What are barrels Keep? Is that a type of watch in the States?Barrels are much worse. Not even consistant.
spec
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What are barrels Keep? Is that a type of watch in the States?Barrels are much worse. Not even consistant.
Sorry Keep- got it. Yes, barrels. Then there are rods, pecks, bushels, carats, hands, acres, hectares, knots ...Related to post 85. 86. 87. 88 I remember the insignificant stuff. So, I guess, it's a KISS/Keep thing.
Sorry Keep- got it. Yes, barrels. Then there are rods, pecks, bushels, carats, hands, acres, hectares, knots ...![]()
Yeah, engine sizes in the UK have always been in cc, but this may because many of the early engines, both car and aeroplane, came from the continent, mainly France.Odd that UK was always CC for engine size, just as N.A. is metric now with Litre.
I really prefer the metric thread/tap drill size, the auto industry has already gone over.
No more BSF, BSW, BA, AF, US Machine screw etc.
Max.
All it means is more money hat i have to spend for tools.
let's see now:
We have the metric adjustible wrench.
The metric Vise Grips
The metric vise
The metric flat blade
The metric pipe wrench.
The metric band wrench.
The metric hack saw.
and the list goes on.
**broken link removed**
(3) The circuit will work OK, but R10 and R11 could be increased to 10K.
Do you mean R4 and R5?Why was it better to increase R10 and R11 on **broken link removed**to 10K?
Q6 with base resistor of 5K6 will have 4.3mA and with a 10K there will be 2.4mA and the transistor will be fully saturated in either case but you mentioned speed at some point and that that was the reason for making the value of R24 on post #88 , for example, a 470Ω which will supply the base with a whopping 17.8mA.
And R10 on the the same **broken link removed**is well within the Ic rating.
I'm still struggling with base resistors and how to determine their values. I'm using them as switches and they will be either fully on or off. I keep in mind the following when considering their values:
Saturation: the minimum current required to saturate the transistor.
Speed: I try to keep the value as small as possible though I still haven't wrapped my head around "parasitic capacitance" and how that affects the discharging of the transistor. It seems to me that if I make the resistor's value small it will turn on the transistor fast but then the high current will slow its turning off.
I've asked those same questions a few times and have read the thread multiple times but the concept is not getting into my thick skull. Not yet anyways.
Do you mean R4 and R5?
If you remember, the 470 Ohm base resistors are not desirable. They are only there so there is no direct path for current to flow from the 24V line and through the ORT into the base of the transistor and to the 0V line.
In this instance the optocoupler receiving transistor (ORT) is the current limiting factor. The opto coupler current transfer ratio, worst case is 10% and as 10mA is being sunk through the opto transmitting LED (OTL), only 1 mA will be available from the ORT, worst case.
Nop. R10 and R11. If you look at post #33 you commented on my modification of your circuit by recommending changing R10 and R11 to 10K which I've actually done and that's what is on the board that I completed and is working quite well.
The opto's If (forward current) has proportional effect on the opto's Ic current as described by the CTR graph in the datasheet.
R35 has zero effect on the Opto's Ic . I changed the value of the resistor and the current remained the same.
R35 has an effect on the current going through R24 and consequently through the base of Q27 . The higher the value of R35 the higher the voltage and current applied to R24 so if I reduce R35 to let's say 2K2 I would get something like 6.7mA (1mA less than with a 5K6) at the base of Q27. No real reason for doing that just playing with the circuit.
Two issues I'm still working on :
1- The voltage regulator will generate heat and will need a heat sink for sure but I'm not sure if the heat is too much to dissipate by a heat sink. I'm expecting to run a motor on about 38 V and let's say I go for the maximum current of 1A then:
38-9= 29 x 1A = 29Watt. Is that too much?