Dear Sirs,
first of all I declare my ignorance in electronic circuits design.
I just need a minimal schema for a single bjt buffer to be used in a unity gain preamp (or directions to where I cand find it).
It should drive easily a 1000 ohm load (so some amount of current is required).
I would like to use a 24V power supply.
I have big problems in choosing the right resistance values.
Thank you to anyone willing to help me.
Kind regards.
beppe61
first of all I declare my ignorance in electronic circuits design.
I just need a minimal schema for a single bjt buffer to be used in a unity gain preamp (or directions to where I cand find it).
It should drive easily a 1000 ohm load (so some amount of current is required).
I would like to use a 24V power supply.
I have big problems in choosing the right resistance values.
Thank you to anyone willing to help me.
Kind regards.
beppe61
Hi
A NPN transistor can be arranged
as follows:
Use a NPN such as the BC547 which should be capable
into a 1000 ohm load.
The collector pin is biased at 1000 ohms for every volt of supply
(so you will need about 24K - 22K or 27K will be fine )
running from your DC 24 positive volt source
( with the transistor writing facing you the collector is on the left)
with a BC547 )
The output of the circuit is on the collector also and is
usually arranged with a coupling capacitor you should make
the voltage rating equal or higher than the DC voltage
supply - its positive leg if using a bipolar cap is connected to the collector leg with the negative available as the circuit output.
The base ( the middle pin ) has the input source attached
also capacitively coupled. ( a 220n cap should do)
From the base back to the collector arrange a resistor of
at least 1Meg ohm up to 3M3 this self biases the base of the transistor.
The emitter the pin on the right - and the only one
available now - is usually just grounded. This forces
the input signal at now a much higher level to be
opposite in phase to appear on the collector.
If you need some control over the level of this little circuit
to assist in it being unity gain you could make the emitter the output - which is then the same phase as the input of course it would have to be ground lifted and this is where you
could control the level with resistance making the capacitive coupling from the emitter leg and resistance - to start with say
470 ohm to ground. The emitter then drives the circuit rather
than the collector and can be very effectively trimmed for
output level. You will however reach a point where the
audio frequency will reduce as it is just too close to ground.
You could of course resistively tailor the output taken from the capacitor attached to the emitter or collector so that it
approaches the unity gain you require. 3K3 in series with about 600 ohm to ground might be a starting point to try
Hope this helps, this circuit is a good performer and hopefully
will provide the result you desire.
Cheers / Chris
A NPN transistor can be arranged
as follows:
Use a NPN such as the BC547 which should be capable
into a 1000 ohm load.
The collector pin is biased at 1000 ohms for every volt of supply
(so you will need about 24K - 22K or 27K will be fine )
running from your DC 24 positive volt source
( with the transistor writing facing you the collector is on the left)
with a BC547 )
The output of the circuit is on the collector also and is
usually arranged with a coupling capacitor you should make
the voltage rating equal or higher than the DC voltage
supply - its positive leg if using a bipolar cap is connected to the collector leg with the negative available as the circuit output.
The base ( the middle pin ) has the input source attached
also capacitively coupled. ( a 220n cap should do)
From the base back to the collector arrange a resistor of
at least 1Meg ohm up to 3M3 this self biases the base of the transistor.
The emitter the pin on the right - and the only one
available now - is usually just grounded. This forces
the input signal at now a much higher level to be
opposite in phase to appear on the collector.
If you need some control over the level of this little circuit
to assist in it being unity gain you could make the emitter the output - which is then the same phase as the input of course it would have to be ground lifted and this is where you
could control the level with resistance making the capacitive coupling from the emitter leg and resistance - to start with say
470 ohm to ground. The emitter then drives the circuit rather
than the collector and can be very effectively trimmed for
output level. You will however reach a point where the
audio frequency will reduce as it is just too close to ground.
You could of course resistively tailor the output taken from the capacitor attached to the emitter or collector so that it
approaches the unity gain you require. 3K3 in series with about 600 ohm to ground might be a starting point to try
Hope this helps, this circuit is a good performer and hopefully
will provide the result you desire.
Cheers / Chris
peranders said:
Dear Mr. Peranders,
thank you very much indeed for your kind and very valuable reply.
Nevertheless I would like to stay as simplest as possible.
Just to see what is possible with a very basic circuit.
With Vcc=24V I have to set :
the Re value
and the value of the two resistors to be placed before the bjt base to make the bjt itself working.
I have read that the voltage differenze Vbase-Vemitter must be around 0,7 V.
Am I wrong?
But I don't know how to calculate Ve.
After that I can calculate the right voltage divider to use in front of the bjt.
I would like to use a BD139. Is it fine?
Thank you very much again.
Kind regards,
beppe61
Chris Daly said:
Dear Mr. Daly,
thank you sincerely for your extremely kind and valuable help.
I am afraid your indications are beyond my understanding capabilities. I would like to stress my ignorance on the topic.
If I understand well, you advice to put a 1M resistor from Vcc to the base, of course coupling with caps the input and output.
Is that correct?
What value for the emitter resistor then?
Thank you greatly for your extremely kind effort to teach me the basics.
I must advise you that it will not be any task indeed.
Eh, eh, eh.
Kind regards,
beppe61
Dear beppe61
No - the 1M to 3M3 resistor is attached from the
collector to the base. Not V+ to the base
The collector is biased at 1000 ohms for every volt of supply
ie it has 24K of resistance between the collector and
the voltage DC source.
You will find that approx .7v will be on the base once the
1M - 3M3 resistor is arranged from the collector to the base.
