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Posts posted by audioguru

  1. Whatever you have does not look like a PIR motion detector because it does not have a Fresnel lens.
    It is anybody's guess what its 3 pins are for. It doesn't even have a manufacturer's logo or a part number printed on it for you to look at its datasheet.

  2. A PIR probably does not have Mosfet pins. Most have an ordinary little transistor at the output that drives an external relay, power transistor or power Mosfet.
    If you put a jumper across its two output pins then it will probably stay turned on.

  3. I made this circuit but i changed Q2 with one Tip41C with 50 hfe.Is it enough?

    The TIP41 has an fT of only 3MHz so it is slow and its delays will probably cause the output amplifier to oscillate and have transient problems. The BD139 that should be used is very fast with an fT of 190MHz.

    The hFE is not important because the minimum hFE of one of the two 2N3055 output transistors at 1.5A is 41 then the collector current of the BD139 must be a maximum of only (1.5A/41) x 2= 73mA where its minimum hFE is about 70 so the maximum output from opamp U2 must be only 73mA/70= 1mA but the minimum output from a TLE2141 opamp is much more at 20 so there is plenty of hFE available. 
  4. Your circuit is missing a resistor from the collector of T2 to the base of the PNP output transistor to limit the base current. T2 will probably overheat without it.
    The PNP output transistor is missing a resistor from its base to its emitter to help turn it off.
    Both transistors add delay and additional voltage gain so the entire circuit will probably oscillate.

    I think you should use a fast driver emitter-follower driving a fast power transistor emitter-follower since they have no voltage gain and their very small delay will not cause oscillation:


  5. Flippityflop,
    I do not know what you are talking about.
    You show two opdinary diodes, not zener (avalanche) diodes. Why does your circuit have two zener diodes?
    There is no such thing as "an ambient voltage".

    Your circuit has no input, no output, no power supply, no biasing for the active device and no load for the active device.

    Nobody knows what you are talking about.

  6. One of the features is not very clear U3 - I know it is responsible for the current protection and opamp works here as a comparator (I think). However, from my calculations in section 6 of the potentiometer P2 maximum voltage of about 1,7V (P2 and R18 are a voltage divider)

    The original circuit has errors. So we fixed it in the forum.
    A maximum voltage from P2 is 1.7V when R18 is WRONG at 56k. When we fixed the original circuit we changed R18 to 33k and added a 100k calibration trimpot in series so when the trimpot is about 46.7k then the total with 33k is 79.7k then the maximum voltage from P2 with a 3A load is exactly 1.41V. 1.41V/0.47 ohms= 3.0A. 

    But another question - if U3 inverting input (-) have less potential than the non-inverting input (+) the output voltage of U3 is potential of leg 7 of U3, and according to that diode D9 does not conduct. However, if power consumption exceeds the voltage at the output of U3 have potential of leg 4 of U3 (so it is about -5V) and diode D9 becomes conductive. But what happens next? Because that's the part I do not understand.

    The voltage across R7 caused by the load current in it goes to the inverting input pin 2 of U3, and the pot P2 voltage goes to the non-inverting input pin 3. Usually pin 3 voltage is higher than pin 2 voltage so the output of the opamp pin 6 is high and D9 is reverse biased and does not conduct.
    If the current in R7 exceeds the setting of pot P2 then the opamp output drops low enough for it to reduce the output voltage through D9 that reduces the output current until the current stays at the setting of pot P2. If the output of U3 went low like a comparator then there will be NO OUTPUT CURRENT! Instead the output voltage of U3 drops only low enough so that the load current is the same as the setting of p2.

    Another problem is the presence of R8 and R9. I think in terms of the setting of gain opamp U2, right? What the presence of C4, C6 and C9?

    No. The voltage gain of the voltage amplifier is 1+ (R12/R11) where R11 has a calibration trimpot to allow you to set the maximum output voltage to exactly 30.0V.
    R8, R9 and C4 simply filter the 11.2V voltage reference. C6 and C9 compensate for phase shifts in the amplifier so it does not ring nor oscillate.

    And the last question - why current limiter isn't linear? I mean when we change output voltage with the same load, the current limiter change value. Why? Maybe you can give me a example power supply we can set output current which doesn't depend of output voltage?

    No. The voltage setting does not affect the current setting and the current setting does not affect the voltage setting EXCEPT when the load current exceeds the setting of the current pot P2 then the output voltage drops low enough that the load current is the same as the pot P2 setting. Then the LED lights to warn that the output voltage is being reduced by the current regulator.
    Here is the fixed voltage amplifier schematic:



  7. Kevin, you need to learn about negative feedback.

    Your simple voltage regulator has no voltage gain. Its voltage gain is a little less than 1.
    Its 200k resistor does almost nothing (it reduces the voltage gain from exactly 1 to a little less than 1).
    The opamp is pretty fast but the darlington connected transistors are probably pretty slow causing phase shift and causing the circuit to oscillate.


