Hi,

This is mine first post

I´m a enthusiastic that don´t have any formation in electronic and use the google as a base of knowledge for the things that i need.

And i need help

I have a BD140 because i need up to 1,5 Amp at 12v, and i need that when the signal (+) is on the power will be cut off and vice verse, when the signal (+) is off that the power is turn on.

Can someone post a systematic for this, please

Regards

 

Your transistor is a PNP so use the pnp circuit

 

hi,

Thanks for the quick reply.

I have tryed that connection and it wont go to 1,5 Amp.

I have found this example of what i need:

I have the plus (+) to activate/deactivate the base of BD140, but if i connect like the example i cant have 12v 1,5 Amp that i need.

Thank you

 

How is the voltage of the (+) activate signal?

 

What is the value of R13, it should be about 1k, also there is no feed for the base resistor to the negative side of the battery, i would put a 1 K resistor across C1 to provide the feed.

 

Welcome to AAC.

The circuit you posted, is this your circuit or simply something close to what you want to do?

What do you mean by can't have the 12V, 1.5A that you need? Do you mean you don't see 12VDC when you measure with you meter? Do you mean you can pull, say 1A, then the voltage drops beyond 1A?

The BD140 datasheet says it can provide a maximum of 1.5A continuously. You will need a nice large heat sink and perhaps a fan to keep it cool if you need to pull a full 1.5A out of it. If you really need that much current, I'd suggest using a larger transistor with a larger current rating.

What is it you're powering that requires 1.5A? If you can post a schematic of your circuit, we can offer a lot more help.

 

How is the voltage of the (+) activate signal?

Hi,

Sorry for the bad English.
What i meant is that in the base of BD140 i have the (+) to deactivate or the lack of (+) to activate the BD140.

 

What is the value of R13, it should be about 1k, also there is no feed for the base resistor to the negative side of the battery, i would put a 1 K resistor across C1 to provide the feed.

Yes is 1K
You are correct

 

Welcome to AAC.

The circuit you posted, is this your circuit or simply something close to what you want to do?

What do you mean by can't have the 12V, 1.5A that you need? Do you mean you don't see 12VDC when you measure with you meter? Do you mean you can pull, say 1A, then the voltage drops beyond 1A?

The BD140 datasheet says it can provide a maximum of 1.5A continuously. You will need a nice large heat sink and perhaps a fan to keep it cool if you need to pull a full 1.5A out of it. If you really need that much current, I'd suggest using a larger transistor with a larger current rating.

What is it you're powering that requires 1.5A? If you can post a schematic of your circuit, we can offer a lot more help.

Hi
Thank You for the welcome part Sir.

The circuit that i post is very close to the one that i made.

The only part that don´t work is when i need the BD140 to act as a switch - Close when it as no power and open when the power is on.

I'm trying to feed one HD disk power by a transformer rated 12v 1,5Amp. If i connect directly to the battery it work fine, but if i connect it to the BD140 as show in the example (instead of the led´s) it light up but don´t work - I suppose that it don´t have enough amp to start

 

Hi,

Sorry for the bad English.
What i meant is that in the base of BD140 i have the (+) to deactivate or the lack of (+) to activate the BD140.

I know what you mean, but I don't know how many volts of the (+)? (5V,9V,12V, ?)

BD140 has 1.5A rating current, it just can be used less then 0.5A, but it better used less then 300mA, that's for the heat and the life of DB140.

How many A(Ampere) do you need for the output?
If you really need 1.5A then you better using a MOSFET to replacing the BD140.

 

I know what you mean, but I do'nt know how many volts of the (+)? (5V,9V,12V, ?)

BD140 has 1.5A rating current, it just can be used less then 0.5A, but it better used less then 300mA, that's for the heat and the life of DB140.

How many A(Ampere) do you need for the output?
If you really need 1.5A then you better using a MOSFET to replacing the BD140.

Ups Noob

The power supplier is 16v because i need the charge a battery of 12v.
In the example they use 9v in the base of BD140

Thanks

 

Hi
Thank You for the welcome part Sir.

The circuit that i post is very close to the one that i made.

The only part that don´t work is when i need the BD140 to act as a switch - Close when it as no power and open when the power is on.

I'm trying to feed one HD disk power by a transformer rated 12v 1,5Amp. If i connect directly to the battery it work fine, but if i connect it to the BD140 as show in the example (instead of the led´s) it light up but don´t work - I suppose that it don´t have enough amp to start

Like i said earlier, you need a base feed resistor from the negative side of the battery, here is where to connect it across the capacitor .

 

Hi

Thank´s to all.

I just draw mine first schematic.
I use LTspice and it´s not difficult

Well,
I know that i probably have to many mistakes in the table
But be gentle and help me true the process

The only thing that don´t work is the BD140

 

I'm not sure how many ampere the HDD needed, but the circuit that I attached is more reasonable and you can see the circuit to think how to design a circuit to suit your need.

 

I'm not sure how many ampere the HDD needed, but the circuit that I attached is more reasonable and you can see the circuit to think how to design a circuit to suit your need.

Thank you Sir for your trouble, work and patience

tomorrow i will try that and report back.

I only have two questions:

- can i use the BC548 only for testing till i buy the transistor 2SC1384?

