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I'm having trouble determine the value of the reference voltage. I don't know what's the beta used for. These is given: Vbe=Vd1=Vd2=0.75V for 1mA and beta=27 for all BJT.

I did the following, but I'm not 100% sure if it's correct:

Vb=-1m*907=-0.907

Vr=Vb-Vbe=-0.907-0.75=-1.65

Any help would be appreciated.

enter image description here

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  • \$\begingroup\$ What reference voltage? \$\endgroup\$ – Andy aka Apr 6 '20 at 15:16
  • \$\begingroup\$ The circuit that gives us the reference voltage is the one in between the vertical dash lines. It’s blurry, but the Vr can be seen in the left side of that circuit below Q1. @Andy aka \$\endgroup\$ – Jaime Miranda Apr 6 '20 at 15:29
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The volt drop across the 907 ohm resistor I calculate at 0.616 volts based on 4 volts being across that resistor and the 4.98 kohm resistor. The 4 volts is 5.2 volts minus 2x 0.6 volts (the two series diodes). So the current through (907 + 4980) ohms is 4/5887 or 0.679 mA.

You calculated 1 mA and this might be a bit excessive even if beta is 27.

But, what this is all telling me is that you should use a simulation tool to properly evaluate what sort of base currents you might be getting. So, if you don't already possess a sim tool, micro-cap 12 is here.

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  • \$\begingroup\$ But the 1mA current is given to me by the problem, I didn’t calculate that \$\endgroup\$ – Jaime Miranda Apr 6 '20 at 15:44
  • \$\begingroup\$ You were given \$I_B\$ as 1 mA for a \$V_{BE}\$ of 0.75 volts. You have used that to calculate the volt drop across the 907 ohm resistor illegally. The current into that transistor won't be 1 mA and I urge you to use a simulator to uncover what it is. If 1 mA were flowing into the base there would be 27 mA flowing through the 6k1 resistor on the emitter and that aint happening. \$\endgroup\$ – Andy aka Apr 6 '20 at 15:47
  • \$\begingroup\$ Thank you! You’ve been really helpful, I appreciate it! \$\endgroup\$ – Jaime Miranda Apr 6 '20 at 15:49
  • \$\begingroup\$ Why did you use 0.6V as the voltage in the diodes instead of the 0.75V given? \$\endgroup\$ – Jaime Miranda Apr 6 '20 at 16:12
  • \$\begingroup\$ It doesn't say anything other than 0.75 volts for 1 mA flowing and there won't be 1 mA flowing through those diodes - the volt drop will be more like 0.6 volts. Use a simulation tool for a much more exact answer. \$\endgroup\$ – Andy aka Apr 6 '20 at 16:17
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beta affects base current supplied thru the top-right resistor.

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  • \$\begingroup\$ So it will affect the 1mA current going through the 907 ohm resistor? How can I calculate that effect though? \$\endgroup\$ – Jaime Miranda Apr 6 '20 at 15:39

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