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'[PIC]: Re: PIC control of PIC power supply'
2002\01\10@103257 by

James Paul wrote, in part:
>[...]
>  Lets say we have an NPN transistor and that transistor has a Beta of
>  100.  Lets further say that the circuit we are powering uses 300 mA of
>  current.  To drive this transistor full on, we'll make the Base current
>  about 10% of the Collector current.  So in this case, the Base current
>  will be ~30 mA.   Now lets say that the PIC pin will output a worst case
>  logic high level of 4.5 volts.  So, if we divide the 4.5 volts by 30 mA,
>  we get a value of 150 ohms.  So, if we use a 150 ohm resistor in the
>  Base, and connect a PIC pin to the other end of the resistor, and make
>  the PIC output a logic high, [...]

The problem with this configuration is that you are asking the PIC
i/o pin to source 30mA of current. According to the 16F84 data sheet,
for example, under "Absolute Maximum Ratings", the most current that
can be sourced by any i/o pin is 25mA. So it sounds to me as though
you are asking for more current than the PIC can comfortably supply,
as well as much more than is actually needed (you only need a few
mA going into the base).

Michael V

Thank you for reading my little posting.

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On 10-Jan-02 Michael Vinson wrote:
> James Paul wrote, in part:
>>[...]
>>  Lets say we have an NPN transistor and that transistor has a Beta of
>>  100.  Lets further say that the circuit we are powering uses 300 mA of
>>  current.  To drive this transistor full on, we'll make the Base current
>>  about 10% of the Collector current.  So in this case, the Base current
>>  will be ~30 mA.   Now lets say that the PIC pin will output a worst case
>>  logic high level of 4.5 volts.  So, if we divide the 4.5 volts by 30 mA,
>>  we get a value of 150 ohms.  So, if we use a 150 ohm resistor in the
>>  Base, and connect a PIC pin to the other end of the resistor, and make
>>  the PIC output a logic high, [...]
>
> The problem with this configuration is that you are asking the PIC
> i/o pin to source 30mA of current. According to the 16F84 data sheet,
> for example, under "Absolute Maximum Ratings", the most current that
> can be sourced by any i/o pin is 25mA. So it sounds to me as though
> you are asking for more current than the PIC can comfortably supply,
> as well as much more than is actually needed (you only need a few
> mA going into the base).

Michael

The problem with this is that James' calculations above are wrong IMHO.

If the transistor  has a beta (Ic/Ib) of 100, and the collector current is
300mA, then the base current is 3mA, NOT 30mA.   I'm not sure where the "10% of
the collector current" rule came from, but I think it's wrong.

The only thing I can think of is the rule that a common emitter bias network
should be designed with a standing current of ~10 times the base current.

Peter.

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E-Mail: Peter Onion <ponionsrd.bt.co.uk>
Date: 10-Jan-02
Time: 16:00:58

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>If the transistor  has a beta (Ic/Ib) of 100, and the collector current is
>300mA, then the base current is 3mA, NOT 30mA.   I'm not sure where the
"10% of
>the collector current" rule came from, but I think it's wrong.

You will not get a transistor to saturate if you use the published beta
value to get the base current. Transistors in saturation always have the
base overdriven to force the device right into saturation. In the case of
bipolar transistors this is known as "forced beta", as the common way of
doing it is to drive the base with a current that is many times what is
normally expected for the same collector current in linear operation.

The general rule of thumb is to use a forced beta of 10 to get a transistor
saturated to the point where the C-E voltage is really at it's minimum
possible. If one worked on a forced beta of 25, I think you would find the
voltage would go up to 0.3 to 0.4. If you relied on the published beta value
the saturation will rise to about the B-E voltage, in other words about 3
times what is achievable.

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On 10-Jan-02 Alan B. Pearce wrote:
>>If the transistor  has a beta (Ic/Ib) of 100, and the collector current is
>>300mA, then the base current is 3mA, NOT 30mA.   I'm not sure where the
> "10% of
>>the collector current" rule came from, but I think it's wrong.
>
> You will not get a transistor to saturate if you use the published beta
> value to get the base current. Transistors in saturation always have the
> base overdriven to force the device right into saturation. In the case of
> bipolar transistors this is known as "forced beta", as the common way of
> doing it is to drive the base with a current that is many times what is
> normally expected for the same collector current in linear operation.

I guess I've been doing too much analogue & RF design work !  Indeed for
saturation the base needs to be overdriven.  In this application I would
probably go for a darlington pair, but if we are going to use two transistors I
would use a NPN with emitter to ground and PNP with it's emitter to the +ve
supply and a couple of current limit resistors in each base.  This also gets
over the problem with PIC output clamp diodes if trying to switch a supply
greater that the PICs supply voltage.

Peter.

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2002\01\10@121839 by
>In this application I would
>probably go for a darlington pair

Be very careful trying to saturate a darlington transistor, as the
saturation voltage is at least one Vbe (for the second transistor) + the Vce
of the first transistor. When trying to do this make sure the two transistor
collectors are not connected together by using separate discrete devices.
Think about the internal circuit of an integrated device.

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You know, there would never be all of this conversation if MOSFET's were
used. They are so much simpler, especially for this switching application.

Cliff

{Original Message removed}
On 10-Jan-02 Alan B. Pearce wrote:
>>In this application I would
>>probably go for a darlington pair
>
> Be very careful trying to saturate a darlington transistor, as the
> saturation voltage is at least one Vbe (for the second transistor) + the Vce
> of the first transistor.

Yes, I've had problems with that in the past when the collector voltage of the
2nd device falls close to ground the 1st device can no longer provide transistor
action and provide the drive current to the 2nd device.  I seem to remember (it
was a few years ago) that something odd happens as the first device starts to
act as just a diode (be junction) and thus the current gain of the pair falls
and thus you need to provide more base current than you expect.

Peter.

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