43oh

high side switching, too simplistic circuit?

Recommended Posts

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?

Share on other sites

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.

Share on other sites

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.

Share on other sites

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

Share on other sites

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...

Share on other sites

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.
Share on other sites

ok.. that is because Q2 is not common base, but common emitter.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
• Blog

• Activity

×
• Create New...