Transistor voltage amplifier

[Allen]

Hi,

I need to "amplify" a dc voltage of 0-5V by 3 so it appears as 0-15V
on a voltmeter linearly. How do people do it before op-amps were
invented?

Does the 0-5V from a battery same as the 0-5V that comes out of a PWM
source (PIC) that appears as 0-5V to an analogue volt-meter?

Thanks for your time reading.

Allen

 

[Ian Bell]

Allen wrote:

Hi,

I need to "amplify" a dc voltage of 0-5V by 3 so it appears as 0-15V
on a voltmeter linearly. How do people do it before op-amps were
invented?

Two ways. First build an op amp out of transistors and resistors. It is

not that hard. The second way is to chop the input dc, ac amplify the
chopped dc then rectify and smooth ot back to dc. Chopper stabilised op
amps were quite common 40 years ago.

Ian

 

[Allen]

"electricked" <no_emails_please> wrote in message news:<nYednRGjxuvC1cfdRVn-sQ@comcast.com>...

"Allen" <sfbong@tm.net.my> wrote in message
news:f91a20cb.0403181531.5a960a55@posting.google.com...
Hi,

I need to "amplify" a dc voltage of 0-5V by 3 so it appears as 0-15V
on a voltmeter linearly. How do people do it before op-amps were
invented?

From my minimal experience with transistors I'm thinking you can drive the
base of a NPN transistor with the small voltage(0-5V) and then drive the
collector/emitter with 15V (might need to add more to drop through collector
resistor and transistor itself). When you apply the varrying tiny voltage to
the base of the transistor, that will open up the collector to emittor path
depending on the beta (amplification constant) and the voltage there would
be between 0 and 15V. You will have to add a suitable resistance before the
base as to limit the current when 5V are applied at base so it doesn't fry
the transistor.

This is exactly what I did. I used BC547B and put a resistor of 680K
between the variable DC input and the base of the transistor, and the
15V DC meter is in parallel with a 20K preset connected to the
collector BC547B. The other end of the meter is conected to +18V and
it worked.

The problem is even BC547B has a HFE between 250 to 450, so when I
duplicated the circuit, each of them had to be calibrated indivisually
and some of them are outside the range of the preset. So that's why I
think there must be a better way to do it !!

Allen

Does the 0-5V from a battery same as the 0-5V that comes out of a PWM
source (PIC) that appears as 0-5V to an analogue volt-meter?

Thanks for your time reading.

Allen

Yes, this is true. The voltage is the same, but the current capability might
be different. You have to read from datasheet if this matters.

--Viktor

 

[Allen]

John Popelish <jpopelish@rica.net> wrote in message news:<405A6260.918F8713@rica.net>...

Allen wrote:

Hi,

I need to "amplify" a dc voltage of 0-5V by 3 so it appears as 0-15V
on a voltmeter linearly. How do people do it before op-amps were
invented?

They succeeded in proportion to how well they invented and opamp.

Sorry for my poor understanding of you above statement. Are you
hinting that a transistorised op amp is the best option for making a
DC voltage amplifier if I do not want to use a readily available op
amp?

But my next problem is I do not want to use +/- Vcc on my design. Can
Op Amps work on single power supply just +15 and 0V?

Does the 0-5V from a battery same as the 0-5V that comes out of a PWM
source (PIC) that appears as 0-5V to an analogue volt-meter?

On average, yes. The PWM output may have an arbitrary amount of AC
ripple voltage riding on the DC average that does not show up on a DC
meter. Whether it is practically the same depends on how much ripple
there is, what its frequency is, and how sensitive your application is
to that ripple.

Thanks for the tips on PWM, I will take a closer look to see whether
the ripple voltage and frequency is having any implications on my
circuit. My current circuit is operating at 3 KHz and the meter seems
quite stable to human eyes. May be it's time for me to dig out my
scope.

Allen

 

[Allen]

Ian Bell <itb@yahoo.com> wrote in message news:<c3ek7u$271317$3@ID-225948.news.uni-berlin.de>...

Allen wrote:

Hi,

I need to "amplify" a dc voltage of 0-5V by 3 so it appears as 0-15V
on a voltmeter linearly. How do people do it before op-amps were
invented?

Two ways. First build an op amp out of transistors and resistors. It is
not that hard. The second way is to chop the input dc, ac amplify the
chopped dc then rectify and smooth ot back to dc. Chopper stabilised op
amps were quite common 40 years ago.

Ian

I have found a transistor op amp using just 4 transistors. 3 for the
differential amplifier and the 4th an emitter follower.

-----+------------+---------+----------------------+----------- +Vcc
| | | |
| | | |
R1 Rc1 Rc2 C Q4
| | |-------------------B
| Q1 | | Q2 E
| C C |
| non ---B B---+ inverting |
| invert E E | |-----||---Vout
| |----+----| | | C1
| | |-Rf1---||-------|
| | | Cf |
| Q3 C Rf2 Re4
|---------------B | |
| E | |
R2 | GND GND
| |
| Re1
| |
|-----------------+-------------------------------------- -Vcc

All transistors are NPN. Do you think this circuit would work if I
connect the required feedback resistors to make it a 3X voltage
amplifier?

As for the 2nd solution, I have very vague idea on how to chop the dc
component of PWM as it appears as spare pulses with amplitude 0-5V on
the scope. A llitle more clue would be appreciated.

Best regards
Allen

 

[Winfield Hill]

Allen wrote...

I have found a transistor op amp using just 4 transistors. 3 for the
differential amplifier and the 4th an emitter follower.

-----+------------+---------+----------------------+----------- +Vcc
| | | |
| | | |
R1 Rc1 Rc2 C Q4
| | |-------------------B
| Q1 | | Q2 E
| C C |
| non ---B B---+ inverting |
| invert E E | |-----||---Vout
| |----+----| | | C1
| | |-Rf1---||-------|
| | | Cf |
| Q3 C Rf2 Re4
|---------------B | |
| E | |
R2 | GND GND
| |
| Re1
| |
|-----------------+-------------------------------------- -Vcc

All transistors are NPN. Do you think this circuit would work if
I connect the required feedback resistors to make it a 3X voltage
amplifier?

The circuit hardly qualifies to be called an "opamp," it's an ac
feedback-controlled amplifier, not usable for dc amplification.
Note the output capacitor, necessary because the quiescent output
voltage is not anywhere near zero volts.

Some of us will chuckle as we remember horrific times, struggling
with the ancient uA703 and uA709 opamps. These were true NPN-only
opamps, although they better designed than the above circuit. This
circuit has several problems, which you can see with some thought.

First, limited output-voltage capability. With Q2 completely off
the output (before the coupling cap) is at Vcc less a diode drop,
and with Q2 fully on the output could be a much as two diode drops
more negative than ground, depending on the values of Re1 and Rc2.
So the output-voltage excursion range is essentially limited to
positve voltages, not very useful. Second, the open-loop gain is
limited, to about 10*Vcc, actually. For example, if Vcc = 15V the
gain would be about 150. That may be enough for a simple G = 3
amplifier, but not for an opamp or an accurate dc amplifier.

Opamps in the early 60s suffered from a lack of high-quality PNP
transistors. But you don't have that problem, so a simple change
in your circuit can make a big difference.

.. ,------------+---------+----------------------+-------- +Vcc
.. | | | |
.. | Rc1 Rc2 |
.. R1 | | E Q4
.. | +-------- | -------------------B
.. | Q1 | | Q2 C
.. | C C |
.. | non -- B B --, inverting |
.. | invert E E | +-------- Vout
.. | '----+----' | |
.. | | +--- Rf1 --------+
.. | | | |
.. | Q3 C Rf2 |
.. |-------------- B | |
.. | E | |
.. R2 | GND Re4
.. | | |
.. | Re1 |
.. | | |
.. '-----------------+---------------------------+-------- -Vee

Now the opamp gain will be *much* higher. If you need a low output
impedance, add an emitter follower after Q4. If a rising transistor
count bothers you, there is a solution. If your supply voltage is
stable, and if your signal is near ground, the Q3 current sink could
be replaced with a resistor to Vee, retaining a four-transistor count.

