# High side vs Low side driver questions...

Discussion in 'General Electronics Chat' started by alank2, Aug 20, 2009.

1. ### alank2 Thread Starter Member

Jul 14, 2009
26
0
Hi,

I was trying to use an NPN transistor as a high side driver - the collector connected to 5V and the emitter connected to my load and the other side of the load to ground. What I found was that when the base was high, a much smaller than expected current would flow.

Then I moved the transistor so that it went from 5V to load to transistor collector, and then transistor emitter to ground and would much more current would flow.

I guess I thought it would be the same either way, but is an NPN transistor capable of switching much more current from near ground to ground than it is from near 5V to slightly lower than 5V using the same amount of base current?

If this is true, then are NPN transistors really only good for a low side driver, emitter to ground. And them PNP transistors would really only be good for a high side driver, collector to 5V, emitter to load?

Thanks,

Alan

2. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
The high side driver you described is an emitter follower (common collector). The emitter voltage will always be ≈0.7V lower than the base. If you have significant resistance in the circuit that drives the base, additional voltage will be dropped across it. The base current will be Ib=Ie/(β+1), where β (beta) is the transistor's current gain.
The common emitter circuit, if driven with adequate base current (standard is Ib≈Ic/10), will saturate (beta is low when the transistor Vce is low). The collector-emitter saturation voltage (Vce(sat) is typically a few hundred millivolts, or less at low currents (milliamps).
You can saturate a high-side NPN, but you still need to provide Ib≈Ic/10. To do this, the base voltage must exceed the collector voltage by ≈0.7V, and it must be current-limited. This means that if you have, for example, a 5V collector supply, you will need a voltage to drive the base current limiting resistor of at least 6V.
This is probably confusing. See the attachment.

It is generally much easier to use a PNP for a high side driver. Keep in mind that the base current flows the other way, so the base drive voltage must be lower than Vcc.

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3. ### alank2 Thread Starter Member

Jul 14, 2009
26
0
Hi Ron,

Thank you - your answer was detailed and spot on - I appreciate it!

Alan

4. ### alank2 Thread Starter Member

Jul 14, 2009
26
0
Hi,

How do you figure out what the voltage will be after the base resistor - after R6 you have 4.6V and after R5 in the 10V example, you have 5.7V.

Great example picture!

Thanks,

Alan

5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
The voltage at the base depends on beta, Vbe (which we are assuming is 0.7V), and the resistor values. I went through the math to come up with the voltages and currents in the example. It's just network analysis, but it is a long process and I don't want to reproduce it here.
That is much simpler. The transistor is in saturation, so the collector-base junction is forward biased. The voltage across a forward-biased diode (which the C-B junction is) is about 0.7V. Maybe 0.6V.

6. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
The standard 555 solved the same problem, but it is a headache to work with. Sill, while it isn't rail to rail, it does source and sink 200ma.

The output of a 555 stops shy around 1.2-1.4V of the positive supply voltage. This has a detrimental effect on a few circuits built with it.

7. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
Bill, I'm confused. Which problem are you referring to?

8. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
The high side drivers, the fact that it isn't rail to rail. I've had to go through some major contortions because of it. CMOS is much cleaner, but has much less drive. Back when the 555 was invented it wasn't practical, but one PNP transistor in the design would have made it so much better with a true push pull output.

For the OP, to understand what I'm talking about visualize a simple common emitter NPN transistor from a 555. Easy, right? Now try the same idea with a PNP common emitter with the emitter on the positive rail. Doesn't work.