Re: International standards (2023 Update)

There are optocouplers available with logic gate outputs which may be easier to use, for example:

I have found that desktop PCs also have 3.3V parallel port outputs, well mine does anyway I haven't carried out a detailed survey.

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gareth.harris

Not a bad idea, but they're bloody expensive, I can get all 3 of my octal chips for less than 4 of these and I'd need 8 of them for all my outputs.

Drive

suppose.

CNC

12.5kHz

Tri-state

from

The IBM PC parallel port spec. says TTL 5V or TTL 0V toggle on output ports, I think the 3.3V comes from using TTL compatible CMOS devices that were not around when the original interface spec. was made up.

I suppose I could use pullup resistors, but that always seems a bodge job to me, think I'll stick to the tri-state buffer with schmitt trigger input for safety > optocoupler > schmitt trigger buffer. That way is a bit belt and braces but should work nicely, I'll let you know.

< B. S.> wrote : "Roger Gt" wrote : >Trying to be funny works only if you ARE funny! : : Ah, now I understand Roger. (Should I bother with the smileys? Naw, : Roger Git doesn't git them...)

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I've got an old modem card here. I'd like to know what the various components do. Have you come across a site that identifies the electronic componets and their functions.

Thanks

SP.

PS. explanations for other boards/cards would also be welcome

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The typical variable voltage power supply consist of a voltage control IC and multiple high power transisors such as 2N3055.

The problem with this design is that it's highly inefficient and requires large power supply for the input.

I'd like a design that can use a 12A ish 24V switch mode power supply for input and output stage made of PWM voltage controller much like CPU voltage controller that can drive external MOSFETs.

Is there such thing as one chip controller that will drive MOSFETs and change output voltage from potentiometer input?

Hi,I have a problem with this circuit

, I try first to calculate and after to measure in a simulation the value of Rin, but the two results are much different!

Where is the mistake?

Data:

Vi= 100mV;

Ii( AC current at the transistor's base ) = 794nA;

hfe = 83.6;

re = 42 Ohm;

Ri ( calculate ) = (hfe+1)*( re + RE ) = 84.6*1042 = 88153.2 Ohm;

Ri ( measured ) = Vi/Ii = 100m/794n = 125944.6 Ohm;

The difference between the two result is too much!

Help me to understand where is my error!

Thanks !

Francesco

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Could someone suggest a simple circuit that would light an led for about 2-4 seconds when power is applied and then turn off and stay off until power is cut and applied again?

Thanks in advance JJ __________________________________________________________________

Figure about 15 watts per pound for conventional transformers, maybe

30 or so for torroids. That assumes all the windings are properly loaded, and resistive loads.

John

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2-4

is

That's not a bad idea! I'll have to think about that... I still have quite a few technical books, though by no means every tech book I used to.

Based on the diagram, I'm assuming R5 is intended to be the output load. For an emitter follower, it's on the wrong transistor terminal. It should be on emitter for an emitter follower - note this would change the RE value to R3 || R5 ( '||' = parallel reduction ). Strictly, as drawn, C2/R5 do nothing right now - for small signal they are not even part of the circuit since all DC sources are effectively grounded for analysis.

Perhaps you are measuing the correctly wired circuit rather than the displayed circuit diagram: the change in RE is ~0.1% but 0.1% * (Hfe+1) is 8.4% difference in the formula. Not enough to explain the whole difference (formula deviates 29% from measured).

More likely: remember that the Ri/Zi formula you are using is only valid for "small signal".

Try measuring with an even lower input signal level, e.g. 1 mV or 100 uV. 0.1 V is probably not "small signal" enough. Roughly speaking with an 0.1V input and hfe of 83.6, your collector current is going to move significantly so your Q-point/bias point is not going to be where you thought it was.

For a 100 mV DC level you've shifted the Q-point, while for 100 mV AC level you're seeing an integral input impedance of all the traversed Q-points. Considering that, the difference of 88K vs 125K seems like a reasonable result.