Thanks,
- Win

whill_at_picovolt-dot-com

 

[Allen]

Winfield Hill <Winfield_member@newsguy.com> wrote in message news:<c3hma60ip3@drn.newsguy.com>...

Allen wrote...

I have found a transistor op amp using just 4 transistors. 3 for the
differential amplifier and the 4th an emitter follower.

-----+------------+---------+----------------------+----------- +Vcc
| | | |
| | | |
R1 Rc1 Rc2 C Q4
| | |-------------------B
| Q1 | | Q2 E
| C C |
| non ---B B---+ inverting |
| invert E E | |-----||---Vout
| |----+----| | | C1
| | |-Rf1---||-------|
| | | Cf |
| Q3 C Rf2 Re4
|---------------B | |
| E | |
R2 | GND GND
| |
| Re1
| |
|-----------------+-------------------------------------- -Vcc

All transistors are NPN. Do you think this circuit would work if
I connect the required feedback resistors to make it a 3X voltage
amplifier?

The circuit hardly qualifies to be called an "opamp," it's an ac
feedback-controlled amplifier, not usable for dc amplification.
Note the output capacitor, necessary because the quiescent output
voltage is not anywhere near zero volts.

Some of us will chuckle as we remember horrific times, struggling
with the ancient uA703 and uA709 opamps. These were true NPN-only
opamps, although they better designed than the above circuit. This
circuit has several problems, which you can see with some thought.

First, limited output-voltage capability. With Q2 completely off
the output (before the coupling cap) is at Vcc less a diode drop,
and with Q2 fully on the output could be a much as two diode drops
more negative than ground, depending on the values of Re1 and Rc2.
So the output-voltage excursion range is essentially limited to
positve voltages, not very useful. Second, the open-loop gain is
limited, to about 10*Vcc, actually. For example, if Vcc = 15V the
gain would be about 150. That may be enough for a simple G = 3
amplifier, but not for an opamp or an accurate dc amplifier.

Opamps in the early 60s suffered from a lack of high-quality PNP
transistors. But you don't have that problem, so a simple change
in your circuit can make a big difference.

. ,------------+---------+----------------------+-------- +Vcc
. | | | |
. | Rc1 Rc2 |
. R1 | | E Q4
. | +-------- | -------------------B
. | Q1 | | Q2 C
. | C C |
. | non -- B B --, inverting |
. | invert E E | +-------- Vout
. | '----+----' | |
. | | +--- Rf1 --------+
. | | | |
. | Q3 C Rf2 |
. |-------------- B | |
. | E | |
. R2 | GND Re4
. | | |
. | Re1 |
. | | |
. '-----------------+---------------------------+-------- -Vee

Now the opamp gain will be *much* higher. If you need a low output
impedance, add an emitter follower after Q4. If a rising transistor
count bothers you, there is a solution. If your supply voltage is
stable, and if your signal is near ground, the Q3 current sink could
be replaced with a resistor to Vee, retaining a four-transistor count.

Thanks,
- Win

whill_at_picovolt-dot-com

Hi Winfield,

I have constructed the circuit below as you advised. I picked Q1 and
Q2 with almost the same HFE (150) and Q4 HFE is 350.

. ,------------+---------+----------------------+-------+-----+Vcc
. | | | | |
. | Rc1 4K7 Rc2 4K7 | |
. R1 56K | | E Q4 |
. | +-------- | -------------------B PNP |
. | Q1 | | Q2 C |
. | C C | |
. | non -- B B -- inverting | C Q5
. | invert E E +-----B NPN
. | '----+----' | E
. | | + |
. | | | |
. | Q3 C | +---Vout
. |-------------- B | |
. | E | |
. R2 10K | Re4 Re5
. | | |4K7 | 1K
. | Re1 1K | |
. | | | |
. '-----------------+---------------------------+-------+---- -Vee

But when I connect the resistors to make it work as a G=3 voltage
amp., it didn't work as intended. I connected it as below:

+10V
|
+\ |
| \|
| \
Vin -----10K-+-------|- \
| | \_________ Vout
| | / |
| +---|+ / |
| | | / |
| | | /| |
| 7K5 +/ | |
| | | |
| GND -10V |
| |
+-------30K-------+

The input is connected to a 5-poles-single-throw switch where the
inputs are 5 10K resistors connected as voltage divder terminated to
GND and +5V. When the Vin is 1V, I am getting -1V at the Vout, and
Vin 2V gets Vout -2V.....

