What is the minimum voltage to turn-on a LED in Darlington circuit?

Discussion in 'Homework Help' started by xchcui, Feb 18, 2016.

  1. xchcui

    Thread Starter Member

    May 12, 2014
    This circuit has been taken from a site that show how can we turn-on
    an ultra-bright red led with radio-waves energy.
    The radio-waves should create a voltage on the 1 wavelength antenna loop(the left square).
    The darlington transistor should amplify the current and turn-on the led.
    My questions are:
    1)What is the minimum voltage that the antenna loop need to have in order to turn on the led?
    I know that generaly there is ~1.4V(0.7VX2)voltage drop on the base-emitter of darlington pair,
    so,it seems like the loop must produced at least 1.4V in order to turn it on,
    but i am not sure.(maybe it might turn on on lower voltage?)
    2)according to the circuit:at the moment that to loop will reach to the minimum voltage
    that it can turn on the led,
    Is the right transistor(second) will be on saturation state,while the left one(first) will be in active state?
    or both are on active state?what are the hFE of each transistor?

    Thanks in advance.
    Last edited: Feb 18, 2016
  2. Wendy


    Mar 24, 2008
    There are other configurations.

    But first, it is vital the loop antenna and cap be tuned to the radio frequency you are trying to respond to. You must have resonance, just like a crystal radio. Matter of fact, the principal is exactly the same. You should probably have the rectifying diode after the tank circuit,.

    You can also use a germanium transistor. It has a 0.3 V B-E drop as opposed to a silicon transistor, which has 0.7V B-E voltage drop.

    A Darlington pair drops both the transistor B-E voltages. So if it is a silicon transistor it will drop 1.4V B-B. Not good.

    There is another high gain configuration called a Sziklia Pair. It is very similar to a Darlington, but only drops one transistor B-E drop. If the front transistor is a germanium then so much the better.

    I have added the link so you can read up on it.
    atferrari likes this.
  3. AnalogKid

    AAC Fanatic!

    Aug 1, 2013
    Your description basically is correct, with one minor error. As a side comment, transistors do not go from a non-conducting state to a conducting state completely at exactly the Vbe of the transistor. Even though the two transistor junctions combine for an overall Vbe of around 1.2 to 1.7 V, the transistor will start to turn on at less than 1 V.

    As for saturation, it is the right-side transistor that does not saturate, and the left-side one that does. The Hfe of the transistor changes as a function of the collector current, and is on the datasheet.

    xchcui likes this.
  4. xchcui

    Thread Starter Member

    May 12, 2014
    Since i am not an expert in planning circuits((i didn't plan it,i usually just build them),
    i think that changing the two silicone transistor to germanium transistors will be
    less complicate for me than using Sziklia pair.
    Can you recommend me,an equivalent germanium transistor that will have the same gain
    as the BC109C silicone ones and fit that circuit?
    I did alot of searching in the net,in order to understand how to calculate the currents and
    voltage in darlington circuit and after that, calculate my circuit,but with no success.
    Can you explain what are going to be the voltages and the currents in the circuit if
    the loop will create,for example 2V?(lets say,for the calculation,that the transistor is silicone)
    My calculation started with:2V(loop)-0.7Vbe-0.7Vbe/100K=6uA.
    Then i did:9V-1.8Vled-___Vce/680Ω=it should be 10mA in order to turn on the led.
    And that means that the Vce should be 0.2V.A condition of saturated transistor with a gain of X20.
    If that is right the Ib of the front(2)transistor should be:0.01mA/20hFE=0.5mA.
    So the first transistor(1)should amplify the current in:0.5mA/0.006mA=83hfe.[Ib(transistor 2)/Ib(transistor 1)=hFE(transistor 1)].But the voltage drop on Vce of transistor 1 seems to be 0.2V(attached photo)no?
    It does not make sence.it is like them both saturated,so it is not right.
    That is the point that i stuck.
    The problem that i dont know if the transistor is on saturation or how
    to determine their Vce.them both Vce.
    Can you help.
  5. Wendy


    Mar 24, 2008
    It may seem less complicated, but it won't work.

    When I have time I'll sketch something up, but there will be no guarantees. A better approach is to tell us what you are after. Is this a school assignment for example? What is the transmitter like? Is all you want to do is light a LED?