Think of it this way -as a feedback path from the collector
back to the base which it also achieves as well as
self biasing.
The BD 139 is a good transistor but is capable up to
about 80Mhz ( i have used them as RF transistors )
though- it should be fine in this application.
Take care with the pinout which is front view
- left pin Emitter
middle pin - collector
and right pin base.
Of course this isnt the only circuit for BJT buffering
but it achieves a good result.
Cheers / Chris
No - the 1M to 3M3 resistor is attached from the
collector to the base. Not V+ to the base
The collector is biased at 1000 ohms for every volt of supply
ie it has 24K of resistance between the collector and
the voltage DC source.
You will find that approx .7v will be on the base once the
1M - 3M3 resistor is arranged from the collector to the base.
Think of it this way -as a feedback path from the collector
back to the base which it also achieves as well as
self biasing.
The BD 139 is a good transistor but is capable up to
about 80Mhz ( i have used them as RF transistors )
though- it should be fine in this application.
Take care with the pinout which is front view
- left pin Emitter
middle pin - collector
and right pin base.
Of course this isnt the only circuit for BJT buffering
but it achieves a good result.
Cheers / Chris
Dear beppe61
Emitter resistor,- try a 470 ohm , you will find the circuit
reaches a point where it decreases its ability with
frequency if you go too much lower but should still
be capable at about 220 ohms.
A 3K3 resistor in series and 600 ohm resistor to ground
after the coupling capacitor should provide a starting point for trimming - assisting your need for unity gain.
Cheers / Chris
Emitter resistor,- try a 470 ohm , you will find the circuit
reaches a point where it decreases its ability with
frequency if you go too much lower but should still
be capable at about 220 ohms.
A 3K3 resistor in series and 600 ohm resistor to ground
after the coupling capacitor should provide a starting point for trimming - assisting your need for unity gain.
Cheers / Chris
peranders said:
Dear Mr. Peranders,
thank you for the interesting link.
Nevertheless I would like to stick with a text book schematic of an emitter follower.
I must confess you that I really tried to understand how to calculate the resistors value but failed miserably.
So I am looking for suggestion on them.
Kind regards,
beppe61
Hello Bepp61,
The text-book emitter follwer is quite a simple circuit.
Supply voltage is 24 v, for a maximum output swing, the emitter must be at 12v. The base-emitter voltage is about 0.7v (diode junction) so the base must be 0.7v higher than the emitter.
Now you know the voltage at the emitter we can calculate the emitter resistor. If we make the emitter resistor 470 ohms, it will pass about 25mA (I=U/R).
Next you can calculate the voltage divider at transistor base. BD139 current gain (hfe) is about 80. For 25mA current ,the base will need 25/80=0.31mA.
We first calculate the upper resister of the divider. The voltage across this resistor is 24v-12.6v=11.4v
Thus R1=11.4V/0.31mA=36K Ohms
R2=12.6/0.31=40K Ohms
DC voltage is present at output and input, so decoupling using a capacitor is needed.
I hope this is of some help.
The text-book emitter follwer is quite a simple circuit.
Supply voltage is 24 v, for a maximum output swing, the emitter must be at 12v. The base-emitter voltage is about 0.7v (diode junction) so the base must be 0.7v higher than the emitter.
Now you know the voltage at the emitter we can calculate the emitter resistor. If we make the emitter resistor 470 ohms, it will pass about 25mA (I=U/R).
Next you can calculate the voltage divider at transistor base. BD139 current gain (hfe) is about 80. For 25mA current ,the base will need 25/80=0.31mA.
We first calculate the upper resister of the divider. The voltage across this resistor is 24v-12.6v=11.4v
Thus R1=11.4V/0.31mA=36K Ohms
R2=12.6/0.31=40K Ohms
DC voltage is present at output and input, so decoupling using a capacitor is needed.
I hope this is of some help.
Attachments
This will work, but generally a solution is taken that takes away the dependency of the base voltage on the hfe. If you make the base voltage divider carry 5 or 10 times the expected base current, it will more look like a voltage source to the base. The hfe can then vary all over the place with only a very small variation in Vb. In the original example you give, even small variations in hfe would immediately change the Vb.
Jan Didden
Jan Didden
beppe61 said:
No, just the opposite!
OK. The given solution doesnot work. In the calculation for the 2nd resistor it is assumed that there will also be 0.31mA through it, but that is not true. That 0.31mA goes into the base. As a result, both the base current AND the current through the two resistors goes through the top resistor. The Vb therefor is much lower, in fact about 8V.
So, first of all, this diagram with the calculated values only works with the 2nd resistor removed. Then you have a voltage drop on the top resistor due to the 0.31mA which is about 12V. So far so good, but what if the transistor hfe is different? The 0.31mA will be different and Vb will be different. Note that it is not uncommon for hfe to vary +/-50% between transistors.
However, if you make a voltage divider with two resistors and make sure that the current through them is much higher than the base current, even with 50% hfe variation, the voltage drop due to base current over the top resistor is only a small part of the total current through this resistor. The Vb will therefor vary much less. For instance, you could use about 5kOhms for both resistors. A disadvantage is that the input impedance becomes lower, because that is (for AC) the parallel resistance of the two resistors, in parallel with the transistor input resistance. That is approximately hfe times Re, or 47k in this example.
So, to keep a high enough input resistance, chose a transistor with a high hfe and use a Re as large as possible. The high hfe not only decreases the transistor input resistance (which is hfe x Re) but also lets you use larger divider resistors because the necessary Ib becomes smaller.
Jan Didden
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