Inner resistance: supplying correct current and voltage?

[Björn]

A chip (a small digital radio reciever chip) should have 5 volts and
4,5 mA according to its specifications. Now, if it had 1111 ohm of
inner resistance, I would only need to connect it to a 5 volt battery,
and the current would be 5/1111 = 0,0045. If the inner resistance had
been smaller, I could've used external resistors to arrive at the
target current and voltage. Not a problem.

However, I measure the inner resistance of the reciever to 18000 ohm.
Since 5/18000 = 0,00028 (0,28 mA) and 18000*0,0045 = 81 volt, I cannot
achieve the specified voltage and current, can I?

This is of course a very basic question. Anyone who has connected
anything to anything must have had the same problem, and found a
clever solution which unfortunately escapes me Should I maybe
ignore the inner resistance I've measured (when the chip is off, of
course) and just put 5 volt between the VCC and GND pins? But then,
what's the point with specifying a current in the first place?

 

[Peter Bennett]

On 30 Jul 2003 06:24:27 -0700, 17538@student.hhs.se (Björn) wrote:

A chip (a small digital radio reciever chip) should have 5 volts and
4,5 mA according to its specifications. Now, if it had 1111 ohm of
inner resistance, I would only need to connect it to a 5 volt battery,
and the current would be 5/1111 = 0,0045. If the inner resistance had
been smaller, I could've used external resistors to arrive at the
target current and voltage. Not a problem.

No - you run the thing from the advertised 5 volts, and let the
current fall where it may.

However, I measure the inner resistance of the reciever to 18000 ohm.
Since 5/18000 = 0,00028 (0,28 mA) and 18000*0,0045 = 81 volt, I cannot
achieve the specified voltage and current, can I?

The "inner resistance" you measure or calculate is not just resistance
- it is actually the overall result of the internal circuit of the
chip, including transistors which may react differently with different
voltages applied to the part.

When you measure the "resistance" of the chip with an ohmmeter, you
will be applying much less than 5 volts to the circuit, so most of the
transistors and diodes inside the chip will not be conducting.

Should I maybe
ignore the inner resistance I've measured (when the chip is off, of
course) and just put 5 volt between the VCC and GND pins?

Yes. The part is designed to operate from 5 volts, and will draw
whatever current it requires. According to your spec, the "typical"
current will be 4.5 mA - but the actual current will vary with
production variations in the devices, and probably also with what the
device is doing at the time, and what it is driving.

But then,
what's the point with specifying a current in the first place?

The current is specified so that the equipment designer who uses this
part can determine how big a power supply he will need.



--
Peter Bennett VE7CEI
GPS and NMEA info and programs: http://vancouver-webpages.com/peter/index.html
Newsgroup new user info: http://vancouver-webpages.com/nnq

 

[Wade Hassler]

17538@student.hhs.se (Björn) wrote in message news:<b118f60c.0307300524.8198a33@posting.google.com>...

A chip (a small digital radio reciever chip) should have 5 volts and
4,5 mA according to its specifications. Now, if it had 1111 ohm of
inner resistance, I would only need to connect it to a 5 volt battery,
and the current would be 5/1111 = 0,0045. If the inner resistance had
been smaller, I could've used external resistors to arrive at the
target current and voltage. Not a problem.

However, I measure the inner resistance of the reciever to 18000 ohm.
Since 5/18000 = 0,00028 (0,28 mA) and 18000*0,0045 = 81 volt, I cannot
achieve the specified voltage and current, can I?

This is of course a very basic question. Anyone who has connected
anything to anything must have had the same problem, and found a
clever solution which unfortunately escapes me Should I maybe
ignore the inner resistance I've measured (when the chip is off, of
course) and just put 5 volt between the VCC and GND pins? But then,
what's the point with specifying a current in the first place?

The current-consumption is specified so that you may size your power
supply appropriately: it must provide enough current to run your
receiver and
any other loads, all at the right voltage.
Wade Hassler

 

[Joshua K Drumeller]

Just hook the component to the specified voltage and it will draw the
current it needs to operate by its self.

Josh

"Björn" <17538@student.hhs.se> wrote in message
news:b118f60c.0307300524.8198a33@posting.google.com...

A chip (a small digital radio reciever chip) should have 5 volts and
4,5 mA according to its specifications. Now, if it had 1111 ohm of
inner resistance, I would only need to connect it to a 5 volt battery,
and the current would be 5/1111 = 0,0045. If the inner resistance had
been smaller, I could've used external resistors to arrive at the
target current and voltage. Not a problem.

However, I measure the inner resistance of the reciever to 18000 ohm.
Since 5/18000 = 0,00028 (0,28 mA) and 18000*0,0045 = 81 volt, I cannot
achieve the specified voltage and current, can I?

This is of course a very basic question. Anyone who has connected
anything to anything must have had the same problem, and found a
clever solution which unfortunately escapes me Should I maybe
ignore the inner resistance I've measured (when the chip is off, of
course) and just put 5 volt between the VCC and GND pins? But then,
what's the point with specifying a current in the first place?