    The circuit you show might work, but it is pretty awful.
  6. xchcui

    Thread Starter Member

    May 12, 2014
    How did you determine that the left one is saturated and the right transistor is not?
    In the data sheet there is only 2 values:IC=10mA, VCE=5V------hFE>100 and
    IC=2mA, VCE=5V------hFE=420-800.thats all.
    How it can help me to know what is the gain on the right transistor?
    If the left transistor is saturated(i hope you may explain me how)
    the hFE is X20(according to the datasheet).
    But if the right transistor will have 5V,4V or 2V Vce,while it is
    on active state,the current will be too small to turn on the led.
    And i can't figure it out how you determine the Vce of the transistor
    with only those two values in the datasheet?
  7. AnalogKid

    AAC Fanatic!

    Aug 1, 2013
    This is a small signal low power circuit, so we can use a typical value like 0.1V for Vcesat, the collector-emitter voltage when saturated, and 0.6 V for Vbe, the base-emitter voltage.

    When an NPN transistor is saturated, the collector is "below" the base. That is, with the emitter tied to GND like the right transistor, the base is at +0.6 V and the collector is at +0.1 V. OK so far?

    But when the left transistor is saturated, it has 0.1 V from its emitter (tied to the right transistor base) to its collector (tied to the right transistor collector). So the minimum voltage from the right transistor's emitter to its collector is 0.7 V , its own Vbe and the right transistor Vcesat. Even if the left transistor is a dead short, that is like like tying the right transistor's base to its own collector. This is called a "diode connection" and can not ever saturate.

    The datasheet probably also has a plot of gain versus collector current to fill in the gap between the listed gain values and conditions. Otherwise you just have to extrapolate between the two values you have.

    xchcui likes this.
  8. xchcui

    Thread Starter Member

    May 12, 2014
    What did you refer to when you said:"It may seem less complicated, but it won't work."
    Did you refer to changing of the silicone transistor to germanium transistor?
    If i will do the change,it won't work?
    But it was your idea?why it will not work?
    Refer to your question,it is a kind of assignment and i try to make it as much as sensitive.
    You don't have to sketch for me any special circuit.If you can help me with the calculation and the value of
    the current and voltage at that circuit(as i described at the other post)it will be more than enough.
  9. dl324

    AAC Fanatic!

    Mar 30, 2015
    What junction conditions are required for a transistor to be saturated? Can the transistor on the right ever satisfy them?
  10. hp1729

    Well-Known Member

    Nov 23, 2015
    The transistors do not have to be in saturation for the LED to light, just conducting enough. How much voltage in depends on the specific transistors and their gain. I don't think your antenna needs to be tuned. You just want to pick up any signal at all. It is going to be a big antenna to output the voltage and current needed. It isn't just voltage.
    The details ... how much current does the LED need? What is the gain of the transistor? How big is the antenna? How strong is the signal being transmitted?
    High power LED? 50 mA? 100 mA? will 10 mA turn it on?
    Gain of the transistors is 1,000? You need about 1.2 V @ 10 microamps. 12 microwatts of power. Not impossible if the transmitter is strong and close enough. Miles away and 10 KW radio station? Maybe not.
  11. Wendy


    Mar 24, 2008
    A Darlington take more voltage than a Sziklia pair on the BE. This is pretty important in this design, given you will be dealing with really low signal strengths (ie voltage).

    I asked this before, but is this homework? If it is you are expected to learn the strengths and weaknesses of various configurations such as Darlington pairs. It goes to how we approach explaining the issue.

    I also asked where the signal is coming from.

    Ball is in your court.
  12. WBahn


    Mar 31, 2012
    MOD NOTE: Moved to Homework Help from General Electronics Chat
  13. xchcui

    Thread Starter Member

    May 12, 2014
    Ok.so if the right transistor will have Vce-1.5V and Vbe-0.6V,
    1)is it mean that the Vce on the left transistor will be 0.9Vce?
    2)When using darlington pair in a circuit,does the left transistor,in general,planned to
    be in saturated state?I mean,Is it,in general, its purpose to be saturated?
    3)Can you recommend me,an equivalent germanium transistor instead of the silicone BC109C transistor?In order that the transistor will turn-on at lower voltage.

    As i mentioned before,This circuit has been taken from a site that shows how can we turn-on
    an ultra-bright red led with radio-waves energy.
    What type of led?10mA led.
    What type of transmitter?It depends of the length of the loop.
    At the site,the transmitter that turn-on the led was a mobile phone,while they used 0.3m
    length of wire for 1GHZ RF,but it shows,also,that by changing the wire length,we can make the loop to resonant on other frequencies(like smaller loop for 3G networks),that will create a voltage to turn-on the led.
    BTW:The 30cm copper wire loop has about 100Ω AC resistance(as it mentioned at the site),
    Can you show me the calculation that determine that 100Ω AC resistance for 30cm copper wire?
  14. Bordodynov

    Well-Known Member

    May 20, 2015

  15. MrAl

    AAC Fanatic!

