Biasing a transistor

[Analog_Fan]

Would this improve?

Normal you emitter would be grounded and you need at least 0.65V at the base to make the transistor start doing something.
If you lower the "potential" on the emitter would you make the led start glowing at 0.05V at the base?

12÷(12000+820) = 0,000936037 Amp.
12000×0,000936037 = 11,2324 voltage drop.
-12 - -11,232444 = -0,7676
Correct?

 

[JohnRoberts]

820k? That polar capacitor looks backwards.

JR

 

[ruffrecords]

Would this improve?

Would it improve what? What is your objective?

Cheers

Ian

 

[abbey road d enfer]

Would this improve?

Improve what? Are you referring to an older post?

Normal you emitter would be grounded and you need at least 0.65V at the base to make the transistor start doing something.
If you lower the "potential" on the emitter would you make the led start glowing at 0.05V at the base?

With 0.05V at the base, the emitter would be sitting at about -0.7V, which would result in about 11.35V across the 12k resistor, so about 1mA, which is enouh to turn on an LED, not very brightly though.
However, even with a very low voltage at the base, such as -10V, the LED would still be on, with a small current.
For an input voltage of -10V, the LED would be feably lit with about 100uA, rising linearly to 0.8mA ar 0V input.
See graph.
I haven't included the capacitor, because it just doesn't do anything DC-wise.

 

[Analog_Fan]

820k? That polar capacitor looks backwards.

JR

it's 825R, a typo from me.

Thnx

 

[Analog_Fan]

Would it improve what? What is your objective?

Cheers

Ian

overcome to 0.65 volt voltage drop if the emitter is tied to GND instead of -12V.
By lowering the floor by 0.65V below GND and have to led "activate" at lower voltage voltage than 0.65V at the base.
thnx

 

[Analog_Fan]

Improve what? Are you referring to an older post?

With 0.05V at the base, the emitter would be sitting at about -0.7V, which would result in about 11.35V across the 12k resistor, so about 1mA, which is enouh to turn on an LED, not very brightly though.
However, even with a very low voltage at the base, such as -10V, the LED would still be on, with a small current.
For an input voltage of -10V, the LED would be feably lit with about 100uA, rising linearly to 0.8mA ar 0V input.
See graph.
I haven't included the capacitor, because it just doesn't do anything DC-wise.
View attachment 101880

the 820K was a typo i accidentally made and should be 820R/825R(with is the one i got at home).
it's just an experiment too see if the led would start to glow at 0.1Volt at the base rather than skipping the first 0.65Volt.
i thought if you lower the emitter by 0.65V approx.

Thank you for the response.

 

[abbey road d enfer]

 

[ruffrecords]

overcome to 0.65 volt voltage drop if the emitter is tied to GND instead of -12V.
By lowering the floor by 0.65V below GND and have to led "activate" at lower voltage voltage than 0.65V at the base.
thnx

What voltage do you want it to activate at?

Just use a comparator and set the threshold to whatever voltage you like.

Cheers

Ian

 

[Analog_Fan]

Thank you, your graph seems to indicate it works.

Since energy is becoming expensive and even BMW a few days ago siad the electric cars wont be answer and will not solve the problem.




Normally, i used the "standard" NPN setup to operate a led (never questioned it), but since i was working on a envelop circuit, i went back a envelop circuit from a known producer and they use this.
I was always intrigued by this circuit using a PNP transistor to operate and LED.
i simulated it with the Falstad simulator and found out it's using 5mA doing nothing.

R3 (100k) sees voltage 0-10V.
the Falstad simulator "suggest" using a 150K on the base.
(i have LTspice, but at the moment it doesn't allow me to draw, using Wine to run windows software on Ubuntu.and i also don't know how to really use it.)




Link to circuit: https://tinyurl.com/2panekdd

i remembered the Analog Devices MAT12 witch has a diode from the emitter to the base.
the Falstad simulator show it lowers the "idle" current usage, slight but it does.

 

[Analog_Fan]

What voltage do you want it to activate at?

Just use a comparator and set the threshold to whatever voltage you like.

Cheers

Ian

Thanx for your suggestion, but i'm using it to "demostrate" the profile of a envelop ad not as indicator.

 

[ruffrecords]

Thanx for your suggestion, but i'm using it to "demostrate" the profile of a envelop ad not as indicator.

Does this mean you want a kind of voltage to brightness circuit?

Cheers

Ian

 

[Analog_Fan]

Does this mean you want a kind of voltage to brightness circuit?

