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Below is a circuit where S1 is the output of a 555 IC timer (S1 switch for simplicity). I need to invert the output which I have done so in another project using the circuit below.

Starting with transistor Q2 which is a 2N3904 (datasheet), how do you calculate the resistor R2 needed if R3 or "Load" is 180Ω. The source will be 5V. I'm having trouble understanding why Collector-Emitter saturation voltage is .3V where IC=50mA, when below it says for hFE4, VCE=1V, IC=50mA

Secondly how do I calculate R1 for Q1, to ensure Q2 is not driven.

I've have this circuit working with 12V and a 350Ω load. I used 1K resistors for both. I can't remember how I calculated it or if I even got it right.

Sample circuit

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The conditions for saturated operation and the various HFE line are various operating conditions. If you operate the transistor with 1V VCE and carrying 50mA you will see a minimum (worst case) HFE of 60. For saturated operation you need to guarantee that any transistor will work in a circuit and usually it is recommended to operate as if the worst case HFE was 10. If you look at the data sheet for saturated operation they show various conditions where they put 1/10 of the current into the base as the collector current (defined by the circuit). This is often referred to as a "forced hfe". The graph at the bottom of page 3 shows this also with forced HFE of 10.

In your circuit you have R3 as 180 ohm which will require ~30mA if Q2 is saturated. Taking the forced HFE as 10 we require 3mA into the base. This will therefore require 1.8k ohm resistor for R2. Similarly since Q1 will be expected to saturate with the current from R2 (~3ma) it needs at least 0.3mA from R1. R1 would need to be about 18k or lower.

The Forced HFE of 10 is a maximum and is not critical. You could use 5 or 15 for this circuit with no problem. A disadvantage of having a higher current into the base is that it consumes more power in the base resistor and also that the higher current will result in more stored charge in the base of the transistor which will take longer to dissipate and slow down the turn-off.

I would also put resistor from the base of Q1 to ground (maybe 10K) to assist in turn off and ensure that any leakage across the switch S1 does not cause Q1 to incorrectly turn-on.

kevin

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  • \$\begingroup\$ Just to clarify, to obtain R2 you simply took the 3mA required at 5V and came to 1666 and rounded? Or is there something I'm missing. Do you subtract the base-emitter saturation voltage from 5V before calculating. Is a gain of 10 a generally known figure to operate a transistor in switch mode or is it derived from the data sheet? And lastly, when you say if you look at the data sheet for saturated operation, where am I looking exactly for that information? Thanks \$\endgroup\$ – JoshNZ Jun 1 '15 at 20:32
  • \$\begingroup\$ You said the load was 180 ohm with a 5v source this gives 27.7mA. Divide that by 10 to give 2.7mA and you need a 1.8k resistor to supply the required current from 5v. This neglects the saturation voltage and the base emitter voltage. 1.5k as suggested by the other person answering would give slightly more current. This is not a critical value however. \$\endgroup\$ – Kevin White Jun 2 '15 at 20:46
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You want to drive it with much more than the hFE would indicate to keep Vce low.

Your load is 28mA. The Vce(sat) specifications are guaranteed at Ic/Ib = 10, so to use those numbers you might want the base current to be 2.8mA, so R2 would be (5-0.7)/0.0028 or 1.5K.

For Q1 we need the Vce(sat) to be low (to fully turn off Q2), so we definitely should not go too much lower than Ic/Ib = 10. In this case, however, the maximum collector current is only 3.3mA so we can calculate a suitable resistor similarly to be 13K for 330uA base current.

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