    Jun 17, 2014

    That's interesting because i did not know that this configuration was named after somebody. I always called it just a "complementary pair". I guess someone somewhere at some time must have invented it :)

    Your circuit has some flaws which make it harder to get it to work. One has been mentioned before here, the forward voltage of the TWO base emitter diodes is a little higher than you should have in order to get this to work well. There is little power available from the antenna, so using that power wisely is a good idea. Of course we cant forget that we still want some signal rejection so that the circuit does not turn on every time somebody turns a motor on or off or somebody in the same room sneezes, but one of the main hurdles is to get high sensitivity, then worry about selectivity later.
    You can also play with the resonance as someone already mentioned and that will get you more selectivity.

    To begin with, the transistors are not biased so they will depend entirely on the signal from the antenna. If you bias the transistors however (which really just means turning them on slightly and keeping them that way even with no signal) then you can get more sensitivity because then even a low power signal can change the conduction state of the transistor(s) and thus you'll see a difference right away. To put it another way, you would be creating an amplifier rather than a switching type circuit. The amplifier can be made very sensitive by using several transistors in cascade and biasing each stage individually and using capacitive coupling to connect them all together and get high gain. You can then probably adjust the bias of one of the stages in order to obtain a squelching effect.

    There are really a lot of ways of doing this, but using a Darlington pair is probably one of the worst ways because of the two diode drops.

    I am assuming the input diode is germanium but i have not checked the part number.
    Last edited: Feb 26, 2016
  16. Wendy


    Mar 24, 2008
    The fewer PN junctions you have on a weak signal the better. For example, you can use a transistor as the detector as shown. Again, a germanium transistor for the 1st one will work better.

  17. xchcui

    Thread Starter Member

    May 12, 2014
    I see that in one circuit you used two silicon transistor and the other one
    you used one silicon transistor and one germanium,while the latter can operate at lower voltage(170mV) and them both work as Sziklia Pair configuration.
    I have some question about the circuits:
    1)I see that you add another component before the diode.
    Is it inductor with 200 nano henry value?why did you add it?must i add it?
    2)Why did you omit the 100KΩ resistor on the upper circuit?
    3)Why can't i use two germanium transistor at darlington configuration,that should give me the same result(even lower voltage drop)as the upper circuit?
    Wendy mentioned before,that it will be good to use the germanium transistor only as the first
    one(the saturated one),the same as you drawn in your circuit.But i don't understand why can't the other one be also germanium?Is it related to the gain?why won't it work?
    4)Can i build the circuits that you drawn as they are?or i need to add,check or change something in order that it will work?
  18. Bordodynov

    Well-Known Member

    May 20, 2015
    1. I do not know what frequency you need to fix.
    For the convenience of modeling (process speed), I used a frequency of 10 megahertz.
    The magnetic antenna at this frequency is an inductive impedance.
    I arbitrarily pictured frame inductance and the voltage source.
    Tell me what frequency you need and what kind of loop antenna you are using.
    Then I may clarify an equivalent circuit.

    2. Resistor 100k greatly reduces the sensitivity of the whole scheme and need for this resistor is not.

    3. darlington configuration worse sensitivity than Sziklia Pair.
    Your first basic question about voltage drop across the transistor open almost does not make sense.
    When the supply voltage 9 volt transistor Von to 1.V or 0.5V the LED current is changed to 7%. I_led=(9V-Vled-Von)/680
    In my circuit PNP transistor may be germanium. The sensitivity depends on the gain of the second transistor.
    I do not know germanium transistor with a gain of 500.
    4. You can repeat my circuit. Pay attention to the additional resistor is 100kohms.
    this is to compensate for the leak of the germanium transistor.
    Last edited: Feb 29, 2016
  19. atferrari

    AAC Fanatic!

    Jan 6, 2004
    Did you bother to check what a Sziklai pair is? I doubt so.

    Would you call this "complicate?


    Depending of those Germanium transistors are available or not, Wendy's suggestion is quite a good one.
  20. Bordodynov

    Well-Known Member

    May 20, 2015
    See Von=250mV