# can base current be far less than collector current?

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?

• Are you aware of Electronics.StackExchange.com? – Transistor Feb 28 '20 at 17:44
• Kirchhoff's Voltage Law determines actual Collector current, so yes. – StainlessSteelRat Feb 29 '20 at 2:30
• Can you please tell us the voltage at the upper end of R220 resistor? – user8055 Feb 29 '20 at 19:10
• @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. – lusher00 Mar 2 '20 at 13:41
• @user8055 the top of R220 is 5.4v as stated in the original question. – lusher00 Mar 2 '20 at 13:42

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.

• 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. – lusher00 Mar 3 '20 at 14:38
• So what are you looking for. It is the math associated with your circuit. There is no need for 100k resistors. – StainlessSteelRat Mar 3 '20 at 14:43
• 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. – lusher00 Mar 4 '20 at 15:46
• No. If your transistor BE is forward biased, transistor is on, just no collector current. – StainlessSteelRat Mar 4 '20 at 20:22

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 $$I_C$$.

References: