simple npn question

Hi,

I'm stuck on a really simple problem here:

I need to drive high-side line driver (ir2104) control pin with a 5V signal. The 2104 is operating at 15V (and is switching 80V on the high-side!), and the input needs to be 15V for the output to turn on.

The control signal is coming from a PIC, at 5V.

I thought, hey, just use an NPN common collector to switch the 2104. Not so fast: I can't seem to get enough base current to drive the collector for 0 to 15V, or something.

My BJT theory is pretty poor, I'm a digital guy :-), can someone enlighten me here? I'm using your generic 2n3904 device as the switch (or 2n3906). Feels like I'm trying to build a level shifter to switch [0,5V] to [0,12V]...?

thanks in advance, tk

Reply to
TK doublee
Loading thread data ...

IIRC you can get 10 or 20mA out of a PIC -- something more impressive than the 2mA from some other parts I've worked with. But even 2mA should be enough base drive to get a reliable 100mA of collector current (that's counting on an HFE of 50, which is a bit much, but not _too_ unrealistic) from a 2N3904, and with a 7.5K resistor to VCC, you should be able to get by with 2mA of collector current -- so I don't know what your problem is.

You are using a base resistor to limit the base current?

BUT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

That's all moot -- the data sheet that I have for the 2104 gives a high- going logic threshold of 3.0V, which is plenty for a CMOS processor running off of 5V.

--
Tim Wescott
Control system and signal processing consulting
 Click to see the full signature
Reply to
Tim

Use the NPN in common-emitter, with the collector connected to a load resistor, maybe 4K7~10K, and the 2104, with that resistor returned to the +15V.

From the output of the PIC, put a 1K resistor to the base, and from that base-resistor junction, put a 10K to ground.

Since the NPN will invert, you'll have to change the polarity of the output from your PIC.

+15V +15V | | | | [4K7] ------------ | | | +-------o 2104 o------ out | | | | ------------ / c [PIC]--[1K]--+-----| 2N3904 | > e [10K] | | | [gnd] [gnd]

Have Fun! Rich

Reply to
Rich Grise

As Tim points out, the datasheet appears to state that 3V is enough, even with Vcc=15V. See V_IL and V_IH on page 3.

Which seems okay for the specs.

A common collector (emitter follower) configuration??? That's not going to get you 15V from 5V. Do you actually have the collector tied to Vcc (15V) and a resistor connected between emitter and ground?

Well, as Tim points out you probably don't need an external BJT. But if you still wanted to use one, you would NOT use it in the emitter follower config.

Interesting device. Although I'm sure it's very old news to electronics designers (I'm not), I'd not considered using a capacitor like that to boost the base drive for an N-channel high side FET. Tricky and cheap. But I think you need to drive the output a solid LOW first to make certain that cap charges up to slightly under Vcc via the external diode.

Jon

Reply to
Jon Kirwan

I missed that common collector part -- hopefully the OP was messed up in his terminology.

That's used everywhere. Tricky, cheap, and ubiquitous. You can't run down to a 0% duty cycle, and you have to maintain a minimum switching frequency, or your upper-side gate drive goes away. But aside from that it doesn't take many components.

--
Tim Wescott
Control system and signal processing consulting
 Click to see the full signature
Reply to
Tim

I think he needs to use a basic Opto-coupler.

Jamie

Reply to
Jamie

I was looking for a reason to explain why the OP felt he wasn't getting enough voltage. Given the light (near zero) current requirements of the IR2104 I had a very hard time understanding why a simple common emitter config wouldn't give close to 15V output. It was obvious once I reinspected his wording -- he'd NEVER get there with an emitter follower and that then had to be the explanation. He actually did use the right terminology! Just used the wrong config.

Thanks. That's about what I'd have guessed about it. I've seen charge pumps to do the same job, but this one has another advantage over those in that you can decide what your Vcc is, which will source your gate drive, and this means you have better control over what gate drive level you want to use in an application. The charge pump devices I've seen for N-channel high sides provided a fixed "lift" that is set by design. Kind of nice, as well as cheap and tricky.

The capacitor can be sized to drive longer (it has to drive those common emitter translators and following stuff as well as leakage and FET gate charge, I guess) but then requires more charge time too. So I guess you decide your basic operating rate and try and stick to it.

What's the name for that capacitor in this configuration? I can see a lot of what it does, but I don't know what to call it so that others in the field would know what I meant without having to write a long sentence about it. It is very much like being 'picked up by your bootstraps', but that's an already overused term I think.

Jon

Reply to
Jon Kirwan

The datasheet says you need below 0.8 for a low, and over 3.0 for a high. Your 5-volt PIC output should be fine.

John

Reply to
John Larkin

I think it's usually boost cap or bootstrap cap -- but I think boost is more common.

--
Tim Wescott
Control system and signal processing consulting
 Click to see the full signature
Reply to
Tim

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.