Looking into the NPN | NPN mirror ...

[jalbers@bsu.edu]

I am beginning to study the basic current mirror circuit consisting of
two matched NPN transistors as described at:
http://www.allaboutcircuits.com/vol_3/chpt_4/12.html

Let I be the current passing through Rb . The transistor on the left
is acting as a diode. The base current Ib should be I/2 since it is
being split between the two identical diodes wired in parallel. The
collector current Ic should be Beta * (I/2) . Then I am stuck trying
to mathematically show that the current through Rb should be identical
to the current flowing through Rload.

Any help would be greatly appreciated. Thanks.

 

[Andrew Holme]

<jalbers@bsu.edu> wrote in message
news:c3799859-5109-4422-ad8e-8db9facc1655@s20g2000prd.googlegroups.com...

I am beginning to study the basic current mirror circuit consisting of
two matched NPN transistors as described at:
http://www.allaboutcircuits.com/vol_3/chpt_4/12.html

Let I be the current passing through Rb . The transistor on the left
is acting as a diode. The base current Ib should be I/2 since it is
being split between the two identical diodes wired in parallel. The
collector current Ic should be Beta * (I/2) . Then I am stuck trying
to mathematically show that the current through Rb should be identical
to the current flowing through Rload.

Any help would be greatly appreciated. Thanks.

Try writing currents normalised to Ib:

1 on each base lead
2 on the wire between the left-hand collector and the bases
beta on each collector lead

(beta + 2) is the current through Rb

The ratio of resistor currents is beta / (beta+2) i.e. not exactly equal.

 

[jalbers@bsu.edu]

On Oct 8, 10:58 am, "jalb...@bsu.edu" <jalb...@bsu.edu> wrote:

I am beginning to study the basic current mirror circuit consisting of
two matched NPN transistors as described at:http://www.allaboutcircuits.com/vol_3/chpt_4/12.html

Let I be the current passing through Rb .  The transistor on the left
is acting as a diode.  The base current Ib should be I/2 since it is
being split between the two identical diodes wired in parallel.  The
collector current Ic should be Beta * (I/2) .  Then I am stuck trying
to mathematically show that the current through Rb should be identical
to the current flowing through Rload.

Any help would be greatly appreciated.  Thanks.

After further thought, the transistor on the left is not acting like a
simple diode. I think that it will always be saturated. If this is
true, then I = (Vcc - .6)/Rb and Ib = Ic / Beta * .5 . But where to
go from here ?

 

[Kevin Aylward]

jalbers@bsu.edu wrote:

I am beginning to study the basic current mirror circuit consisting of
two matched NPN transistors as described at:
http://www.allaboutcircuits.com/vol_3/chpt_4/12.html

Let I be the current passing through Rb . The transistor on the left
is acting as a diode. The base current Ib should be I/2 since it is
being split between the two identical diodes wired in parallel. The
collector current Ic should be Beta * (I/2) . Then I am stuck trying
to mathematically show that the current through Rb should be identical
to the current flowing through Rload.

Any help would be greatly appreciated. Thanks.

Forget the base current, it's essentially irrelevant to basic transistor
operation. The bipolar transistor is fundamentally a voltage controlled
device (e.g
http://www.kevinaylward.co.uk/ee/voltagecontrolledbipolar/voltagecontrolledbipolar.html),
so the above paper does nothing but confuse the issue.

First, pretend that the hfe/beta is infinite. There is a current in Q1,
given by I=(Vs - Vbe)/Rbias. Vbe doesn't vary much over a wide current
range, so one can have a fair approximation by initially assuming, say 0.7.
For a 5V supply, we would then have I= 4.3/Rbias,. If in fact at that
current, Vbe was 0.65, then the error would be only be 50m/4.3, not a lot,
so next...

This sets up a voltage, Vbe= Vt. Ln(I/Is), where Vt=KT/q

Since this is the same voltage that is applied to the vbe of Q2, the current
in Q2 must be the same as Q1, as Is for each transiter is the same..

The base current is just a next order error term. If hfe was 100, the error
would only be around 2%, hence ignoring base current, is the way to go...