Please tell me what is wrong with the circuit?

Allen.

 

[Allen]

John Popelish <jpopelish@rica.net> wrote in message news:<405BD9FE.84AE4C0E@rica.net>...

Allen wrote:

John Popelish <jpopelish@rica.net> wrote in message news:<405A6260.918F8713@rica.net>...
Allen wrote:

Hi,

I need to "amplify" a dc voltage of 0-5V by 3 so it appears as 0-15V
on a voltmeter linearly. How do people do it before op-amps were
invented?

They succeeded in proportion to how well they invented and opamp.

Sorry for my poor understanding of you above statement. Are you
hinting that a transistorised op amp is the best option for making a
DC voltage amplifier if I do not want to use a readily available op
amp?

I apologize for being flippant. The point is that an opamp is a well
developed solution to the problem of DC amplification. If you don't
use one, you essentially have to reinvent it to do a good job at this
task.

You don't have to apologize and I was not offended at all. The
misunderstanding is caused by my poor English. But at least I have
leant a new word "flippant". I am going to use it in future ;-)

But my next problem is I do not want to use +/- Vcc on my design. Can
Op Amps work on single power supply just +15 and 0V?

Sure. Opamps work over a wide range of voltages, depending on the
specific model. The problem is to keep the inputs and output within
their operating range and that is somewhere between the supply rails.
Some units work with the inputs anywhere between the rails (called
rail-to-rail opamps, amazingly enough).

The cheap and dirt common LM324 quad opamp works with supplies as low
as 5 volts total, though its inputs and output work only from the
negative rail to 1.5 volts below the positive supply rail. So don't
expect a 15 volt output swing from it with a 15 volt supply.

I took a look at the datasheet of the LM324 and I was amazed as it was
designed for single supply from the ground up. Is there an LM324 with
only 2 op-amps in dip-8 foot-print?

What is the difference between a comparator from an op-amp. Can LM393
(2 comp) be used to replace the LM324 (4 op-amp)?

John, I must thank you for being so helpful in this NG and I admire
your wide scope of knowledge in electronics!!

Cheers.


Allen

 

[John Popelish]

Allen wrote:

John Popelish <jpopelish@rica.net> wrote in message news:<405BD9FE.84AE4C0E@rica.net>...
Allen wrote:
....
But my next problem is I do not want to use +/- Vcc on my design. Can
Op Amps work on single power supply just +15 and 0V?

Sure. Opamps work over a wide range of voltages, depending on the
specific model. The problem is to keep the inputs and output within
their operating range and that is somewhere between the supply rails.
Some units work with the inputs anywhere between the rails (called
rail-to-rail opamps, amazingly enough).

The cheap and dirt common LM324 quad opamp works with supplies as low
as 5 volts total, though its inputs and output work only from the
negative rail to 1.5 volts below the positive supply rail. So don't
expect a 15 volt output swing from it with a 15 volt supply.

I took a look at the datasheet of the LM324 and I was amazed as it was
designed for single supply from the ground up. Is there an LM324 with
only 2 op-amps in dip-8 foot-print?

That would be the LM 358:
http://cache.national.com/ds/LM/LM158.pdf

There are lots of other opamps designed ot work from single or low
voltage supplies, many with better high frequency capability than the
LM324 and LM358.

For example (just from National Semiconductor)
low voltage amplifiers:
http://www.national.com/parametric/0,,2812,00.html
low power amplifiers:
http://www.national.com/parametric/0,,696,00.html
rail-to-rail input voltage range:
http://www.national.com/parametric/0,,2801,00.html

What is the difference between a comparator from an op-amp. Can LM393
(2 comp) be used to replace the LM324 (4 op-amp)?

