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I'm building a transistor switch circuit and below is the schematic.

schematic

simulate this circuit – Schematic created using CircuitLab

I calculated R2 resistor (base resistor) value from this formula:

$$R_{R2}=\frac{V_{BAT2}-V_{be}}{(V_{BAT1}-V_{ce})/R_{R1}/h_{fe}*2}.$$

Is this enough for this circuit to operate normally in most situation?

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    \$\begingroup\$ A manufacturer's datasheet characterises Vce(sat) vs Ic, for Ic/Ib = 1000. I interpret that as a 'hint', by the manufacturer, for the device's intended envelope. So, without any deeper insight, make R2 less resistance than R1*1000 = 70kΩ. Then scale that based on the ratio of BAT2/BAT1. If the circuit is a simplification, and the base will be driven by electronics at some non-trivial frequency, not a human operated switch, then drive it harder, i.e. a lower resistance. \$\endgroup\$ – gbulmer Sep 10 '14 at 16:44
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The formulation is correct. You must be careful with the \$h_{fe}\$ value you use. Search the device data sheet curve \$h_{fe}\$ vs. collector current, to the value of \$h_{fe}\$ in saturation state.

EDIT: from the datasheet

enter image description here

You must use the lower curve to size the output current and by the ratio $$ \dfrac{I_C}{I_B}=1000 $$ obtain the required value of base current.
For the saturation condition, the value of \$h_{fe}\$ is lower than for the condition of linear operation, so may the value selected for the base resistance is rather high.

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  • \$\begingroup\$ Thank you for advice. The datasheet curve of \$h_{fe}\$ vs \$I_c\$ is at \$V_{ce}=5V\$, and how can I get \$h_{fe}\$ at my situation? \$\endgroup\$ – user1448742 Sep 10 '14 at 16:19
  • \$\begingroup\$ The fig 10 of the datasheet from ONsemi, shows the collector-emitter voltage for saturation, at \$\frac{I_C}{I_B}=1000\$ vs collector current. This is the curve you must use. \$\endgroup\$ – Martin Petrei Sep 10 '14 at 16:28

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