me? think you can shoot me a PM with your Email address?:eek:
 
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Resqueline

Jul 31, 2009
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Darlingtons are not for switching but for current driving capbablities. if you use a switch to drive slenoid better think of a force be beta of 10 not 1000. for .5 amps that becomes 50ma of base current now you know

I'm trying hard to keep my tongue here, but I'm utterly confused by your statements neon...


Bill is correct though, if the PIC doesn't have open-collector outputs and is run from 5V then it'll have a hard time switching off the PNP..
 
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