transistor NOT gate

[Michael]

I am trying to build a NOT gate similar to this one:

http://www.kpsec.freeuk.com/trancirc.htm#inverter

and was wondering how the resistor values of 1k and 10k were
determined.

 

[krw@att.bizzzzzzzzzzzz]

On Fri, 29 Oct 2010 17:44:52 -0700 (PDT), Michael <mrdarrett@gmail.com> wrote:

I am trying to build a NOT gate similar to this one:

http://www.kpsec.freeuk.com/trancirc.htm#inverter

and was wondering how the resistor values of 1k and 10k were
determined.

Makes the arithmetic easy?
Beta(min)=10?

 

[Michael Black]

On Fri, 29 Oct 2010, Michael wrote:

I am trying to build a NOT gate similar to this one:

http://www.kpsec.freeuk.com/trancirc.htm#inverter

and was wondering how the resistor values of 1k and 10k were
determined.


They found the circuit somewhere else.

They found a circuit but used resistor values that they had.

From experience they had a general feel for things, so they guessed.

They used trimpots and adjusted the values until things worked.

They did some or all of the above and then made sure it worked, maybe
making adjustments to the values to ensure it worked.

They got a calculator and formulas and planned it all out.

Michael

 

[Michael Robinson]

"Michael" <mrdarrett@gmail.com> wrote in message
news:5c6413f3-85fb-4e54-8e1d-ea4601033388@x42g2000yqx.googlegroups.com...

I am trying to build a NOT gate similar to this one:

http://www.kpsec.freeuk.com/trancirc.htm#inverter

and was wondering how the resistor values of 1k and 10k were
determined.

Consider the 1k load or "pullup" resistor.

You want high enough resistance that you don't waste a lot of power, but low
enough resistance that you have enough current to drive whatever stage comes
after the inverter (aka not gate). 1k is a pretty good value. At 5 volts
it only dissipates a fortieth of a watt, so you know you won't have to use a
big power resistor. Even at 12 volts, 1k only dissipates about a seventh of
a watt.
The base resistor just needs to deliver enough current to turn the
transistor on. It's a function of the gain. In this circuit, if the input
signal has the same voltage as Vs, then you would have about a 10 to 1
ratio of collector to base current. Just about any bipolar transistor you
can find, of the sort used in a circuit like this, will have much more gain
than that. So with a 10k base resistor you know the circuit will work.
They could have used a bigger base resistor. I guess they picked 10k
because it's a power of ten.

 

[Tim Wescott]

On 10/29/2010 07:43 PM, Michael Robinson wrote:

"Michael"<mrdarrett@gmail.com> wrote in message
news:5c6413f3-85fb-4e54-8e1d-ea4601033388@x42g2000yqx.googlegroups.com...
I am trying to build a NOT gate similar to this one:

http://www.kpsec.freeuk.com/trancirc.htm#inverter

and was wondering how the resistor values of 1k and 10k were
determined.

Consider the 1k load or "pullup" resistor.
You want high enough resistance that you don't waste a lot of power, but low
enough resistance that you have enough current to drive whatever stage comes
after the inverter (aka not gate). 1k is a pretty good value. At 5 volts
it only dissipates a fortieth of a watt, so you know you won't have to use a
big power resistor. Even at 12 volts, 1k only dissipates about a seventh of
a watt.
The base resistor just needs to deliver enough current to turn the
transistor on. It's a function of the gain. In this circuit, if the input
signal has the same voltage as Vs, then you would have about a 10 to 1
ratio of collector to base current. Just about any bipolar transistor you
can find, of the sort used in a circuit like this, will have much more gain
than that. So with a 10k base resistor you know the circuit will work.
They could have used a bigger base resistor. I guess they picked 10k
because it's a power of ten.

When the transistor goes into saturation, the beta goes down -- down
below 1, if you push it hard enough. Something like a 2N2222 or a
2N3904 could be expected to need that much base current to really work
reliably.

Note that this thing will be slooooooow at turning off unless there's
something positive to clear charges from the base. A resistor from
transistor base to ground would do wonders for faster turn-off, at the
cost of needing a lower series resistor for turn-on.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Michael]

On Oct 29, 7:43 pm, "Michael Robinson" <nos...@billburg.com> wrote:

"Michael" <mrdarr...@gmail.com> wrote in message

news:5c6413f3-85fb-4e54-8e1d-ea4601033388@x42g2000yqx.googlegroups.com...>I am trying to build a NOT gate similar to this one:

http://www.kpsec.freeuk.com/trancirc.htm#inverter

and was wondering how the resistor values of 1k and 10k were
determined.