Opamps not only have high differential gain, but are slowed down
enough that they can react ot their own outputs, mid swing to produce
stable output voltages based on negative feedback.

Comparators have similar differential gain, but are optimized to swing
their outputs as fast as possible between two logic states. Most
comparators do not even have a pull up- pull down output stage, but
just a single transistor connected to the negative supply rail, so you
need a pull up resistor added to the output to get any voltage swing
out of them.

In effect, comparators are 1 bit analog to digital converters.

John, I must thank you for being so helpful in this NG and I admire
your wide scope of knowledge in electronics!!

Wide, but not so deep, perhaps. But it is nice to be appreciated.

--
John Popelish

 

[Winfield Hill]

Allen wrote...

Winfield Hill wrote ...

. ,------------+---------+----------------------+-------- +Vcc
. | | | |
. | Rc1 Rc2 |
. R1 | | E Q4
. | +-------- | -------------------B
. | Q1 | | Q2 C
. | C C |
. | non -- B B --, inverting |
. | invert E E | +-------- Vout
. | '----+----' | |
. | | +--- Rf1 --------+
. | | | |
. | Q3 C Rf2 |
. |-------------- B | |
. | E | |
. R2 | GND Re4
. | | |
. | Re1 |
. | | |
. '-----------------+---------------------------+-------- -Vee

Now the opamp gain will be *much* higher. If you need a low output
impedance, add an emitter follower after Q4. If a rising transistor
count bothers you, there is a solution. If your supply voltage is
stable, and if your signal is near ground, the Q3 current sink could
be replaced with a resistor to Vee, retaining a four-transistor count.

Hi Winfield,

I have constructed the circuit below as you advised. I picked Q1 and
Q2 with almost the same HFE (150) and Q4 HFE is 350.

,------------+---------+----------------------+-------+-----+Vcc
| | | | |
| Rc1 4K7 Rc2 4K7 | |
R1 56K | | E Q4 |
| +-------- | -------------------B PNP |
| Q1 | | Q2 C |
| C C | |
| non -- B B -- inverting | C Q5
| invert E E +-----B NPN
| '----+----' | E
| | + |
| | | |
| Q3 C | +---Vout
|-------------- B | |
| E | |
R2 10K | Re4 Re5
| | |4K7 | 1K
| Re1 1K | |
| | | |
'-----------------+---------------------------+-------+---- -Vee

But when I connect the resistors to make it work as a G=3 voltage
amp., it didn't work as intended.

OK, let's analyze your circuit, running from +/-10V supplies: First
we see 3.0 volts on Q3's base, so the longtail-pair current is about
2.4mA, which means each side should have 1.2mA when balanced. But
1.2mA across 4.7k is 5.6 volts!, whereas we want only 0.65V across
the Q4 base-emitter junction. Instead, with 4.7k, the circuit will
try to stabilize with Q1 running at about 0.138mA, which forces Q2
to run at about 2.26mA, creating a 10.6V drop across its resistor.

This looks like it'll saturate Q2 and wreck the circuit's operation.
But if it fails to saturate Q2, it'll mean Q2 operates at about 16x
the current of Q1, so the input voltage offset will be 70mV more
(see AoE page 91) than the Q1 - Q2 mismatch, which is pretty bad!

So Rc1 should be about 560 ohms, and Rc2 can be eliminated entirely!
Alternately with Rc1 = 4.7k, drop the input-stage current to 275uA
by increasing Re1 to about 8.7k. Now try the circuit again.

Thanks,
- Win

whill_at_picovolt-dot-com

 

[Allen]

Winfield Hill <Winfield_member@newsguy.com> wrote in message news:<c41rki0972@drn.newsguy.com>...

Hi Mr. Winfield,

I have constructed the circuit below as you advised. I picked Q1 and
Q2 with almost the same HFE (150) and Q4 HFE is 350.