Another source of error: is the Hfe measured or from the datasheet? If the latter, there is statistical variance between any actual transistior and what the data sheet says (which is only the average value of Hfe). If measured, how? In reality Hfe = Hfe(Ic) so it's Q-point dependent also. Again gets back to not being small signal. BTW the datasheet value of Hfe is always Hfe(Icmax) where Icmax is the Ic that gives maximum(Hfe) - obvious marketing prefers listing *that* value.

A remote possibility is Rb: the "full form" is Zin(CE/CC|s.s.) = (rb + RB) + (Hfe+1)*(re + RE). Usually Rb is ignorable but not always.

Small signal can be a tricky concept: small signal is small only in the sense of bias point shifting (as in Taylor expansions and the resulting shift of y (or in this case Ic) ), but *not ever* in the human intuition of when signal level seems "small" sense. Thus small-signal really means "*small enough* that I *feel* comfortable ignoring the *actual deviations* from my *assumption* of linear operation". It's an important subtlety.

MM

Namely, look for "one shot" or monostable 555 circuits.

Jon

Try this:

John

I have a Kenmore electric dryer that went on the blink this weekend. I set the timer (which I can hear running) and push the START button. I hear a humming sound whilst pushing the button but nothing happens. I took the back off and wiggled the wires. Still doesn't work. I couldn't see the belt very good from the back but it looked attached. I can move the drum one way manually.

I have limited tools (my trusty swiss army knife is alls I have used thus far if this gives you an idea). Should I just bite the dust and call a repair person, or are there a few things I can still check.

Thanks in advance.

I have a salvaged transformer and I'd like to estimate its max current rating. I've just installed it in a home-brew power supply whose transformer I burnt out the other day (after about 20 years usage). This 0-30V supply includes a current limiter with various settings, including 2A and 4A. So I need the estimate in order to be confident I can retain one or both of those top end settings.

Physically the transformer is a bit smaller than its predecessor, with dimensions of about 7 x 3.5 x 6 cm (LxWxH of metal outer) and weighing about 1.0 kg.

Unloaded Sec pair Ohms AC V

-------- ---- -------- Grey 8.0 9.6 ~ 10 Blue 2.8 17.6 ~ 18 Red 0.6 10.6 ~ 11 Orange 0.9 26.2 ~ 26

I assume that the grey winding is much thinner wire than the other three pairs, and that the orange pair, with 0.03 ohms/V, is the heaviest duty? That's the one I've used for the DC power supply.

I've also made the red and blue secondaries accessible, to give me roughly 10, 18 and 28 V AC. The red pair gave me this data:

Load current A AC V

-------------- ---- 0 10.6 1.45 9.3 1.60 9.1 1.77 9.0 2.58 8.1 2.7 8.0

So presumably I could also use the red secondary at a fairly high current. But what - 1.5A, 2A, 3A...?

For the main orange secondary (disconnected from the bridge and all subsequent DC circuitry), I measured these:

Load current A AC V

-------------- ---- 0 25.9 0.84 24.8 0.88 24.6 1.76 23.0 2.52 22.2 3.70 20.1

Does that provide an estimate of the 'max safe current' I can use over a long period (assuming the DC circuitry remains robust, which it has appeared to be for the last couple of decades)?

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So from the way you talk the motor does not run at all? Gas or electric? Did you find a schematic? Sometimes taped to the back plate inside. sometimes on the back of the controll panel. Check the door interlock switch. It's usually activated by a small button operated when the door is closed. You can bypass this switch and see if it operates. This is the most common problem.

Did you check to see if the belt is continuous? Properly engaging the idler pulley?

Richard

I read in sci.electronics.design that Terry Pinnell wrote (in ) about 'Estimating transfomer current rating?', on Mon, 9 Aug 2004:

As a rough guide, for that mass of transformer, you can take the load regulation to be between 5% and 10%. Using just one secondary at once, you can probably go for 10%. So the load current that causes the voltage to drop by 10% from the no-load voltage is the maximum permissible current.

But don't forget that the AC current from the transformer is much larger than the DC output current - 1.6 to 1.8 times for a bridge rectifier.

The final test is 'how hot does it get?', which depends on where you have put it as well as its own properties. It should not get hotter than you can touch and stay touching, as, again, a rough guide.

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This should be OK:

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Hobbyist, West Sussex, UK