Why would this simple switching circuit now work.

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AbhimanyuSingh

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Signal comes from a microcontroller which is able to drive BC109 Q1 in schematic at 3.3 volts. But why would R1 which is load not switch with this circuit. What am I missing.
 

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Q2 takes 12/7 ~ 1.5A (1500 mA), so if its Hfe is 50, then it needs 30mA base current at .8V. If Q1 is to deliver 30 mA, then it needs 30/100(Hfe) mA ~ .3 mA at 1.6 V (two Vbes in series). Your processor can deliver 3.3 V, so at .3 mA, you have to drop 3.3 - 1.6 @ .3 mA = 1.7/.3 ~ 6K. You use 1K so it should work. Perhaps your working Hfes are too low.
What is R3 for?
Frank
 
What is R3 for?
Click to expand...
It's deadly necessary. And R3 must be increased to limit the Q1 current to the maximum permitted Q2 base current respectively 100 mA maximum BC109 current. Otherwise both transistors might be burned.
 
@chuckey: I have not connected R4 at all. The current supply rating for my microcontroller is 25 mA. Although the voltage at base of Q1 falls to 2V. I did not know the use of R3. The voltage at emitter if Q1 is good 12V But no voltage across R1.


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@FvM: But if R3 is increased, the current through Q2 decreases. And the voltage falling across becomes very low, so cannot increase R1.
Its still not working anyway.

FvM said:
It's deadly necessary. And R3 must be increased to limit the Q1 current to the maximum permitted Q2 base current respectively 100 mA maximum BC109 current. Otherwise both transistors might be burned.
Click to expand...

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@Schmitt: I sure will try that next.
schmitt trigger said:
Why don't you connect both transistors in REAL Darlington configuration?
Tie both collectors together.
Click to expand...

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More ever, I don't know how to switch on a load of (just) 7 ohms. Theoretically these BJTs have 0 resistance when they are switched on, but that's not true practically. All the voltage falls across BJTs. There working resistance is much high as compared to 7 ohms. I m dealing with this subject after years and have no clue how to go.
 
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But if R3 is increased, the current through Q2 decreases. And the voltage falling across becomes very low, so cannot increase R1.
Click to expand...
Your circuit has to be designed to keep the maximum transistor ratings. The original circuit doesn't.
Its still not working anyway.
Click to expand...
Are you talking about the simulation or a real circuit? If it's the latter case, what's the Q2 transistor type?
 

FvM said:
Your circuit has to be designed to keep the maximum transistor ratings. The original circuit doesn't.
Click to expand...

Maximum transistor rating, you mean maximum current to be flowing through these transistors?

FvM said:
Are you talking about the simulation or a real circuit? If it's the latter case, what's the Q2 transistor type?
Click to expand...

Real circuit. Simulation works fine. Both Q1 and Q2 are NPN transistors, BC109C.
 

BC109 has maximum continuous collector current of 100 mA, in contrast to 1.5 A required in your circuit. So it can't work and will be mosr likely destroyed if you try to switch the load.
 
BC109 is NOT rated for over an amp of collector current, you have killed it (Probably both of them).
Replace Q2 with something actually rated for that much current (TIP31A, or whatever comes to hand), remove R2 & R3 and connect the collector of Q1 to the collector of Q2 making a darlington and it should work.

Actually I would use a logic level mosfet for this (Less components, usually less losses in the device), but darlingtons are perfectly acceptable if you rate the parts correctly.

Regards, Dan.
 
Thanks to all of you. It works now. But there is just 7 V across R1. Can I increase this to anywhere close to 12?
 

BC547 has 300 mA maximum collector current rating, still not right.

LTSpice is about to foolproof by showing current and voltage ratings in the "pick-up transistor" selection. Don't ignore it.
 
FvM said:
BC547 has 300 mA maximum collector current rating, still not right.

LTSpice is about to foolproof by showing current and voltage ratings in the "pick-up transistor" selection. Don't ignore it.
Click to expand...

I am using ZTX649 which has current capacity of around 2 A. Not using BC547. Please ignore BC547 in the image.
 

I am using ZTX649 which has current capacity of around 2 A. Not using BC547. Please ignore BC547 in the image.
Click to expand...

Which other differences between your real circuit and the LTSpice schematic should we expect?

Although ZTX649 saturation voltage is specified for Ic/Ib=10, it should be typically fine with Ic/Ib=20, minimum 2A current gain is 75. So there's no reason why the circuit shouldn't work, provided the driving voltage doesn't collapse. Checking all node voltage with a multimeter will easily tell why the circuit doesn't work as expected.
 

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