Replacing a relay with a Transistor

[sazerac99us]

I know this is pretty basic stuff, but i want to do it right...this circuit, uses the output of an LM358 op amp to drive a BC548 transistor, which either drives a small load, or a relay, for a bigger load.

My questions is this. The load I want to drive is in the range of 1 amp DC, and I'd rather drive it with a transistor than a relay. The problem, is that I dont think the out put of the LM358 is enough to drive a big enough NPN (with their much lower gain) to handle 1 amp hard enough to saturate it. Is it bad practice to drive a pair of the smaller VC548s, and use them in parrallel to handle larger amounts of current?

Any ideas or circuit diagrams will be greatly appreciated.

 

[philba]

a lot depends on your power transistor. a gain of 10 would mean you need to give it 100 mA which, depending on voltage, could be supplied by a small signal device. Or use a darlington with the right current rating.

On the other hand, a properly sized mosfet would not require a driver tansistor and would likely run cooler.

a lot depends on the nature of your load - duty cycle, resistive vs inductive, ...

 

[Hero999]

Also not that the gain of the transistor varies with the current. It will be low at low currents, then reach a peak and decline at high currents, see the datasheet.

By the way a MOSFET it probably more suited to your application.

 

[philba]

yes, that's right. you need to use a reasonable minimum for the Hfe (gain) if using an NPN, based on the likely current draw in your application.

see the datasheet

probably the sagest advice in the whole thread.

 

[sazerac99us]

Thanks for the advice so far...

A few questions though. I thought that a MOSFET worked just the opposite of and NPN...namely, current flows only when the gate voltage drops. Also, I was under the impression that an NPN transister will have less resistance then a an FET while its conducting. I found an equation:

hFE(min) > 5 × load current Ic
_______________
max. chip current

I think the LM358's will output 20ma, so, according to this, I'd need an NPN with an hFE of about 250, which I dont think is available for a 1A collector current. Would a darlington pair have too much gain, and therefore be too noisey?

The loads in this case, will either be LED's, incandescent bulbs, piezo buzzer, or possibley even drive a relay.

Thanks for your help!!!!!!!!!!!
Maurice

 

[philba]

I don't think you will care much about noise here if you are driving the transistor into saturation. shouldn't be a problem.

bipolar transistors don't have "resistance". It does have a voltage drop (Vcesat fromthe datasheet) when in saturation which can be used to determine heat dissipation. Darlingtons often have a much higher Vcesat so will generate a lot more heat.

An N channel mosfet will start conducting when its gate-source voltage (Vgs) reaches the threshold (see datasheet). this is the difference between the source and gate voltages. So, no, in an NMOSFET, the current flows when the gate voltage rises. Maybe you are thinking of P Channel MOSFETs. As the Vgs rises above the threshold, the Rds goes down. You want to get Rds as low as possible since that will reduce the voltage drop in the MOSFET and thus reduce the generated heat. The datasheet will often tell you an Rds at a specific Vgs. Select a MOSFET with a Vgs your circuit can deliver. If your LM358 is run with a 5V supply, then you need to select a MOSFET with a good Rds at 4V or 4.5V. These are often call sensitive or logic MOSFETs. If you are running at 12V, then most NMOSFETs will work pretty well.

I'd class your loads by current - up to 20 mA (like a typical LED), you could drive directly, up to about 200 mA, you could use a small signal transistor like the 2N3904. much above that, I'd either use a darlington or a Mosfet. Above around 1A, MOSFETs rule though a relay would be a reasonable choice if you aren't switching it very frequently. Note that a lot of this is about heat generation and the ability of the package to handle it.

 

[Hero999]

An N channel mosfet will start conducting when its gate-source voltage (Vgs) reaches the threshold (see datasheet). this is the difference between the source and gate voltages. So, no, in an NMOSFET, the current flows when the gate voltage rises. Maybe you are thinking of P Channel MOSFETs.

That depends on whether it's a depletion or enhancement MOSFET.

 

[sazerac99us]

Thanks for all your help!!!! The reasons I wanted to use a solid state device in the first place, was cost and reliability, but there didnt seem to be any bipolar transistors that would be satisfactory. I did come accross one last night, a PBSS4220V, but i'd rather go with the MOSFET...since it seems the "tried and true" way to go. This will also save about $2.00-$3.00 per unit ove the cost of relays, which will add up in the long run.

Thanks again!!!!!!!!!!
Maurice

 

[philba]

That depends on whether it's a depletion or enhancement MOSFET.

true but then what's readily available?

 

[sazerac99us]

So I got this MOSFET from Radioshack (yeah, i just love their markup), the only one they had was a IRF510. Its bigger then I need, 4 amps continous, but there is a parameter on the packaging called Igm which is listed at 1.6A. Could that be a gate current of 1.6 amps? I thought the gates on mosfets had very high impedence, and the gate currents were minimal. I've looked online at the IRF510 data sheets, and could find no reference to this.