  8. Hi Redwire,
    On the original and fixed version your diode D2 was a resistor with almost NO voltage drop. Now the D2 has a voltage drop that changes when the temperature changes then the output voltage changes when the temperature changes.

    Without a negative supply for the current control opamp the output current will not be regulated when it is set to less than about 0.8V/0.27 ohms= 2.96A. The output voltage of the current control opamp must be able to go to -0.7V so that regulation at low currents works. Then its negative supply must be at least -0.9V so use two diodes like I did.

    The value of your R9 is MUCH too low.

    The current sertting pot should have a low value resistor in series with its lower end so that any input offset voltage of the current control opamp does not have a huge effect.

  9. When there is no transistor connected and the switches are set to "Tr" then the CD4093 IC must have plenty of output current or the LED load stops it from oscillating.
    But with a 9V supply two paralleled gates of a CD4093 have a minimum output current that is too low to work in this circuit so it is a bad design.

    Try increasing the value of the 470 ohm resistor at the bottom of the schematic.

    A hex Schmitt Trigger inverters IC (74C14 or MC14584) should have been used. Then the oscillator can be separate from the LED drivers, and the oscillator will not be loaded down by the LEDs.
    But I still think the circuit destroys the transistors it is testing by causing avalanche breakdown when it reverse-biases the base-emitter junction with a voltage that is too high.

  10. I turned off the video because I could not understand the very strong Engrish accent combined with the echoes in the room he is in. I think in part 1 he mentioned a lousy old LM358 opamp (it has poor high frequency response).

    One of the comments at the end of the video mentions poor transient response and/or oscillation.

    The 100k series base resistor for the 2N2222 adds tremendous phase shift (it is feeding stray and transistor capacitance) that you do not want. The series 1k base resistor feeding the BD140 transistor also adds phase shift. But the opamp is designed so that its gain is less than 1 at a frequency where its internal phase shift causes negative feedback to become positive feedback.
    The phase shifts from the two added transistors add to the opamp's internal phase shift and causes the opamp to oscillate at a lower frequency where the entire amplifier still has some gain.

  11. But I didn't add another driver transistor

    Your circuit has Q1 with high voltage gain and a phase shift from R15. You also have Q2 with high voltage gain and phase shift from R13. Then you feedback the high gain and extra phase shifts to the inverting input of the opamp so it oscillates.
    The original circuit used only a single driver transistor as an emitter-follower (a voltage gain of only 1) with a very small phase shift.
  12. The link works when it is not crammed together with "newbielink". This link works:  http://na7kr.us/in-circuit_transistor_tester.htm .

    I think the tester slowly destroys a transistor when it makes the base-emitter reverse biased higher than its 5V maximum allowed rating which causes the fragile base-emitter junction to have avalanche breakdown like a zener diode. Then the LED that is supposed to be turned off glows a little.

    I have used thousands of transistors and all of them have a part number printed on them so I knew if they were NPN or PNP. Also I have never found a defective new transistor so I never needed to tested them. 

  13. The new circuit oscillates because the output amplifier transistors provide too much voltage gain.
    The original circuit and all the improvements until recently had only one driver transistor as an emitter follower with a voltage gain of only 1 like a piece of wire. Your new circuit has Q1 and Q2 as common emitter amplifiers feeding series resistors that add with the following capacitance which causes phase shift. 

  14. Your camera is horrible. Why don't you sketch the schematic using Microsoft Paint program or another program?

    Your opamp has no power supply so it will not work. If it has only a single positive supply and ground then it must have inputs that have a common-mode voltage range that includes ground like an opamp in an LM358 dual opamp.

    It looks like you are heating the load resistor by discharging the battery into it. The battery current also includes the base current of the PNP transistor.
    Connect the sensing resistor for the ZXCT1107 between the positive terminal of the battery and the emitter of the PBZXCT110NP transistor.

  15. Good point.
    The current regulating opamp has its negative supply pin at the input of the 0.47 ohms sense resistor and the voltage regulating amplifier has its negative supply pin at the output of the current sense resistor which is 1.41V higher than the current regulator negative pin when the output is shorted and the current is set to 3A. Then the diode between them causes the output voltage to drop until the current is 3A. But when the output is shorted and the current is set lower than 3A then the current regulator will not work properly below a certain setting.
    So the negative supply is needed after all.

  16. The negative supply is needed for the current regulating opamp to cause the output voltage to drop to zero when the current into a short circuited output exceeds the setting of the current-setting pot. Without the negative supply then the output of the current regulator opamp cannot cause the output voltage to go down to zero volts because it feeds the voltage regulator amplifier though a diode. Short circuit current will be unlimited.

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