Cheers

Ian

Yeah and as efficient as possible for eurorack synthesizers using 78/7912 regulators.
(internet: The 7812 is a commonly used linear regulator. Input voltage can range from 14 - 35VDC and it outputs a fixed 12V at over 1A of current)
Like i do.

Many eurorack modules consume apparently 20/30 mA on both poles, consisting merely of a few opamps or have an aditional ota.
(i really do not want a bigger transformer, i'm using a 2 x 18V AC, the regulators although having quite big coolers i grabbed from a old 486 motherboard, cut in half, much bigger than you buy from store and they get very hot, i need a smaller trafo.)
but still would like to have many modules.

i need to pay like 0.77€ / kilowatt now, let's see how this turn out.

the circuit i got now, look up, seems ok, doubt it can be anymore efficient.

 

[ruffrecords]

Yeah and as efficient as possible for eurorack synthesizers using 78/7912 regulators.

Since LED brightness is proportional the LED current (to a first order) what you need is a voltage to current convertor. Easy to do with any op amp.

Cheers

Ian

 

[Analog_Fan]

Does this mean you want a kind of voltage to brightness circuit?

Cheers

Ian

that's the space i got.

now to reroute a lot, this was an afterthought.

 

[Mr. Moscode]

Try this:
Replace the 825R resistor with a diode, anode (unbanded end) to ground, cathode (band) to emitter. That will clamp the emitter at -.6 volts which means anything positive on the base will turn on the xstor and light the led. If that isn't close enough you can replace the diode with another 2sc945, base to ground, emitter to emitter, collector open.

Good luck.

 

[Admiral-dk]

There is a very good reason why the Transistor is in Parallel to the LED in the PNP example.
Here you have a (close to) constant Current in the circuit - no matter if the LED is On or Off ...!!!
This helps avoiding a Multitude of Noise problems in the Circuit ...!!

If you really want to lover the Current - then raise the series resistor in value and find a newer more efficiant LED.
You should be able to get one where 1mA gives you a lot more light than the old ones did a 5mA.

Per

 

[abbey road d enfer]

There is a very good reason why the Transistor is in Parallel to the LED in the PNP example.
Here you have a (close to) constant Current in the circuit - no matter if the LED is On or Off ...!!!

The problem with the PNP circuit, as it is, is that the threshold for lighting the LED is difficult to manage, since it depends on the transistor's Beta and the LED's on voltage.
The variable-shunt connection is perfectly adequate for definite on/off states, much less for proportionality.

 

[Admiral-dk]

Sorry - I should probably mention that I have used NPN's (and Logic Gates - w. a diode between) as the switch/shunt in different production designs ....
I didn't mean that it needed to be a PNP.

I also agree that this methode isn't particularly good for proportionality .... but I haven't read that this was part off the original question (not very application specific) ....

Per

 

[Bo Deadly]

the circuit i got now, look up, seems ok, doubt it can be anymore efficient.

It sure can. Realize that an LED is going to require at least 10mA but powering it from +12 to -12 means 24V * 0.01A = 0.24W. If you only used one half of the supply, that would be 12V * 0.01A = 0.12A. But Eurosynth usually has +5V I believe in which case you could use that instead and now power is 5V * 0.01A = 0.05W.

Technically you could do even better than this because you could, in theory, make a DCDC converter that converted whatever voltage into just enough voltage required to power the LED which has a voltage drop between 2V and 4V. But just using 5V or a DCDC converter to make 5V from 12V (which is a very cheap part an a tiny package that makes 10's of mA) would be 0.05W which, again, is 5 times more efficient than your circuit.

Now, if you want the LED-on threshold to be near 0V, then you could use a PNP with a large emitter resistor from +12V to -12V and then drive the base of an NPN with LED and Rmax resistor from +12 to 0V. The PNP serves three purposes here:

1) The diode drop of the PNP shifts the NPN bias to compensate for it's diode drop (meaning the LED will start to turn on when the control voltage is just above 0V and not at 0.6V).
2) The PNP stage would consume very little power with it's large emitter resistor (47K would be 24V * 0.0005A = 0.012W) and the NPN LED driver would only consume power between +12 and 0V which would double the efficiency of your circuit with only one additional transistor and resistor. If 5V is available, again, much less still.
3) If Rmax is the NPN emitter resistor, it will provide feedback to yield a fairly linear voltage to current converter. This is good because LED brightness is proportional to current. With just the NPN, you will get an exponential converter which is almost certainly not what you want.

If you restated your objectives more clearly, I think you would get much better answers in this case. You would probably also benefit greatly from using LTSpice. This is precisely the sort of circuit that LTSpice will help you understand.