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Following is the circuit I have been analyzing. My intention is to calculate the turn on and off condition of the transistor Q1. R1 and R2 (10 kΩ each) are internal to the transistor.

schematic

simulate this circuit – Schematic created using CircuitLab

Following is my understanding, whenever there is voltage at the input V1, it will equally divide across R1 and R2. When it reaches 0.7 at Vbe, Q1 will be turned on. If I want to increase the turn on voltage of the transistor, introducing voltage divider at the input (before resistor R1) will be sufficient.

I want to on and off the transistor above to a fixed voltage, say 5 V. Please advise my understanding.

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  • \$\begingroup\$ I'm confused. The BJT you reference is a digital BJT with both \$R_1\$ and \$R_2\$ internalized within the device, already. (A fact you added to the schematic, but which may not be clear to readers of your question.) This BJT is designed for ON/OFF use, which means the collector is intended to be either "high impedance" or else "acting like a voltage source very close in value to the emitter voltage." When you mention \$5\:\text{V}\$ in the text, what exactly is that referring to? Where in the schematic is it expected? \$\endgroup\$ – jonk Jan 6 at 19:14
  • \$\begingroup\$ My aim was to control the bjt on and off for particular user defined input voltage. Just for explanation purpose I took it as 5 V. \$\endgroup\$ – vt673 Jan 6 at 19:35
  • \$\begingroup\$ This digital switch must be designed for worst case. Ice=100uA OFF @Vce=5V and Ib is not specified but could be as high as 33uA with 10% of hFE being 30 or much lower for Ib \$\endgroup\$ – Sunnyskyguy EE75 Jan 6 at 19:39
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    \$\begingroup\$ You should probably use a comparator and voltage reference instead of a single transistor. \$\endgroup\$ – mkeith Jan 6 at 19:46
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    \$\begingroup\$ Transistors are logarithmic responders, in that the base-emitter voltage is the log of the collector current. Every 0.058 volts of base voltage should cause another 10:1 more collector current. Thus "off" is an uncertain point, until you take the input voltage to ZERO volts. Even at 0.6 volts input, or 0.3 volts on the base, you'll have nanoAmperes of collector current. Does that make sense? \$\endgroup\$ – analogsystemsrf Jan 6 at 21:31
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This makes a lousy comparator R1/R2 0.8min 1.0typ 1.2max

Thus +/-20% of 0.6V becomes 20% of 5V is best case and Leakage current affects R3 Ambient temperature variation adds another 10%.


General design not for this Q1

  • Assuming hFE=100 (bad assumption) and Vce=Vce(sat) = 0.2V at 10% of hFE or Ic:Ib=10:1
  • Ic=15.8V/0.1MΩ = 158uA, Ib=16uA

    • At high temp Leakage current can be 20uA edge of graph not spec'd

    - so 100k load is a poor choice for this Q1 and hFE can be as low as 30

  • if Vin =5V and Vbe=0.6 ( not 0.7 at this low current)
  • Rb=4.4V/16uA=275k

  • Next the threshold for turn-on needs to be 5V input.

    • thus 0.6V/5V= 12%

    • thus adding external R1 to internal R1= 75k-10k= 48k or **external R1' ~65k

Threshold at room temp may be 4.4 to 5.7V for 80% output swing. Simulated with triangle wave. enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

Suggestion

Define Cost limit, Ambient range, V threshold tolerance and go back to drawing board

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  • \$\begingroup\$ Rb will include internal 10 kΩ resistor too, right? \$\endgroup\$ – vt673 Jan 6 at 18:15
  • \$\begingroup\$ If V1 is a 5V voltage source, R1=275k and R2 is not needed. But if PNP open collector then a pull down is needed. Or you can make the threshold symmetrical with a Thevenin equivalent divider \$\endgroup\$ – Sunnyskyguy EE75 Jan 6 at 18:17
  • \$\begingroup\$ I think you have forgotten the R2 resistor that adds \$60 \rm \mu A\$ to the calculated \$I_B\$. \$\endgroup\$ – user2233709 Jan 6 at 18:19
  • \$\begingroup\$ You didn't bother to look at the datasheet, did you? \$\endgroup\$ – Elliot Alderson Jan 6 at 18:39
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    \$\begingroup\$ Read again. under General Design, if load must be 100k and threshold of 5V +-20% if 75k is 0, threshold is 1.2V \$\endgroup\$ – Sunnyskyguy EE75 Jan 7 at 17:24
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No need for a second voltage divider before the first; you just need to add a second resistor in series with R1. Anyway, beware that the “switching” of a transistor in not ideal. Don’t expect it to behave like an open switch at \$4.9 \rm V\$ and as as closed one at \$5.1 \rm V\$.

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  • \$\begingroup\$ But on voltage is shown as 2.5 V in the datasheet.!! (Page-4) \$\endgroup\$ – vt673 Jan 6 at 18:21
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    \$\begingroup\$ I just had a look at the datasheet. There is a “typical” on voltage of \$1.8 V\$ with a “minimum” of \$2.5 V\$‽ They must have messed minimum and maximum values… \$\endgroup\$ – user2233709 Jan 6 at 18:27
  • \$\begingroup\$ The number in the datasheet is the maximum input voltage required to produce a given output current. Note that the typical voltage is only 1.8V. According to the datasheet the transistor could be ON for any voltage greater than 0.8V (the minimum OFF voltage), but it is guaranteed to be ON for voltages greater than 2.5V. \$\endgroup\$ – Elliot Alderson Jan 6 at 18:28
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    \$\begingroup\$ Hence, you can only be sure that the “switching” happens between \$0.8 V\$ and \$2.5 V\$ and, if you are lucky, it should happen between \$1.2 V\$ and \$1.8 V\$. \$\endgroup\$ – user2233709 Jan 6 at 18:29
  • \$\begingroup\$ Can I use zener diode at 5.1 V at the input to confirm the switching? \$\endgroup\$ – vt673 Jan 6 at 18:40

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