Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
Given this circuit that is attached, I'm assume with the voltage divider law, the votlage to the base of the transistor should be 2V, but when simulating it, it shows 2.4V. What causing the value to be non-ideal?
The base of Q1 takes to work. The current gain of Q1 appears to be about 136. The emitter resistance is 220 ohms. The base resistance is 136 x 220= 30k ohms.
The voltage at the base is about 2.5V
For 2.5V to happen you need R2=10k and (R1 parallel with the base resistance of Q1)=10k.
Think of Q1 looking like a 30k resistor from base to (+5v).
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.
