Bias Supply Circuit Question

[fordlvr1]

Hi All,

I'm an electrical that's coming from mostly experience in software and trying to learn some schematics of our current products.

Can't show it all but had a question about this circuit. We have an IC being fed by a 15V supply, it drives the gate of 2 MOSFETS in sync. Looking at the datasheet I understand all the resistors and caps on each input but I don't understand the purpose of the circuit. If we already have 15V, and we're driving both the driver circuit, and the MOSFETS, then why don't we just pass along that same voltage to the rest of the circuit where it gets bumped up to different values via bias transformer?

Are we using this to make sure the 15V is isolated, regulated, and "clean" from the original source? In that case why not just use 1 switch in a buck boost topology?

Appreciate it,

 

[VEBill]

Presumably, the two transistors provide much higher current than is available from the driver IC.

 

[fordlvr1]

Thx for response.

Not sure though because the schematic has a +15V supplying the driver circuit (going to each gate)...but the same 15V supply also goes to the drain of Q4. Q5 source goes down to the return terminal.

Why not just go straight from the original 15V? its only going to be able to source so much current regardless of some switches in line with it no?

 

[VEBill]

Although you've only provided a small section of the schematic, it's a safe bet that they're (basically) making an AC type waveform (switched DC). You mentioned a transformer, so plain old DC isn't going to work.

The circuit exerpt you've provided is a very common architecture. It's essentially one half of an H-bridge. Search Google Images for H-bridge.

 

[IRstuff]

The implication is that whatever is driving the two transistors does not have sufficient current driving capability to drive the intended load, i.e., the IC's output impedance is too high. The two transistors have low output impedance. This is essentially a definition of an AMPLIFIER, which allows a small amount of current or voltage to produce a large amount of current or voltage.

TTFN
faq731-376

 

[fordlvr1]

@Ve1bll - Good catch, I just assumed that all the other voltage supplies on here were DC...since we've got a xfmr this is got to be AC. Did some searching on the H-Bridge and your right that's what the circuit is. Most of the stuff I find on H-Bridge has to do with motor direction, but my application looks like a single phase half bridge inverter. The transformer bumps the voltage up to 25V & helps provide some isolation I assume.

Having a hard time finding literature on that though. I get results saturated with motor rotation from the full-H topology. Much less on this being used as a DC to AC source. Its probably going to give a pretty rough AC waveform out no?

Thanks for the feedback

 

[IRstuff]

These circuits are rarely intended for sinusoidal outputs, since that would dump truckloads of wasted parasitic current through the driver totem pole itself.

TTFN
faq731-376

 

[fordlvr1]

Seemed a little strange to me also. Maybe if I understood the IC driving the two MOSFETS better. Its a IR2085S. Its datasheet calls it a "self-oscillating half bridge driver w/50% duty cycle.

Seems strange to me that we'd generate AC this way through the board...but as one of the comments mention - there is a transformer immediately after the two mosfets. In order for that to work, it would have to be AC...

Sorry to be a pest on this circuit. I've gone ahead and worked forward on all the others with no problems but this one really confuses me.

 

[IRstuff]

My bad, I misread; "AC" has specific connotations, so you need to be clear what you actually mean. Your circuit is apparently cranking square waves into the transformer, then.

TTFN
faq731-376

 

[VEBill]

I chose my words carefully: "...an AC type waveform ([highlight #FCE94F]switched DC[/highlight]). "

 

[iop995]

It's part of driver inverter; it's used to assure correct switching currents for main switch devices. If a xfmr is added will be assured isolation also. All inverter switch DC bus voltage to create AC voltage; any switched DC voltage (pulse / PWM controlled or any other method) it's composed by an infinite sinusoidal waveforms with different voltages and frequency in integer muliple relation (Fourier series); one of inverter goal is to find switching method to have as small harmonics and a clean sinusoidal fundamental. AC "pure" sinusoidal amplifier is used in audio and radio area not in power because bad efficiency (a class A amplifier have 50% efficiency, meaning 50% loses, so aprox. 50% heating - very difficult to remove from a power device).

 

[IRstuff]

"a class A amplifier have 50% efficiency, meaning 50% loses, so aprox. 50% heating - very difficult to remove from a power device"

That's precisely why notation and clarity is critical to such discussions. Switched DC does NOT require decomposition into Fourier components. Totem pole drivers do have 50% efficiency because their transition regimes are generally designed to be a small portion of the duty cycle. The OP did not show enough of his schematic to conclude that his design is a sinusoidally driven class A output.

TTFN
faq731-376