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| | #1 |
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I was adviced by an engineer that in addition to a resistor one of its legs connected to Vsource and second to Base, to connect another resistor from Base to ground. 1. Why is it recommended? 2. Should the resistor be very large so the Vsource (MCU GPIO pin) wont source large current? Thank you. | |
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| | #2 |
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It's to help ensure the transistor turns off, values depend on the exact circumstances.
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| | #3 |
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How exectly does it ensure that the transistor is in cut-off? The present Base resistor is 300ohm (The base-Vsource resistor). Thanks. | |
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| | #4 |
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In two ways, first it discharges the base/emitter capacitance, and it forms a potential divider with the series resistor - not all outputs feeding it may go low enough to switch it off, by adding a potential divider this ensures it will.
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| | #5 | |
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I very understand the C_BE discharging issue. I dont quite understand the reisitve divider issue. In cut-off, the BE resistance is very large, so connecting in paralel to BE (E goes to ground) another resistor, would only decrease the total resistance from Base to ground. So therefore less voltage will drop on base? I think i probably have a mistake on that, so i want to figure this issue out. | ||
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| | #6 |
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300 ohm resistor to base ONLY - logic feed to resistor is 4V when HIGH, and 1V when LOW - so transistor is always ON. 300 ohm resistor to base, 300 ohm resistor base to emitter, input voltage is divided by two, so transistor turns OFF when LOW is supplied. | |
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| | #7 |
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But the resistor-to-base and base-to-emitter resistors are not connected in series, so how can you say that when both used then base voltage is divided by two? And what will happen when source is HIGH? Aslo here the base voltage is divided by two? | |
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| | #8 | ||
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Quote:
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| | #9 | |
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the current that flows through base resistor, splits to two: 1. BE current 2. Base-GND current. Last edited by alphadog; 24th February 2009 at 04:58 PM. | ||
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| | #10 |
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Below is a simplified view of when Rbe is required. The fact that the base is connected to the junction of the two resistors does not mean the resistors are not in series. They are. Last edited by Roff; 24th February 2009 at 05:18 PM. | |
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| | #11 |
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In series = same current goes through them. When HIGH and when LOW, IB is still not zero, so how can you say that both when HIGH and when LOW, they are connected in series? What about IB? By the way, adding this resistor means that when HIGH, the VBE will take its maximum value thats written on datasheet - 1.3V ? (becaues VCC/2 is 1.65, so VBE will aspire to be 1.65V?). Last edited by alphadog; 24th February 2009 at 05:31 PM. | |
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| | #12 | ||
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Quote:
No. And where did Vcc/2=1.65 come from? EDIT: What datasheet are you referring to? Post a link, please. Last edited by Roff; 24th February 2009 at 06:40 PM. | |||
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| | #13 | ||
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So when input is HIGH, therefore Vbe > 0.7, they are not connected in series, right? And at this state (input HIGH) the voltage on the 'base-to-GND' resistor is forced to be the around 0.7V? (depend on the transistor) Quote:
By the way, What set the V_Forward of B-E? lets say that Vbe_on_max is 1V (accordint to datasheet of http://www.fairchildsemi.com/ds/PN/PN2222A.pdf ), then what happens if i apply 2V on B-E? | |||
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| | #14 | |
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Otherwise, generally, a voltage divider is used for biasing of the transistor as in an amplifier, or as stated earlier, for turn-on/off level control as needed. | ||
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| Tags |
| base, bjt, ground, resistor |
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