- In your drawing the HD disk will be always on? If that the case you are a genius!

Thanks again for your patience.

 

- can i use the BC548 only for testing till i buy the transistor 2SC1384?

The rating current of 2SC1384 is 1A, and I make it flows through about 150mA, so you better find some other bjt is similar or bigger.

- In your drawing the HD disk will be always on? If that the case you are a genius!

The key is the Q3, normally the Q3 was turned on when the power in on.

When the output voltage of LM317 is drops down until the b of Q3 lossing the current and then the Q3 will be turn off, and the Q4 turn on and Q2 turn on too, and the HDD got the power, this is what you want, if you want the b of Q3 cutoff quickly then you can in series with a zener as below and adjust the values of 10K resistor.

5.1V-1N4733A
5.6V-1N4734A
6.2V-1N4735A
6.8V-1N4736A
7.5V-1N4737A
8.2V-1N4738A
9.1V-1N4739A
10V-1N4740A

1. To find out the current for HDD.
2. How much current the battery could offer?

 

The rating current of 2SC1384 is 1A, and I make it flows through about 150mA, so you better find some other bjt is similar or bigger.


The key is the Q3, normally the Q3 was turned on when the power in on.

When the output voltage of LM317 is drops down until the b of Q3 lossing the current and then the Q3 will be turn off, and the Q4 turn on and Q2 turn on too, and the HDD got the power, this is what you want, if you want the b of Q3 cutoff quickly then you can in series with a zener as below and adjust the values of 10K resistor.

5.1V-1N4733A
5.6V-1N4734A
6.2V-1N4735A
6.8V-1N4736A
7.5V-1N4737A
8.2V-1N4738A
9.1V-1N4739A
10V-1N4740A

1. To find out the current for HDD.
2. How much current the battery could offer?

Once more - Thank you !!

Well I just test it and it work just fine ...

It turn off the LED when the power is on, and it turn off the LED when the power is off.

I just replace the resistor R4 because it was very hot when working for 820R resistor.

But... the HD disk still not working... when the white LED is on the HD Disk light up, but still not working.

I made a test just like the drawing in this reply and the HD disk work just fine. The BD140 it takes the load without any problem. I just need to cut the negative signal to the base when the power is on.

 

The circuit was designed to turn off automatically for the HDD when the LM317 is on, you can measuring the Vce of Q3 and trying to figure out the theory and where the problem is, and to see the description on #16 carefully.

 

R4 should have been 82 Ohms instead of 820 Ohms. However, when the circuit is operating as designed, R4 will be dissipating TWO WATTS power, so it will need to be rated for at least 1.6 times that much, or at least 3.3 Watts. That's really not an acceptable level of power dissipation for a base resistor when devices like MOSFETs are available.

The transistors' absolute maximum continuous collector current rating is 1.5A. The load current has been stated as 1.5A. It is poor design practice to use a semiconductor device at it's maximums, as its' service life will be short. You should have a considerable safety factor added in there. I suggest de-rating semiconductors by 50%, particularly when you are new to electronics, to help to ensure that a project will have a decent chance at survival over time.

The 1N4007 is rated as a 1A 1KV silicon rectifier diode. The circuit load is 1.5A. Thus, the 1N4007 is not suitable for this circuit. Power dissipation would be excessive, as would the voltage drop across it.

I don't know why there is a 13v Zener being used, as that portion of the circuit was originally intended to limit the charge current. As it sits now, the Zener prevents the LM317's internal regulation from operating correctly, and the only current limiting is via the LM317's own internal limiting. This will probably cause the regulator to fail rather quickly, as it will run very, very hot - and in shutdown much of the time.

 

R4 should have been 82 Ohms instead of 820 Ohms. However, when the circuit is operating as designed, R4 will be dissipating TWO WATTS power, so it will need to be rated for at least 1.6 times that much, or at least 3.3 Watts. That's really not an acceptable level of power dissipation for a base resistor when devices like MOSFETs are available.

The transistors' absolute maximum continuous collector current rating is 1.5A. The load current has been stated as 1.5A. It is poor design practice to use a semiconductor device at it's maximums, as its' service life will be short. You should have a considerable safety factor added in there. I suggest de-rating semiconductors by 50%, particularly when you are new to electronics, to help to ensure that a project will have a decent chance at survival over time.

The 1N4007 is rated as a 1A 1KV silicon rectifier diode. The circuit load is 1.5A. Thus, the 1N4007 is not suitable for this circuit. Power dissipation would be excessive, as would the voltage drop across it.

I don't know why there is a 13v Zener being used, as that portion of the circuit was originally intended to limit the charge current. As it sits now, the Zener prevents the LM317's internal regulation from operating correctly, and the only current limiting is via the LM317's own internal limiting. This will probably cause the regulator to fail rather quickly, as it will run very, very hot - and in shutdown much of the time.

Hi,

Thank you Sir for your advice.

That´s the kind of advice i need.

The Zener is being use to stop the charge of the battery once is charged. I tested this part and by rather strange the LM317 don´t run very hot only warm.

I read about the MOSFET´s but i´m confuse. If you be so kind and offer some advice (part n and where do i put it in the circuit)

Thank´s
Regards