Consider the 1k load or "pullup" resistor.
You want high enough resistance that you don't waste a lot of power, but low
enough resistance that you have enough current to drive whatever stage comes
after the inverter (aka not gate).  1k is a pretty good value.  At 5 volts
it only dissipates a fortieth of a watt, so you know you won't have to use a
big power resistor.  Even at 12 volts, 1k only dissipates about a seventh of
a watt.
The base resistor just needs to deliver enough current to turn the
transistor on.  It's a function of the gain.  In this circuit, if the input
signal has the same voltage as  Vs, then you would have about a 10 to 1
ratio of collector to base current.  Just about any bipolar transistor you
can find, of the sort used in a circuit like this, will have much more gain
than that.  So with a 10k base resistor you know the circuit will work.
They could have used a bigger base resistor.  I guess they picked 10k
because it's a power of ten.

Ok, thanks. I will have to ponder this information.

Michael

 

[Rich Grise]

On Fri, 29 Oct 2010 17:44:52 -0700, Michael wrote:

I am trying to build a NOT gate similar to this one:

http://www.kpsec.freeuk.com/trancirc.htm#inverter

and was wondering how the resistor values of 1k and 10k were
determined.

Pretty much by the seat of one's pants. ;-)

Cheers!
Rich

 

[Jon Kirwan]

On Fri, 29 Oct 2010 23:34:37 -0700, Tim Wescott
<tim@seemywebsite.com> wrote:

snip
When the transistor goes into saturation, the beta goes down

Yes.

-- down below 1, if you push it hard enough.

Yes, I suppose. Technically, it is possible to drive that
much current into the base. No one will argue that. But for
practical purposes, in that arrangement, I never see beta
"down below 1" as a design input.

Something like a 2N2222

.... would exhibit a typical beta of 100 at Vce=100mV and a
beta of 30 at Vce=50mV. That's a "more than practical" Vce,
and still way over 1 beta.

or a 2N3904 could be expected to need that much base current
to really work reliably.

I'm not an expert. You are. No question. Could you explain
this in more detail for a hobbyist type?

What do you mean by suggesting that one would need to plan
Ib=Ic in that topology for it to "work reliably." I am
mystified. I'd like to understand what you see here that
makes you comment in that way.

Jon

 

[Tim Wescott]

On 10/30/2010 03:52 AM, Jon Kirwan wrote:

On Fri, 29 Oct 2010 23:34:37 -0700, Tim Wescott
tim@seemywebsite.com> wrote:

snip
When the transistor goes into saturation, the beta goes down

Yes.

-- down below 1, if you push it hard enough.

Yes, I suppose. Technically, it is possible to drive that
much current into the base. No one will argue that. But for
practical purposes, in that arrangement, I never see beta
"down below 1" as a design input.

Something like a 2N2222

... would exhibit a typical beta of 100 at Vce=100mV and a
beta of 30 at Vce=50mV. That's a "more than practical" Vce,
and still way over 1 beta.

or a 2N3904 could be expected to need that much base current
to really work reliably.

I'm not an expert. You are. No question. Could you explain
this in more detail for a hobbyist type?

What do you mean by suggesting that one would need to plan
Ib=Ic in that topology for it to "work reliably." I am
mystified. I'd like to understand what you see here that
makes you comment in that way.

I'm not saying that you'd need Ib = Ic -- the beta < 1 example was
intentionally extreme, to show that you can't count on anything.

The answer, in all the detail you need, is -- look at the data sheet,
and understand what it means. In fact, had _I_ done that, I'd have had
my memory prompted about the 2N2222 being better than I remembered at
low Vce (or newer ones are better).

While you're looking at the data sheet you also need to consider if the
numbers are typical or guaranteed, and if they're for the whole
temperature range of the device or just at 25C. Then ask if you're
going to be happy with one prototype that works but can't be duplicated,
and if you're going to be happy with a device that only works at room
temperature or if you're ever going to take it outside.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

 

[Tom Biasi]

"Michael Black" wrote in message
newsine.LNX.4.64.1010292225290.14627@darkstar.example.net...

On Fri, 29 Oct 2010, Michael wrote:

I am trying to build a NOT gate similar to this one:

http://www.kpsec.freeuk.com/trancirc.htm#inverter

and was wondering how the resistor values of 1k and 10k were
determined.


They found the circuit somewhere else.

They found a circuit but used resistor values that they had.

From experience they had a general feel for things, so they guessed.

They used trimpots and adjusted the values until things worked.

They did some or all of the above and then made sure it worked, maybe
making adjustments to the values to ensure it worked.

They got a calculator and formulas and planned it all out.

Michael

Or, they posted their problem on the net and asked for help.

 

[Michael]

On Oct 31, 8:36 am, "Tom Biasi" <tombi...@optonline.net> wrote:

"Michael Black"  wrote in message

newsine.LNX.4.64.1010292225290.14627@darkstar.example.net...

