olivluca 12 Posted June 26, 2013 Share Posted June 26, 2013 Hello, googling for high side switching, I found this circuit. The canonical circuit is a little more complex, however I simulated the simpler circuit in LTSpice (substituting the transistor pair with a bc337/bc327, using a 500 Ohm resistor and omitting the zener), and it seems to fit the bill (for a 500mA output current, power dissipation in the transistors is well below the limit, current on the microcontroller pin is in the uA range). What's the catch? Quote Link to post Share on other sites
oPossum 1,083 Posted June 26, 2013 Share Posted June 26, 2013 It is a very reasonable design. Be aware the that voltage being switched must be about one volt higher than the microcontrollers's supply voltage. So if you are running a MSP430 at 3.3V, you could switch a 5V or higher rail, but not the 3.3V. Also be aware that if the higher voltage supply is absent then the microcontoller's output may be overloaded. R1 can be calculated with this formula: R1 = (Vcc - 0.7) * 10 / I So for a microcontroller running at 3.6V and a 100 mA load... R1 = (3.6V - 0.7) * 10 / 0.1 = 290 ohms (use 270 or 330) There will be some small leakage current thru Q2 when it is off. This can be reduced will a resistor between the base and emitter. Using the same value as R1 is fine. The circuit as show with D1 is potentially unstable. There is a high gain uncompensated feedback loop that could become unstable with a reactive (capacitive and/or inductive) load. timotet and olivluca 2 Quote Link to post Share on other sites
olivluca 12 Posted June 26, 2013 Author Share Posted June 26, 2013 Thank you! The supply is coming from a car cigarette lighter socket (nominal 12V but I considered 14). I calculated the load at 500mA (though it will not exceed 200mA), and I used 500Ohm (the minimum HFe of a bc327 at 300mA is 40). I didn't consider the case with the 14V supply missing but LTSpice gives me a current of 5mA from the output, so it should be safe I think. Quote Link to post Share on other sites
spirilis 1,265 Posted June 26, 2013 Share Posted June 26, 2013 Being a bit of a noob when it comes to the analog world, I am confused what the purpose of D1 is in there. Why would that be added to the circuit? edit: nm, didn't see the link at the top of the post, that explains it Quote Link to post Share on other sites
rockets4kids 204 Posted June 26, 2013 Share Posted June 26, 2013 If I am understanding this circuit correctly, it should be quite a bit faster than the canonical one. In the canonical circuit, both transistors are configured as common-emitter and will be subject to the miller effect. In this circuit, I believe Q1 is common collector (emitter-follower) and Q2 is common base, neither of which are subject to the miller effect. I think this also means that current through Q1 will actually be greater than the current through Q2. Off to the simulator myself now to see what is really going on... Quote Link to post Share on other sites
olivluca 12 Posted June 26, 2013 Author Share Posted June 26, 2013 Being a bit of a noob when it comes to the analog world, I am confused what the purpose of D1 is in there. Why would that be added to the circuit? edit: nm, didn't see the link at the top of the post, that explains it If you aren't interested in overvoltage protection you can simply omit it (as I did in my simulation). I think this also means that current through Q1 will actually be greater than the current through Q2. Not really, with 500mA through Q2 I see only 5mA on Q1. The only catch is, as noted by oPossum, if the high level voltage is 0, all those 5mA will come from the microcontroller, with 14V the microcontroller has only to source 10uA. Quote Link to post Share on other sites
rockets4kids 204 Posted June 26, 2013 Share Posted June 26, 2013 ok.. that is because Q2 is not common base, but common emitter. Quote Link to post Share on other sites
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