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PNP transistor not working

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venkates2218

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pwr.jpg

When the power supply is give to the connector means the PNP transistor have to switch and +5V is given to the PIC controller's pin for reference.
But when i connected the PNP transistor means I can't get any output from the regulator.If I removed the PNP transistor means i can get +5V from the regulator.R4 is resistor is connected to ground,which used for PIC power supply.

This regulator circuit should active when the power supply is given to it.
 

I'm not sure what you are trying to achieve with this circuit but I would expect it to produce about 0.7V output from the regulator and consume a lot of current!

Hint: look at Vbe for the transistor and what it is connected across.

Brian.
 

Hi,

V_BE is about 0.6V... but you apply 5V to it, thus there will be (too) high I_BE, causing
* either the supply voktage to drop
* the transistor to be killed
* or both.

You need to limit I_B .. usually this is done by a resistor.

Btw: GND symbols are for free. Use them.

Klaus
 

Probably not. R10 should be grounded. Then you could use emitter as a reference voltage, but it will be less than you want.
Why don't use the regulator itself for reference voltage?
 

Hi,

The NPN is backwards, and the 1k R10 should be on the PIC enable pin side to limit input current, surely? If so, it could be a larger value, say 10k. The emitter should be facing the PIC input, not the regulator (see how you have a diode's cathode facing the output it uses?). It might have been better to use a PNP in the signal path instead, to get the most out of the 5V - 1k signal, but now that would mean using two transistors and it would "be wasteful", comparatively speaking: a PNP on/off switch for the enable path and an NPN to turn the PNP on/off.

- - - Updated - - -

Hi again,

You could be better served by sticking to your original circuit and just put an e.g. 10k or even better still 100k from the PNP base to ground and make R4 100k or better yet 1M and R1 10k. There's no power available to the PNP emitter unless the regulator is on, right?

You'll never get 5V with things in the way. A resistor drops a finite voltage based on current passing through it, a PNP or an NPN drops Vce voltage based on its internal resistance... If the PIC doesn't need over e.g. 4.5V, at a guess, the PNP version should be okay. Maybe forget the BJT switch completely if on reflection of what's happening there it is superfluous to the 5V being presented or not to the PIC enable pin. Hope I've understood what you're doing.
 

enable_2.png

This is the block diagram of my circuit.An INPUT of +12VDC to +18VDC is used as power supply.Positive and negative from the supply is give to the control unit and another point from the power supply is given to the switch unit.

The switch unit will enable the control circuit only if the both positive and negative supply is given to the 7805 regulator.the output +5VDC is given to the PIC controller to enable the circuit.The Input pin at PIC controller is pull down with 1K resistor.

At initial i tried this circuit without connecting an NPN or PNP transistor between controller and regulator.At this condition when we give positive supply from the regulator means also the circuit is working without negative from the switch side.

So I tried to introduce an transistor.At this condition,if we give both positive and negative supply means only the transistor will enable and circuit also work as per my requirement.
 

Hi,

It would help to understand what this is with a complete schematic drawn in a normal manner.

I think you would need far more complicated circuitry to accomplish this goal. Downside is it would need an additional voltage to supervise the presence of +V and 0V.
You seem to be describing a combinational logic function: 1 + 0 = 1.

For that you could use a "reverse polarity protection" PMOS in the power path with the gate tied to ground/negative supply, accepting a small voltage drop across the PMOS.

If the full 5V is needed, you'd get about 4.95V from a NAND gate output, but while a NAND would fulfil the basic premise of 0 + 1 = 1, it is too simplistic: 0 + 0 = 1, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 0...leading to problems if input leads are reversed.

Try the "reverse polarity protection" PMOS, maybe, it doesn't need a supervisor voltage and works.
 

Hi,

I agree.

A bit more detailed description is needed.
I see you are no friend of GND (symbols). Please use them. It helps to avoid misunderstandings.

Example: You talk about "negative" voltage. But with "negative" voltage one usually means negative with respect to GND.
Thus I'm not sure if you really have negative voltage or you mean the minus terminal of the supply used as GND.

You talk about 7805 regulator, but in your sketch we don't see where it is located.

Your purple line: is it a single wire and GND comes from the supply, or are ther two wires: control signal plus it's return path (usually GND)?

Klaus
 

Enable_3.png

The Controller will in on state,when the power(by using an switch.When switch is released both positive +12VDC and GND will be disconnected) is given in the J2 means RC0 pin have to go to high state and enable the operation in the controller.Th

Is this my requirement.Is this circuit for my requirement..
 

LED won't operate. And still don't undestand why you add a series transistor with the regulator?
 

LED won't operate. And still don't undestand why you add a series transistor with the regulator?

Without transistor If i give only positive supply to regulator means also it working.
I need to avoid that one.The RC0 pin shoukd enable if we give both positive and gnd in J2 means only
 

Could do with a clue here - is 'ENABLE' an input to this circuit or an output from it?
Are you perhaps trying to gate the signal on RC0 so it only gets passed onwards if the supply at J2 is present?
In either case, the LED will never light up.

Brian.
 

It's apparently voltage supervisory application. You would specify threshold voltages for all monitored voltages, e.g. +5V >= 4.75V, -5V <= -4.75 and use appropriate circuitry.
 

Hi,

The circuit should be like this:

**broken link removed**

The problem, and it's longwinded rather than complicated to explain, is this:

**broken link removed**

...because there's nothing to stop a positive voltage getting into the regulator ground pin or the RC0 pin through the resistor (besides the additional current that might be generated, as the simulation shows), to be honest, the post-regulator PNP switch is unnecessary in this circuit, in the real world all it will do is "steal" a bit of output voltage from the regulator and so provide a little less to the RC0 pin as a consequence.

Putting a diode in the positive path and a diode in the ground path is possible but horrible as the ground line diode lifts the entire ground reference for every device by e.g. 300mV to 700mV; using reverse polarity protection MOSFETs instead of the diodes is the same because it would shift the ground reference in a perhaps unacceptable fashion and I think would mean using back-to-back devices on each line (positive and ground) from what I was just doing.

Using a PNP connected as a shunt across the supply rails has a few issues, namely it does its job to a reasonable degree, so long as you don't mind it having to sink a large amount of current to divert it away from the rest of the circuit, but if you add a resistor to reduce the "PNP short to ground" then it won't be any use anymore as it will be the path of most resistance rather than the desired path of least resistance...

I guess an optocoupler is over-engineering and pointless too as the ground line is shared by all devices (regulator and MCU)?

It almost makes you think that a fast-blow fuse would be an option if the RC0 pin needs guaranteed protection from reverse polarity events.

I think I have run out of ideas for this application.

I'd just put a fuse and a (blocking) diode in the 12V to 18V path, or use a connector that makes it impossible to connect the power supply wrong way around, and perhaps omit the PNP which really doesn't provide any useful functionality.
 

d123, both attachments are invalid unfortunately.
 

As soon as we upload an attachment, I believe we should submit the post quickly because the attachment is preserved only a short time in limbo.
 

Hi,

Okay. Nice to know about how long attachments are kept.

This is with correct polarity:

pnp problem positive polarity.JPG

This is with reverse polarity:

pnp problem negative polarity.JPG
 

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