Kevin Aylward
kevin@kevinaylward.co.uk
www.kevinaylward.co.uk

 

[Kevin Aylward]

ggherold@gmail.com wrote:

On Oct 8, 10:58 am, "jalb...@bsu.edu" <jalb...@bsu.edu> wrote:
I am beginning to study the basic current mirror circuit consisting
of two matched NPN transistors as described
at:http://www.allaboutcircuits.com/vol_3/chpt_4/12.html

Let I be the current passing through Rb . The transistor on the left
is acting as a diode. The base current Ib should be I/2 since it is
being split between the two identical diodes wired in parallel. The
collector current Ic should be Beta * (I/2) . Then I am stuck trying
to mathematically show that the current through Rb should be
identical to the current flowing through Rload.

Any help would be greatly appreciated. Thanks.

"The base current Ib should be I/2 since it is
being split between the two identical diodes wired in parallel."

I think this might be your mistake. The bias current from Rb flows
almost entirely through the collector of the transitor.

Yes.

Almost no
current flows along the wire connecting the collector to the base.

Yes.

I
know this looks wierd to me too, but that's the way it works. This
bias current "programs" the base-emmiter voltage of the first
transistor

No it doesn't. The base current is pretty much irrelevant to Vbe. Vbe is set
by the applied voltage at the Rbias node. We might argue, that Ic indirectly
"programs" Vbe. Consider, the transistor as "a diode connected transistor",
i.e. a diode.

From, http://www.kevinaylward.co.uk/ee/widlarlambert/widlarlambert.html

I can be calculated as Vs/R W((is.R/vt).(exp(vs/Vt))), W being the lambert W
function

and this voltage is transfered to the second transitor.
This forces the second transitor to have the same collector current as
the first.

Yes, its the fact that the Vbe's are equal.

If we had too identical transistor, except that one had a hfe of 100, and
the other had a hfe double at 200, the collector currents of each transistor
would still be, essentially, the same. Its the "is" of the transistor that
determines matching, and this current, has, essentially, nothing to do with
base current.

Note:

Ie = Is.exp(Vbe/Vt)

has no base current term in it, hence trying to explain current mirror
operaton based on base current, is futile.


Kevin Aylward
kevin@kevinaylward.co.uk
www.kevinaylward.co.uk

 

On Oct 8, 10:58 am, "jalb...@bsu.edu" <jalb...@bsu.edu> wrote:

I am beginning to study the basic current mirror circuit consisting of
two matched NPN transistors as described at:http://www.allaboutcircuits.com/vol_3/chpt_4/12.html

Let I be the current passing through Rb .  The transistor on the left
is acting as a diode.  The base current Ib should be I/2 since it is
being split between the two identical diodes wired in parallel.  The
collector current Ic should be Beta * (I/2) .  Then I am stuck trying
to mathematically show that the current through Rb should be identical
to the current flowing through Rload.

Any help would be greatly appreciated.  Thanks.

"The base current Ib should be I/2 since it is
being split between the two identical diodes wired in parallel."

I think this might be your mistake. The bias current from Rb flows
almost entirely through the collector of the transitor. Almost no
current flows along the wire connecting the collector to the base. I
know this looks wierd to me too, but that's the way it works. This
bias current "programs" the base-emmiter voltage of the first
transistor and this voltage is transfered to the second transitor.
This forces the second transitor to have the same collector current as
the first.

George Herold

 

[Robert Monsen]

On Wed, 8 Oct 2008 07:58:54 -0700 (PDT), "jalbers@bsu.edu"
<jalbers@bsu.edu> wrote:

I am beginning to study the basic current mirror circuit consisting of
two matched NPN transistors as described at:
http://www.allaboutcircuits.com/vol_3/chpt_4/12.html

Let I be the current passing through Rb . The transistor on the left
is acting as a diode. The base current Ib should be I/2 since it is
being split between the two identical diodes wired in parallel. The
collector current Ic should be Beta * (I/2) . Then I am stuck trying
to mathematically show that the current through Rb should be identical
to the current flowing through Rload.