On Fri, 29 Oct 2010, Michael wrote:
I am trying to build a NOT gate similar to this one:

http://www.kpsec.freeuk.com/trancirc.htm#inverter

and was wondering how the resistor values of 1k and 10k were
determined.

They found the circuit somewhere else.

They found a circuit but used resistor values that they had.

From experience they had a general feel for things, so they guessed.

They used trimpots and adjusted the values until things worked.

They did some or all of the above and then made sure it worked, maybe
making adjustments to the values to ensure it worked.

They got a calculator and formulas and planned it all out.

    Michael

Or, they posted their problem on the net and asked for help.

I suppose they could have plugged it into a SPICE program too... but I
was trying to determine the proper way to find these values, since I
know very little about transistors other than that they can be used as
electronic switches, sometimes.

Michael

 

[Rich Grise]

On Sun, 31 Oct 2010 16:11:24 -0700, Michael wrote:

On Oct 31, 8:36 am, "Tom Biasi" <tombi...@optonline.net> wrote:
"Michael Black"  wrote in message
On Fri, 29 Oct 2010, Michael wrote:
I am trying to build a NOT gate similar to this one:

http://www.kpsec.freeuk.com/trancirc.htm#inverter

and was wondering how the resistor values of 1k and 10k were
determined.

They found the circuit somewhere else.

They found a circuit but used resistor values that they had.

From experience they had a general feel for things, so they guessed.

They used trimpots and adjusted the values until things worked.

They did some or all of the above and then made sure it worked, maybe
making adjustments to the values to ensure it worked.

They got a calculator and formulas and planned it all out.

Or, they posted their problem on the net and asked for help.

I suppose they could have plugged it into a SPICE program too... but I
was trying to determine the proper way to find these values, since I
know very little about transistors other than that they can be used as
electronic switches, sometimes.

http://www.google.com/search?hl=en&source=hp&q=transistor+tutorial

Have Fun!
Rich

 

[Bill Bowden]

On Oct 29, 4:44 pm, Michael <mrdarr...@gmail.com> wrote:

I am trying to build a NOT gate similar to this one:

http://www.kpsec.freeuk.com/trancirc.htm#inverter

and was wondering how the resistor values of 1k and 10k were
determined.

They may be assuming use with the old TTL logic chips that require
about 2 milliamps for a low input, while the output is around 0.5
milliamp in the high state.
The TTL output voltage can also be a lot less than 5 volts, maybe only
2.5.

The transistor e/b junction will drop around 0.7 volts, so that leaves
2.5 - 0.7 = 1.8 volts across the 10K resistor, which is 180 microamps
of base current. The 1K resistor will require 5 milliamps to drop the
5 volt supply to near zero. So, the minimum gain for the transistor
is .005 / .00018 = 27 with no load.

Now, if you connect the transistor output to another TTL input, you
need about 2 milliamps to pull it low, so the transistor now has to
provide 5 milliamps to the 1K resistor and another 2 milliamps to the
load, so the minimum gain is now is .007 / .00018 = 39. Probably still
in the ballpark for a single TTL load.

-Bill

 

[krw@att.bizzzzzzzzzzzz]

On Thu, 4 Nov 2010 22:37:28 -0700 (PDT), Bill Bowden
<bperryb@bowdenshobbycircuits.info> wrote:

On Oct 29, 4:44 pm, Michael <mrdarr...@gmail.com> wrote:
I am trying to build a NOT gate similar to this one:

http://www.kpsec.freeuk.com/trancirc.htm#inverter

and was wondering how the resistor values of 1k and 10k were
determined.

They may be assuming use with the old TTL logic chips that require
about 2 milliamps for a low input, while the output is around 0.5
milliamp in the high state.

Yes, the standard TTL input(low) current was 1.6mA (16mA output low drive, for
a fanout of 10).

The TTL output voltage can also be a lot less than 5 volts, maybe only
2.5.

Certainly. TTL is far from symmetric as we've gotten used to with CMOS. The
output-high is current limited to contain smoke (output low, not so much) if
outputs are shorted to ground or other outputs. The output LPUL is 2.4V, with
at the input. The threshold is between 1.6V and 2.0V, IIRC for a 400mV noise
margin. Wow, those are old memories. ;-)

The transistor e/b junction will drop around 0.7 volts, so that leaves
2.5 - 0.7 = 1.8 volts across the 10K resistor, which is 180 microamps
of base current. The 1K resistor will require 5 milliamps to drop the
5 volt supply to near zero. So, the minimum gain for the transistor
is .005 / .00018 = 27 with no load.

Now, if you connect the transistor output to another TTL input, you
need about 2 milliamps to pull it low, so the transistor now has to
provide 5 milliamps to the 1K resistor and another 2 milliamps to the
load, so the minimum gain is now is .007 / .00018 = 39. Probably still
in the ballpark for a single TTL load.

But my bet is still that it works, and that 1K and 10K are nice looking
numbers.