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I'm using an NPN transistor to pull the gate of a P-channel FET low. I'm driving the base with 3.3v and a 10k series resistor. I have a 100k resistor in series with the collector and the gate of the P-channel FET which is then pulled up to 5.4v with another 100k resistor. So my base current is about 20x my collector current. Any trouble with that?

enter image description here

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  • $\begingroup$ Are you aware of Electronics.StackExchange.com? $\endgroup$ – Transistor Feb 28 '20 at 17:44
  • $\begingroup$ Kirchhoff's Voltage Law determines actual Collector current, so yes. $\endgroup$ – StainlessSteelRat Feb 29 '20 at 2:30
  • $\begingroup$ Can you please tell us the voltage at the upper end of R220 resistor? $\endgroup$ – user8055 Feb 29 '20 at 19:10
  • $\begingroup$ @StainlessSteelRat I’m not sure how KVL answers my question, I think you mean KCL but that doesn’t tell me whether or not the transistor is conducting. That’s all a function of the transistors parameters. $\endgroup$ – lusher00 Mar 2 '20 at 13:41
  • $\begingroup$ @user8055 the top of R220 is 5.4v as stated in the original question. $\endgroup$ – lusher00 Mar 2 '20 at 13:42
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Kirchhoff's Voltage Law (KVL) must be satisfied.

You are fixiated on:

$$ \beta = \frac {I_C}{I_B}$$

So $I_C$ must be larger than $I_B$, but this implies the transistor is operating in the active region.

KVL input side:

$$V_{BB} - V_{R_B} -V_{BE} = 0$$

$$I_B = \frac {V_{BB} - V_{BE}}{R_B} = \frac {3.3V - 0.7V}{10k\Omega} = 260 \mu A$$

KVL output side:

$$V_{CC} - V_{R_{C_1}} - V_{R_{C_2}} - V_{CE} = 0$$

$$I_C = \frac {V_{CC} - V_{CE}}{R_{C_1} + R_{C_2}}$$

If we assume $V_{CE} = 0$, we can get $I_{C_{Max}}$.

$$I_C = \frac {V_{CC}}{R_{C_1} + R_{C_2}} = \frac {5.4V}{100k\Omega + 100k\Omega} = 27 \mu A$$

So based upon biasing resistors, the maximum $I_C$ is 1/10 of $I_B$. Transistor is in the cut-off region.

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  • $\begingroup$ This is not at all what I'm looking for. All you did was verify what I said to begin with, that the base current is much greater than the collector current. I also did that math and that's why I'm here asking the question. Your math just assumes the transistor is conducting which was my only question. $\endgroup$ – lusher00 Mar 3 '20 at 14:38
  • $\begingroup$ So what are you looking for. It is the math associated with your circuit. There is no need for 100k resistors. $\endgroup$ – StainlessSteelRat Mar 3 '20 at 14:43
  • $\begingroup$ Just a yes or no answer. Do NPN BJT’s behave as expected with very small collector current in comparison to the base current? I did the same math but it assumes the BJT is on. $\endgroup$ – lusher00 Mar 4 '20 at 15:46
  • $\begingroup$ No. If your transistor BE is forward biased, transistor is on, just no collector current. $\endgroup$ – StainlessSteelRat Mar 4 '20 at 20:22
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As stated by @StainlessSteelRat the transistor is operating in the cut off region. Below is graph from the datasheet showing the relationship between $V_{CE}$, $I_B$ and $I_C$.

Vce Ib and Ic relationship for 2SC4617


References:

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