Any help would be greatly appreciated. Thanks.

The main parameter here is the voltage between the base and the
emitter of the transistors. Note that it is equal. Since it is equal,
the current, which depends on the Vbe of the transistor, is equal.

I = Is * (exp(Vbe/Vt) - 1)

OTOH, there are effects which make the voltage to current relationship
described here less true than one would like. For example, the
collector voltage of the two transistors can be quite different. Since
the collector voltage affects the current that will come out of
identical transistors, given a fixed base voltage, then your current
won't be exactly equal (this is called the "Early Effect"). You can
fix the problem with a clever circuit like this:

o-----o-----------------.
| |

| |
|---------o---|

/| | |\
| | |
| '-----o V = Vcc - Vbe
| |
|V = Vcc - 2*Vbe |
| |<
o---------------|
| |\
| |
| |
| '--------------- out
/|\
( ~ ) ctrl
\_/ current
|
|
===
GND
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

The collector voltage of both transistors in the mirror is fixed by
this circuit to a value close to Vcc. So, as the ctrl current varies,
the output current deviates from the ctrl current much less than
before.

I think this is called a wilson mirror. A spice run with a 10V Vcc and
a ctrl current from 1mA to 20mA shows that the original current mirror
has about an 8% variance in current between the ctrl and output,
whereas the wilson mirror has less than 1% difference (more like
%0.7). Unfortunately, it has a smaller output range (compliance).

Regards,
Bob Monsen

 

Kevin Aylward wrote:

ggherold@gmail.com wrote:
On Oct 8, 10:58 am, "jalb...@bsu.edu" <jalb...@bsu.edu> wrote:
I am beginning to study the basic current mirror circuit consisting
of two matched NPN transistors as described
at:http://www.allaboutcircuits.com/vol_3/chpt_4/12.html

Let I be the current passing through Rb . The transistor on the left
is acting as a diode. The base current Ib should be I/2 since it is
being split between the two identical diodes wired in parallel. The
collector current Ic should be Beta * (I/2) . Then I am stuck trying
to mathematically show that the current through Rb should be
identical to the current flowing through Rload.

Any help would be greatly appreciated. Thanks.

"The base current Ib should be I/2 since it is
being split between the two identical diodes wired in parallel."

I think this might be your mistake. The bias current from Rb flows
almost entirely through the collector of the transitor.

Yes.

Almost no
current flows along the wire connecting the collector to the base.

Yes.

I
know this looks wierd to me too, but that's the way it works. This
bias current "programs" the base-emmiter voltage of the first
transistor

No it doesn't. The base current is pretty much irrelevant to Vbe. Vbe is set
by the applied voltage at the Rbias node. We might argue, that Ic indirectly
"programs" Vbe. Consider, the transistor as "a diode connected transistor",
i.e. a diode.

From, http://www.kevinaylward.co.uk/ee/widlarlambert/widlarlambert.html

I can be calculated as Vs/R W((is.R/vt).(exp(vs/Vt))), W being the lambert W
function

and this voltage is transfered to the second transitor.
This forces the second transitor to have the same collector current as
the first.

Yes, its the fact that the Vbe's are equal.

If we had too identical transistor, except that one had a hfe of 100, and
the other had a hfe double at 200, the collector currents of each transistor
would still be, essentially, the same. Its the "is" of the transistor that
determines matching, and this current, has, essentially, nothing to do with
base current.

Note:

Ie = Is.exp(Vbe/Vt)

has no base current term in it, hence trying to explain current mirror
operaton based on base current, is futile.


Kevin Aylward
kevin@kevinaylward.co.uk
www.kevinaylward.co.uk

OOPS!!

"No it doesn't. The base current is pretty much irrelevant to Vbe. Vbe
is set
by the applied voltage at the Rbias node. We might argue, that Ic
indirectly
"programs" Vbe. "

Thanks Kevin, my mistake. I meant to say that the "collector current"
programs Vbe.

It still is a bit "weird" that almost no current flows through the
wire connecting the collector and base.

George