The junction field-effect transistor action. It was previously emphasized that one of the main properties of the bipolar transistor is that it is a current-controlled amplifying device; the output current is controlled by a small input current. In the case of the field-effect transistor (FET) it is the input voltage which controls the output current. The current drawn by the input is usually negligible. This is a great advantage where the signal comes from a device such as capacitor microphone or piezoelectric transducer, which is unable to supply a significant current. FET’s are basically of two types: the junction field-effect transistor (JFET) and the insulated gate field-effect transistor (IGFET). The latter is more commonly known by a name metal-oxide semiconductor field-effect transistor (MOSFET).
At a point along the bar a region of p-type silicon forms a p-n junction. In normal operation, the junction is reverse-biased. The lower contact on the bar is called the source and the upper contact the drain. The electron current flows from source to drain and is controlled by the voltage applied to the p-region called the gate.
An alternative type of construction is the p-channel device where the gate is made of n-type material.
The operation of the JFET depends upon variations in the size of the depletion layer at the reverse-biased gate junction. The p-type gate is much more heavily doped than the n-type bar, so that the depletion region exists almost entirely in the bar. The gate carries a negative bias voltage relative to the source which give rise to the particular shape of the depletion region: this is wider at the top than the bottom. The wider the depletion layer, the narrower the channel there is available for the flow of electrons from source to drain, since the depletion region itself being devoid of current carries, behaves like an insulator.
Unlike the bipolar transistor, the FET employs only majority carriers for its operation. It is therefore sometimes called the unipolar transistor and is less susceptible than the bipolar type to temperature changes and nuclear radiation.
поклажа О ?узлов, но сравняется с М, если 1 возьмет у М;↓ поклажа М ? узлов, но будет в два раза >О, если возьмет 1 узел у О.↑ Решение.
О + 1 = М - 1 запись первого условия; М = О + 2 следует из первого условия; 2*(О - 1) = М + 1 запись второго условия; 2О - 2 = (О +2) + 1; подстановка выражения для О во второе условие; 2О - О = 2 + 2 + 1 перегруппировка выражения; О = 5 (узлов) поклажа осла; М = 5 + 2 = 7 (узлов) поклажа мула. ответ: 5 узлов тащил осел, 7 узлов тащил мул. Проверка: 5+1 = 7-1; 6=6; Решение отвечает первому условию. 7+1 = 2(5 -1); 8 = 8 Отвечает второму условию.
1). 1 + 1 = 2 (узла) разница в узлах между М и О, так как для равенства у М нужно 1 отнять, а О 1 добавить; 2). 2 + 1 +1 = 4 (узла) будет разница если мул возьмет у О еще один узел, а у того станет на 1 узел меньше; 3). 4 * 2 = 8 (узлов) будет поклажа М с одним "лишним" узлом, взятым у О, так как при этом по условию М будет тащить в два раза больше О. Т.е. разница в 4 узла будет составлять половину его поклажи. 4). 8 - 1 = 7 (узлов) первоначальная поклажа М; 5). 7 - 2 = 5 (узлов) первоначальная поклажа О. ответ: Мул тащит 7 узлов, Осел тащит 5 узлов. Проверка: 5+1 = 7-1; 6=6; 7+1 = 2(5-1); 8 = 8.
The junction field-effect transistor action. It was previously emphasized that one of the main properties of the bipolar transistor is that it is a current-controlled amplifying device; the output current is controlled by a small input current. In the case of the field-effect transistor (FET) it is the input voltage which controls the output current. The current drawn by the input is usually negligible. This is a great advantage where the signal comes from a device such as capacitor microphone or piezoelectric transducer, which is unable to supply a significant current. FET’s are basically of two types: the junction field-effect transistor (JFET) and the insulated gate field-effect transistor (IGFET). The latter is more commonly known by a name metal-oxide semiconductor field-effect transistor (MOSFET).
At a point along the bar a region of p-type silicon forms a p-n junction. In normal operation, the junction is reverse-biased. The lower contact on the bar is called the source and the upper contact the drain. The electron current flows from source to drain and is controlled by the voltage applied to the p-region called the gate.
An alternative type of construction is the p-channel device where the gate is made of n-type material.
The operation of the JFET depends upon variations in the size of the depletion layer at the reverse-biased gate junction. The p-type gate is much more heavily doped than the n-type bar, so that the depletion region exists almost entirely in the bar. The gate carries a negative bias voltage relative to the source which give rise to the particular shape of the depletion region: this is wider at the top than the bottom. The wider the depletion layer, the narrower the channel there is available for the flow of electrons from source to drain, since the depletion region itself being devoid of current carries, behaves like an insulator.
Unlike the bipolar transistor, the FET employs only majority carriers for its operation. It is therefore sometimes called the unipolar transistor and is less susceptible than the bipolar type to temperature changes and nuclear radiation.
Пошаговое объяснение:
поклажа М ? узлов, но будет в два раза >О, если возьмет 1 узел у О.↑
Решение.
О + 1 = М - 1 запись первого условия;
М = О + 2 следует из первого условия;
2*(О - 1) = М + 1 запись второго условия;
2О - 2 = (О +2) + 1; подстановка выражения для О во второе условие;
2О - О = 2 + 2 + 1 перегруппировка выражения;
О = 5 (узлов) поклажа осла;
М = 5 + 2 = 7 (узлов) поклажа мула.
ответ: 5 узлов тащил осел, 7 узлов тащил мул.
Проверка: 5+1 = 7-1; 6=6; Решение отвечает первому условию. 7+1 = 2(5 -1); 8 = 8 Отвечает второму условию.
1). 1 + 1 = 2 (узла) разница в узлах между М и О, так как для равенства у М нужно 1 отнять, а О 1 добавить;
2). 2 + 1 +1 = 4 (узла) будет разница если мул возьмет у О еще один узел, а у того станет на 1 узел меньше;
3). 4 * 2 = 8 (узлов) будет поклажа М с одним "лишним" узлом, взятым у О, так как при этом по условию М будет тащить в два раза больше О. Т.е. разница в 4 узла будет составлять половину его поклажи.
4). 8 - 1 = 7 (узлов) первоначальная поклажа М;
5). 7 - 2 = 5 (узлов) первоначальная поклажа О.
ответ: Мул тащит 7 узлов, Осел тащит 5 узлов.
Проверка: 5+1 = 7-1; 6=6; 7+1 = 2(5-1); 8 = 8.