Any thoughts? I tend not to believe what radioshack says, but just want to be sure.

 

[philba]

https://www.electro-tech-online.com/custompdfs/2006/08/irf510.pdf

no mention of Igm in the DS. Note that this mosfet is spec'd for an Rds of .4 ohms when Vgs is 10V.

Whether you can use this depends on your power supply voltage. If you are running the LM358 at > 10V supply, then it should be fine but if you are using 5V, it's not likely to work well as the mosfet will be in the linear region and generating lots of heat.

 

[Nigel Goodwin]

The reasons I wanted to use a solid state device in the first place, was cost and reliability, but there didnt seem to be any bipolar transistors that would be satisfactory.

If it's cheaper, then it's cheaper (bear in mind though the cost of heatsinking that might be required as well?) - but reliability is unlikely to be as high as a relay, relays are VERY reliable - and very sturdy devices.

 

[Tarsil]

I know this is pretty basic stuff, but i want to do it right...this circuit, uses the output of an LM358 op amp to drive a BC548 transistor, which either drives a small load, or a relay, for a bigger load.

My questions is this. The load I want to drive is in the range of 1 amp DC, and I'd rather drive it with a transistor than a relay. The problem, is that I dont think the out put of the LM358 is enough to drive a big enough NPN (with their much lower gain) to handle 1 amp hard enough to saturate it. Is it bad practice to drive a pair of the smaller VC548s, and use them in parrallel to handle larger amounts of current?

Any ideas or circuit diagrams will be greatly appreciated.

U can use this circuit.
Q2 is a BD244C with a small heatsink. Q1 is a BC547C,547 etc.
I bougt a BD244C & a small heatsink for under 1$ so it's quite cheap.
Eaven at 1A Q2 doesn't disipate to much heat so it will be very reliable.
The relay is a good ideea also. U can drive some of them using the 358, but the price is @10$ or more.

 

[sazerac99us]

This is what i tried with the IRF510. (attached drawing). Q1 is the IRF510 MOSFET (a little larger then needed, but all that was available). R2 is 100,000K. The load was a piezo buzzer, about 200ma. The problem, was that the load was always on, whether or not the LM358 output was high or not (on-state voltage on the output pin was about 9.5 V, off state was measured at about 3.5mV).

I was thinking maybe the FET was fried. I've read where FET's are very senstive to static (though I was careful), so maybe driving a small power transistor with a smaller one will be the best. This circuit will be in a high vibration environment, which would make a "cheap" relay more prone to failure.

 

[Hero999]

That isn't the symbol for a MOSFET, it's the symbol for a JUGFET. You've connected up as a source follower so you wan't get any voltage gain, you'll just get the gate threshold voltage drop. Connect it up common source style, with the load connected from the drain to +V.

 

[philba]

That isn't the symbol for a MOSFET, it's the symbol for a JUGFET. You've connected up as a source follower so you wan't get any voltage gain, you'll just get the gate threshold voltage drop. Connect it up common source style, with the load connected from the drain to +V.

yup yup yup. to clarify, that means move the load between +V and the transistor.

 

[sazerac99us]

Hey, you guys are pretty sharp!!! The MOSFET symbol was the only one i could find in PCB123 that was close, so I used it. I've got it hooked up, as in the drawing, and it works. There is a small chirp from the buzzer, when the circuit is powered up, but its not objectionable ("power on diagnostics, an undocumented feature").

What I dont understand, is why it works though. I thought that the Source of a MOSFET was connected to the most positive voltage, and the drain to ground. This is opposite, and works!!!

 

[audioguru]

What I dont understand, is why it works though. I thought that the Source of a MOSFET was connected to the most positive voltage, and the drain to ground. This is opposite, and works!!!

You are thinking about how a P-channel Mosfet works. But the IRF510 is an N-channel Mosfet.

 

[Hero999]

You've still using the wrong symbol for the MOSFET, the correct symbol is **broken link removed**.
**broken link removed**

 

[sazerac99us]

As far as the MOSFET symbol goes, will have to find one in a library for the drawing package i use.

What has finally sunk in, is that using an N-channel, or an NPN transistor, interrupts the ground circuit of the load. While this works, its not what someone installing a device would expect...they'd expect their device (buzzer lets say), to be grounded, and get the positive voltage from this circuirt im making.

By using an PNP bjt or a P channel MOSFET, would that enable me to build a circuit where the positive voltage going to the load would be interrupted, instead of the ground? Any disadvantages to this?

This would be very similar to the circuit that Tarsil posted, except that in his, I didnt see any way that the base of Q2 would be kept high to shut off Q2.

There is no drawing of the P-channel MOSFET circuit, as it is the one I uploaded in error before.