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schematic

simulate this circuit – Schematic created using CircuitLab

The transistor in this circuit is supposed to act as a switch, with the LED operating with 2V at 10mA. I do not know how to calculate the collector-emitter voltage and base-emitter voltage of the transistor in such a case where transistors arent used for amplification. I am interested in finding theses calculations when SW2 allows current to flow to the base of the transistor (Q1). This will be greatly appreciated since I cannot find anything on the internet that explains how to calculate for the voltages of the transistor in this scenario. Much thanks!

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  • \$\begingroup\$ Base-Emitter voltage doesn't change much, so it's not important (assume 0.7V and it will be close enough). Do you know how to calculate Base current and Collector voltage when the transistor is used as an amplifier? \$\endgroup\$ – Bruce Abbott Nov 30 '19 at 1:50
  • \$\begingroup\$ Is this homework? If not, this is not a good way to switch the LED... \$\endgroup\$ – MadHatter Nov 30 '19 at 1:52
  • \$\begingroup\$ Yes, I know how to calculate base and collector voltage when it is used as an amplifier @BruceAbbott. No, this is not homework, but I am confused that the collector-emitter voltage drops less than 0.3V in this case, so the collector current is not 100 times more, and the base-emitter voltage is not 0.7V. I was hoping someone knew why this happens and how to find those values. \$\endgroup\$ – microhex Nov 30 '19 at 1:57
  • \$\begingroup\$ If you didn't see 0.7 V across the base-emitter junction, what did you see? \$\endgroup\$ – The Photon Nov 30 '19 at 3:03
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You only have 2 states here, assuming that battery voltage is at 4.5 volts.

In state 1 the switch is OFF and the CE voltage rises to the LED turn-ON voltage, about 3 volts. R1 prevents the LED and Q1 current from becoming dangerous. Now the LED is ON and the BE voltage should be much less than 700 mV, which is internal leakage in Q1 due to an absence of a resistor from base to emitter.

In state 2 the switch is ON, the BE voltage is now ~700 mV. The CE voltage drops to just hundreds of mV, based on the ON resistance of Q1. The low CE voltage causes the LED to go out, as its turn-on voltage is about ~2.9 volts.

Assuming the battery is plugged in these are the only 2 valid states for this circuit.

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I do not know how to calculate the collector-emitter voltage and base-emitter voltage

In this switching circuit the exact values of those voltages are not important, only that the transistor be fully turned on under all expected conditions. So long as it always gets enough Base current to keep it in saturation the circuit will work. But will it?

First calculate the approximate Base current. Assuming a Base-Emitter voltage of ~0.7V, (4.5V-0.7V)/5kΩ = 0.76mA.

Next calculate the Collector current that would occur if the transistor was turned fully on (Collector voltage less than 0.3V). (4.5V-0.3V)/250Ω = 16.8mA.

For this to happen the transistor must have a current gain of at least 16.8/0.76 = 22.1. Now look up the datasheet to see if it can meet this requirement. We note that at 25°C the guaranteed minimum gain at 50mA Collector current and 1.0V Collector-Emitter voltage is 60, well above what we need. So we can safely assume that the Collector voltage will be well below the turn-on voltage of the LED.

Transistor characteristics vary from unit to unit, so we can't calculate an exact value for a particular specimen. However we can get some idea of the typical values by examining the datasheet and extrapolating from the data supplied. The graph of Base-Emitter Saturation Voltage vs Collector Current shows ~0.75V at 0.7mA Base current, so our assumption was close. The graph of Collector-Emitter Saturation Voltage shows less than 0.15V out to 100mA at a current gain of 10, so our assumed 0.3V is probably conservative. The transistor should indeed be fully turned on as intended, and our calculated currents and assumed voltages should be close to reality.

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In comments you explained,

I am confused that the collector-emitter voltage drops less than 0.3V in this case, so the collector current is not 100 times more,

So your base current is somewhere in the neighborhood of 800 uA.

100 times that would be 80 mA.

80 mA through the 250 ohm resistor would drop 20 V.

So to get 100x the base current through the collector, the collector voltage would have to be about -15 V.

And to get current flowing in to the collector at a negative voltage, the transistor would have to generate power. But transistors are not a source of free energy, so this can't happen.

Instead, the collector is stuck at 0.2 or 0.3 V, and the collector current is less than you'd get in forward active operation, because there is only ~4.3 V across the 250 ohm resistor. But at least conservation of energy is maintained and the universe doesn't go up in smoke.

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The LED will be turned off when Vce is less than 1V, so you can use the hFE figures for the transistor in question (which are typically specified at Vce = 1V). As the transistor saturates, the forced beta drops.

At 1.0V Vce, the collector current will be (4.5V-1V)/250 ohms = 14mA. Referring to the datasheet, the gain will be, at a minimum, between 60 and 100 over the range of 10-50mA, probably a lot closer to 100 since 14mA is much closer to 10mA than 50mA. So let's use 80. That means a base current of 14mA/80 or 175uA is just adequate to turn the LED off in a guaranteed fashion (assuming room temperature).

The base current will be approximately (4.5V-0.7V)/Rbase since the Vbe is about 0.7V or so with the transistor on. So the maximum Rbase would be 21.7K. In reality we'd use a lower value because hFE drops at temperature extremes and the datasheet values are for 25°C. At -55°C (fig. 15) the hFE is about half, so 10K would be safe.

In order to drive the transistor well into saturation, meaning a collector current approaching 18mA, we can use Ic/Ib of around 10 or 20. Using 20, we would want a base resistor of 4.2K. (The datasheet only guarantees Vce for Ic/Ib =10).

So the given value for the base resistor is certainly in the correct range for the circuit to work (turn off the LED without the transistor necessarily saturating) and are about right at the higher end of where saturation is guaranteed.

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