OK So if I get this right you want to switch a 12v circuit on and off. It appears that you want to switch the positive supply, which means you need more parts than if you just switch ground. I'm not in the mood for drawing ASCII art so I'll describe this instead. Low side switch (NPN transistor). You're controlling the transistor from a 5v logic signal. You need a resistor between the signal and the transistor base, more on that later... The emitter connects to ground. The collector connects to the negative side of a load. The positive side of the load connects to the supply. High side or relay-like switch (PNP transistor). This transistor needs a negative signal to turn it on so we'll drive it from the collector of the NPN circuit. It needs a base resistor between the collector of the NPN transistor and the base of this transistor. An extra resistor connects between the base and the emitter of the PNP transistor. The emitter connects to the positive supply. The collector connects to the positive end of the load and the negative end connects to ground. Resistor values... Decide what current you're going to switch and try to find a transistor with a quoted minimum hfe value for that current. The base current needs to be at least the load current/minimum hfe or the transistor may not turn on fully. For a low voltage drop (saturation) the base current may need to be as much as a tenth of the collector current. Lets say you want to switch 200mA so you decide you want a base current of 20mA. Supply voltage was 12v, knock off 2v for Vbe and drive transistor Vce and you get 10v and 20mA, so the base resistor is 2K. To ensure the transistor turns off you fit an extra base to emitter resistor. If you aim to dump 5% of the base current then assuming Vbe of 0.6v and current of 1mA you need 600R. Anything between 470R and 1K should be fine. When OFF this resistor conducts away any leakage. For the NPN transistor repeat the calculation but the hfe will probably be much higher so 1mA base current should be more than enough. Assuming a 5v signal and allowing 1v for drops means a 4K resistor. The base-emitter resistor is usually unnessecery as a logic signal is actively grounded when off so leakage is not an issue. Other methods: There are variations on this circuit that may use one less resistor, may be supply voltage independant... Substitute a P-type mosfet for the PNP transistor and use very high resistor value between gate and source for minimum current drain in a battery application. Use a "High side switch" IC. These are handy for currents >1A. ----- Original Message ----- From: "Quah" To: Sent: 06 May 2003 18:23 Subject: [EE] Transistor As Switch (REPLACE RELAY) > I am new to electronic, pls bear with me this simple question. > > Any transistor circuit that I can make as RELAY SWITCH that normally is > OPEN (not nearly an Open but real open circuit, Not Ground) > And when base is turn on, it is CLOSED for Vcc (12 v ), the base is turn > on not on 12 V. > > Thanks. > > -- > http://www.piclist.com hint: PICList Posts must start with ONE topic: > [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads > -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads