Tip31c / Pic16f628

PeterDove · Dec 27, 2006

Hi All,

Could some one tell me what the minimum voltage/current one needs to apply to the base of the TIP31C to allow current to flow through the Collector Emitter? Also is there a minimum voltage for the CE part?

Just having trouble getting my PIC to activate the transistor

Cheers

Peter

Oznog · Dec 27, 2006

Like any bipolar NPN transistor, you need about 0.5 to 1v (or more) from base-to-emitter to turn on the transistor. The higher the base current, the more collector current it will pass. Higher base currents require slightly higher voltages.

The base-emitter junction is only a diode. As such, you just put +5v across it and it will draw way too much current. You must use a current-limiting resistor on the base or emitter (two very different configurations btw).

Google for general info on how to use transistors.

Papabravo · Dec 27, 2006

Your request is a bit vague. If the emitter is grounded and you are using the TIP31C as a low side switch, then small amounts of base current will produce some amount of collector current. This happens when the base is about 0.6 to 0.7 volts above the emitter. Now depending on the beta or Hfe of the tansistor a given amount of base current results in beta times the base current as the collector current. Normally the collector will be connected to one side of the load and the other side of the load will be connected to a supply voltage.

In order for the transistor to operate the BE junction must be forward biased and the BC junction must be reverse biased. For practical purposes I think this means that the supply must be greater than about a volt although I could believe any other number in the range of 0.7 to 1.0 Volts

bananasiong · Dec 27, 2006

Depends on how much current do u want to switch. If just small current, u don't need this big transistor, if high current, u need to make a darlington with additional another transistor.
Last time I've tried to connect the base of the npn transistor straight away from the microcontroller, it didn't work, i.e. can't switch.

audioguru · Dec 27, 2006

The max allowed output current from a PIC is 25mA. A TIP31C transistor at minimum current gain (about 30 but it won't be saturated) can have a max collector current of 25mA x 30= 750mA, or about 300mA if it is saturated well.

The output of a PIC has some resistance that is in series with the series base resistor of the TIP31C transistor, so a base resistor of about (5V-0.7v-1V)/25mA= 133 ohms (150 is the closest standard value) shoud be used for max output current.

bananasiong · Dec 27, 2006

Why -1v? is it the for voltage drop of the internal resistance?

audioguru · Dec 27, 2006

bananasiong said:
Why -1v? is it the for voltage drop of the internal resistance?

Yes. The output voltage of a PIC drops as current flows through its internal resistance.

bananasiong · Dec 27, 2006

Oh, I don't include this. When I switch a relay with a transistor, I just use the relay resistance * the current gain to get the base resistor. I should include this and the 0.7 Vbe.

PeterDove · Dec 28, 2006

bananasiong said:
Depends on how much current do u want to switch. If just small current, u don't need this big transistor, if high current, u need to make a darlington with additional another transistor.
Last time I've tried to connect the base of the npn transistor straight away from the microcontroller, it didn't work, i.e. can't switch.

This is the problem I am having - the transistor doesnt switch and I cant for the life of me work out why. The image below shows the desired set up - the pins are connected to the 5V source just to simulate the PIC. The 0.45V is supposed to be an example voltage out of the pin 5 of the L200C - I think I am going to have to strip it down to a simpler test

**broken link removed**

Peter

audioguru · Dec 28, 2006

Hi Peter,
1) The L200C regulator has a minimum output voltage of 2.85V, not 0.45V, so most of your load resistances will try to cause a current of more than the 2A max from the L200C.
2) The 180 ohm resistors will create a base current of about only 20mA. If the TIP31C transistors have the minimum current gain of 30, then the max collector current will be only 600mA and the collector voltage will be 4V.

PeterDove · Dec 28, 2006

audioguru said:
Hi Peter,
1) The L200C regulator has a minimum output voltage of 2.85V, not 0.45V, so most of your load resistances will try to cause a current of more than the 2A max from the L200C.
2) The 180 ohm resistors will create a base current of about only 20mA. If the TIP31C transistors have the minimum current gain of 30, then the max collector current will be only 600mA and the collector voltage will be 4V.

The 0.45V is simulating the equations used to work out the value of R3. R3 = 0.45 / Amps and it is R3 which dictates the current output. The purpose of having all the switches and resistors in parralell is to be able to change the value of R3 in order to control the current.

Peter

PeterDove · Dec 28, 2006

audioguru said:
Hi Peter,
1) The L200C regulator has a minimum output voltage of 2.85V, not 0.45V, so most of your load resistances will try to cause a current of more than the 2A max from the L200C.
2) The 180 ohm resistors will create a base current of about only 20mA. If the TIP31C transistors have the minimum current gain of 30, then the max collector current will be only 600mA and the collector voltage will be 4V.

Maybe I ought to stick with switching relays

audioguru · Dec 28, 2006

Hi Peter,
If you are trying to use the transistors to program the output current of the L200 then you shoud have said so in the 1st post.
Notice in the datasheet that the sensing resistor isn't grounded like your transistors would do. Your resistors are the load on the L200.

Transistors won't work anyway because with a load of 2A and even with a very high base current of 200mA, their max saturation voltage is about 1.0V. If you are lucky and get some "typical" transistors then the saturation voltage is 0.32V so the current setting will be much too low.
But your base current is only 20mA, not 200mA.

You need some kind of floating high current switches, like relay contacts.

PeterDove · Dec 28, 2006

audioguru said:
Hi Peter,
If you are trying to use the transistors to program the output current of the L200 then you shoud have said so in the 1st post.
Notice in the datasheet that the sensing resistor isn't grounded like your transistors would do. Your resistors are the load on the L200.

Transistors won't work anyway because with a load of 2A and even with a very high base current of 200mA, their max saturation voltage is about 1.0V. If you are lucky and get some "typical" transistors then the saturation voltage is 0.32V so the current setting will be much too low.
But your base current is only 20mA, not 200mA.

You need some kind of floating high current switches, like relay contacts.

Thanks for that - can you explain what 'saturation voltage' is please?

Peter

audioguru · Dec 28, 2006

Saturation voltage is the voltage between the collector and the emitter of a turned-on transistor at a specified amount of collector current. Usually a base current of 1/10th the collector current is used.

The max saturation voltage for a TIP31 transistor with Ic= 3A and Ib= 375mA is 1.2V. The graph in the datasheet shows typical saturation voltages with the base current 1/10th the collector current.

PeterDove · Dec 28, 2006

audioguru said:
Saturation voltage is the voltage between the collector and the emitter of a turned-on transistor at a specified amount of collector current. Usually a base current of 1/10th the collector current is used.

The max saturation voltage for a TIP31 transistor with Ic= 3A and Ib= 375mA is 1.2V. The graph in the datasheet shows typical saturation voltages with the base current 1/10th the collector current.

Ah... so does this mean because the voltage on the emitter side ( which will be about 7V ) is much higher than the voltage at the collector side, it wont work?

Peter

audioguru · Dec 28, 2006

The emitters of the transistors and the 0V of the PIC will be at about +7V (the output voltage of the L200). So you need two separate power supplies.

The transistors need a base current that is much higher than the max current from a PIC.

The saturation voltage of the transistors is much too high.

PeterDove · Dec 28, 2006

audioguru said:
The emitters of the transistors and the 0V of the PIC will be at about +7V (the output voltage of the L200). So you need two separate power supplies.

The transistors need a base current that is much higher than the max current from a PIC.

The saturation voltage of the transistors is much too high.

Thanks again, I really appreciate the time you put in to helping others

Peter

legrotalia · Jan 28, 2007

Sos

Could someone tell me where can I find assembly codes for PIC16F628? I can't even find how to initialize it... To be more specific, I have to make a circuit which has a Light Dependent Resistanse and sends output to the PC by serial port. Any examples including assembly code would help me very much.

Nigel Goodwin · Jan 28, 2007

legrotalia said:
Could someone tell me where can I find assembly codes for PIC16F628? I can't even find how to initialize it... To be more specific, I have to make a circuit which has a Light Dependent Resistanse and sends output to the PC by serial port. Any examples including assembly code would help me very much.

You can't have looked very hard, try reading my tutorials!.

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