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schematic

simulate this circuit – Schematic created using CircuitLab

I want to drive a totem-pole structure with a logic gate. The problem is, the logic gate can only supply 8.8mA output sink/source current. Can I use cascaded totem-poles to supply more current to the MOSFET gates? I googled for it, but there was no result.

(I tried to do a simulation, but I couldn't observe the results I was looking for, because the logic gates are ideal (can give infinite current) and they supply 5V output high level voltage.
The double inverters at the low sides are for buffering the output of TL494.
I hesitated to use something which I have never seen before in other people's circuits. So, I decided to ask it here. Would it cause any problem?)

Larger image of the schematics:
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Waveforms:

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First, what you show isn't really totem pole but bi-directional emitter follower.

Your basic concept makes sense in that the emitter followers will have current gain equal to the gain of the transistors. However, a problem is that each bi-directional emitter follower stage will lose two junction drops of voltage swing.

Using a single emitter follower stage to drive a FET gate can be appropriate, and I have done exactly that in production designs. However, you have to make sure the FET is still driven to the necessary min/max voltage for good switching. When you're only starting with logic level voltages, losing one junction drop on each side could be a challenge. Losing 4 junction drops due to two cascaded stages most likely won't work. Even with a 5 V logic signal, you'd be left with only around 2.2 V swing on the FET gate.

However, I don't see the point to two current gain stages. You can easily find transistors with a minimum guaranteed gain of 50 for your current and voltages. Actually 100 shouldn't be hard to find with minimal searching. Given that kind of gain available in a single stage, you shouldn't need two stages. 8 mA from your logic gate turns into 800 mA at the FET gate. If you need more than that, you should be using FET driver chips that are intended just for that purpose.

You should also ask yourself whether the logic gate by itself is enough. The gate will be driven a bit slower due to the limited current to charge or discharge the gate capacitance, but does that really matter in your design? I've done lots of designs with microcontroller pins directly driving "logic level" low side N channel switches. If you're only switching a solenoid with 24 kHz PWM, for example, you probably don't need any current gain at all.

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  • \$\begingroup\$ 1) The logic gates work with 12V supply voltage. I didn't mention this detail in the text, but wrote it on the schematic. 2) High gain transistors are a lot more expensive. And those MOSFET driver chips are even more expensive. I'm trying to do the same thing with cheap low-gain transistors. That's the point of my question. 3) I'm trying to do a high frequency switching application, so directly driving the MOSFETs is not an option. \$\endgroup\$ – hkBattousai Jun 17 '14 at 6:41
  • \$\begingroup\$ @hkbat: 1) If the gates are driving 0-12 V, that's good. 2) Nonsense. Most transistors are dirt cheap, whether high gain or not. I still think a single emitter follower current gain stage per FET gate is all you need. I can't read the FET model on your schematic, but surely those cost a lot more than the additional 5-20 cents for a pair of bipolar transistors. \$\endgroup\$ – Olin Lathrop Jun 17 '14 at 11:17
  • \$\begingroup\$ Then, can you please suggest me a low cost, high gain NPN-PNP pair for this project? The best one I could found was [PBSS4160DPN](www.nxp.com/documents/data_sheet/PBSS4160DPN.pdf), which costs 0.3$ per piece while 2N2222 and its PNP pair cost 0.03$ together. \$\endgroup\$ – hkBattousai Jun 17 '14 at 11:46
  • \$\begingroup\$ @hkbat: Finding parts is your job. If I needed to do this I'd start by spending quality time on the Mouser site. Or, I'd just go with my usual jellybean 4401/4003. Those are just a few cents each and have at least a gain of 50 for most applications. Sounds good enough to me. \$\endgroup\$ – Olin Lathrop Jun 17 '14 at 12:40

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