,------------+---------+----------------------+-------+-----+Vcc
| | | | |
| Rc1 4K7 Rc2 4K7 | |
R1 56K | | E Q4 |
| +-------- | -------------------B PNP |
| Q1 | | Q2 C |
| C C | |
| non -- B B -- inverting | C Q5
| invert E E +-----B NPN
| '----+----' | E
| | + |
| | | |
| Q3 C | +---Vout
|-------------- B NPN | |
| E | |
R2 10K | Re4 Re5
| | |4K7 | 1K
| Re1 8K7 | |
| | | |
'-----------------+---------------------------+-------+---- -Vee

But when I connect the resistors to make it work as a G=3 voltage
amp., it didn't work as intended.

OK, let's analyze your circuit, running from +/-10V supplies: First
we see 3.0 volts on Q3's base, so the longtail-pair current is about
2.4mA, which means each side should have 1.2mA when balanced. But
1.2mA across 4.7k is 5.6 volts!, whereas we want only 0.65V across
the Q4 base-emitter junction. Instead, with 4.7k, the circuit will
try to stabilize with Q1 running at about 0.138mA, which forces Q2
to run at about 2.26mA, creating a 10.6V drop across its resistor.

This looks like it'll saturate Q2 and wreck the circuit's operation.
But if it fails to saturate Q2, it'll mean Q2 operates at about 16x
the current of Q1, so the input voltage offset will be 70mV more
(see AoE page 91) than the Q1 - Q2 mismatch, which is pretty bad!

Where can I find AoE page 91?

So Rc1 should be about 560 ohms, and Rc2 can be eliminated entirely!
Alternately with Rc1 = 4.7k, drop the input-stage current to 275uA
by increasing Re1 to about 8.7k. Now try the circuit again.

I didn't try the 560 ohms approach but instead I changed Re1 from 1K
to 8.7K and it was improving. I found that the voltage on my voltage
divider constructed using 5 10K resistors was effected by the input of
my op-amp so I changed the 10K to 1K and then it worked like a charm.

The output is 3 times the input voltage in the negative direction. ie
Vin = 1V then Vout= -3V; Vin=2V then Vout= -6V with resp to GND. the
voltage drop across Re4 and Re5 are as follows:

Vin Vre4 Vre5
1V 8.76V 8.09V
2V 5.86V 5.33V
3v 2.89V 2.27V

I forgot to mention that during experimenting, I acidentally shorted
the +10V & -10V and fried my power supply. After I get another power
supply (from old pc) The Vcc and Vee becomes +12 and -11.14V resp.
How do I calculate the input impedance of my Op-Amp?

The next thing I am going to experiment is to modified it to work rail
to rail like a LM324. I will ask more questions when I run into rocks
;-)

Thank you very, very much for your assistance and I appreciated it
very much.

Allen.

Thanks,
- Win

whill_at_picovolt-dot-com

 

[Winfield Hill]

Allen wrote...

Where can I find AoE page 91?

Possibly at a local library. Learning and self help is
always a good idea. http://www.artofelectronics.com

Thanks,
- Win

whill_at_picovolt-dot-com

 

[Watson A.Name - \"Watt Su]

"Allen" <sfbong@tm.net.my> wrote in message
news:f91a20cb.0403260408.35c49d8d@posting.google.com...
[snip]

I took a look at the datasheet of the LM324 and I was amazed as it was
designed for single supply from the ground up. Is there an LM324 with
only 2 op-amps in dip-8 foot-print?

The LM358 is half of a LM324, in an 8-pin package.

Cheers.

Allen

 

[Watson A.Name - \"Watt Su]

"Allen" <sfbong@tm.net.my> wrote in message
news:f91a20cb.0403270838.51d05b68@posting.google.com...
[snip]

(see AoE page 91) than the Q1 - Q2 mismatch, which is pretty bad!

Where can I find AoE page 91?

Art of Electronics - go to this URL 'www.everybookstore.com' and then
click on compare prices.
http://ebs.allbookstores.com/search?type=title&q=Art+of+Electronics&Go.x
=0&Go.y=0&Go=Go

[snip]

Allen.